Solution :

Fig. 2 shows masses $m _{1}$ and $m _{2}$ distance $A B=r$, apart. Choose line $A B$ as $x$-axis and $A$ as origin of coordinates. Co-ordinates of $A$ and $B$ are $(0)$ and $(r)$ respectively. Let $C$ be the center of mass of the two particle system. Then

Fig. 2

$\mathrm{AC}=\mathrm{X}=\frac{\mathrm{m} _{1} 0+\mathrm{m} _{2} \mathrm{r}}{\mathrm{m} _{1}+\mathrm{m} _{2}}=\frac{\mathrm{m} _{2} \mathrm{r}}{\mathrm{m} _{1}+\mathrm{m} _{2}}$

$\mathrm{BC}=\mathrm{r}-\mathrm{X}=\frac{\mathrm{m} _{1} \mathrm{r}}{\mathrm{m} _{1}+\mathrm{m} _{2}}$

$\therefore \quad \frac{\mathrm{AC}}{\mathrm{BC}}=\frac{\mathrm{m} _{2}}{\mathrm{~m} _{1}} \quad$ or $\quad \mathrm{AC}: \mathrm{BC}:: \frac{1}{\mathrm{~m} _{1}}: \frac{1}{\mathrm{~m} _{2}}$

Solution :

We know,

$\mathbf{R}=\frac{m \mathbf{r} _{\mathrm{A}}+2 \mathrm{~m} \mathbf{r} _{\mathrm{B}}}{\mathrm{m}+2 \mathrm{~m}}$

or

$$ \hat{i}+\hat{j}-2 \hat{k}=\frac{1}{3}\left[(3 \hat{i}-2 \hat{j}+\hat{k})+2 \mathbf{r} _{B}\right] $$

$\therefore \quad 2 \mathbf{r} _{\mathrm{B}}=(3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})-(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$

or $\quad \mathbf{r} _{B}=\frac{5 \hat{j}-7 \hat{k}}{2}$

Solution :

Let $\mathrm{m} _{1}, \mathrm{~m} _{2}$ and $\mathrm{m} _{3}$ be masses of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ respectively. Obviously.

$$ \mathrm{m} _{1}=(3 \ell)^{2} \sigma=9 \ell^{2} \sigma ; \mathrm{m} _{2}=(2 \ell \times 4 \ell) \sigma=8 \ell^{2} \sigma ; \mathrm{m} _{3}=(\ell \times 3 \ell) \sigma=3 \ell^{2} \sigma $$

Let $\mathrm{O} _{1}$ and $\mathrm{O} _{2}$ and $\mathrm{O} _{3}$ be the mid point of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. These points are C.M. of each block. The system shown is equivalent to point masses $\mathrm{m} _{1}, \mathrm{~m} _{2}, \mathrm{~m} _{3}$ placed at $\mathrm{O} _{1}, \mathrm{O} _{2}$ and $\mathrm{O} _{3}$ as shown in Fig. 4. Co-ordinates of $\mathrm{O} _{1}, \mathrm{O} _{2}$ and $\mathrm{O} _{3}$ are

Fig. 4

$\mathrm{O} _{1}\left(\frac{3 \ell}{2} ; \frac{3 \ell}{2}\right), \mathrm{O} _{2}(5 \ell ; \ell), \mathrm{O} _{3}\left(8.5 \ell ; \frac{\ell}{2}\right)$

Let $(\mathrm{X}, \mathrm{Y})$ be the co-ordinates of C.M. of system

$$ \begin{aligned} & X=\frac{\left(9 \ell^{2} \sigma\right)\left(\frac{3 \ell}{2}\right)+\left(8 \ell^{2} \sigma\right)(5 \ell)+3 \ell^{2} \sigma(8.5 \ell)}{20 \ell^{2} \sigma}=\left(\frac{79}{20}\right) \ell \\ & Y=\frac{\left(9 \ell^{2} \sigma\right)\left(\frac{3 \ell}{2}\right)+\left(8 \ell^{2} \sigma\right)+\left(3 \ell^{2} \sigma\right) \frac{\ell}{2}}{20 \ell^{2} \sigma}=\left(\frac{23}{20}\right) \ell \end{aligned} $$

Solution :

$\mathrm{m}=$ mass of circular (complete) disc of radius $\mathrm{R}=\pi \mathrm{R}^{2} \sigma _{1}$

$\mathrm{m} _{1}=$ mass of circular disc cut-out $=\pi \mathrm{r}^{2} \sigma _{1}$

$\mathrm{m} _{2}=$ mass of square $=\mathrm{a}^{2} \sigma _{2}$

The arrangement is equivalent to mass $\mathrm{m},-\mathrm{m} _{1}$ and $+\mathrm{m} _{2}$ placed at $\mathrm{O}, \mathrm{O} _{1}$ and $\mathrm{O} _{2}$ respectively. $(\mathrm{X}, \mathrm{Y})$ are co-ordinates of C.M.

$$ \begin{aligned} & X=\frac{m \times 0+\left(-m _{1}\right) b+m _{2}(0)}{m+\left(-m _{1}\right)+m _{2}}=-\frac{r^{2} b}{\left[\left(R^{2}-r^{2}\right)+\left(\frac{\sigma _{2}}{\pi \sigma _{1}}\right) a^{2}\right]} \\ & Y=\frac{m \times 0+\left(-m _{1}\right) x _{0}+m _{2} \times c}{m+\left(-m _{1}\right)+m _{2}}=-\frac{a^{2} c}{\left[\frac{\pi \sigma _{1}}{\sigma _{2}}\left(R^{2}-r^{2}\right)+a^{2}\right]} \end{aligned} $$

Solution :

Initially C.M. of bomb is at rest at its center. The bomb explodes on its own i.e. there is no external force. The equation of motion of C.M. of fragments is

$$ \left(\mathrm{m} _{1}+\mathrm{m} _{2}+\mathrm{m} _{2}\right) \frac{\mathrm{d}^{2} \mathbf{R}}{\mathrm{dt}^{2}}=0 $$

or

$$ \mathrm{MA}=0 $$

Since $\mathrm{M} \neq 0 ; \mathbf{A}=\frac{\mathrm{d} \mathbf{v}}{\mathrm{dt}}=0$. Since $\frac{\mathrm{d} \mathbf{v}}{\mathrm{dt}}=0 ; \mathbf{v}=\mathbf{v} _{0}=$ Initial speed of center of mass $=0$

The center of mass remains at the center of the bomb, i.e. in its original position.

Solution :

Let vertical downward direction be the direction of $\mathrm{z}$-axis and initial position of ball as origin of co-ordinates. Let $\mathrm{z}$ be the instantaneous position co-ordinate of C.M of ball, before explosion. The equation of motion of C.M is

$$ \begin{equation*} \mathrm{M} \frac{\mathrm{d}^{2} \mathrm{z}}{\mathrm{dt}}=\mathrm{Mg} \quad \text { or } \quad \frac{\mathrm{d}^{2} \mathrm{z}}{\mathrm{dt}^{2}}=\mathrm{g} \tag{1} \end{equation*} $$

The center of mass moves along $\mathrm{z}$-axis with uniform acceleration $\mathrm{g}$.

When ball breaks up, on its own; there is no other external force. The equation of motion of C.M. of fragments is

$$ \left[\sum _{j=1}^{n} m _{j}\right] \frac{d^{2} z}{d t^{2}}=\sum _{n=1}^{m} m _{j} g \quad \text { or } \quad \frac{d^{2} z}{d t^{2}}=g $$

This is same as Eqn. (1). Hence C.M of fragment continues moving along $z$-axis with uniform acceleration $=\mathrm{g}$

Solution :

The instantaneous position vector $\mathbf{r}$ is

$$ \mathbf{r}=\mathbf{r} _{0}+\mathbf{v} _{0} \mathbf{t} $$

The instantaneous linear momentum, $\mathbf{p}$; is

$$ \mathbf{p}=\mathbf{m} \mathbf{v} _{0} $$

The instantaneous angular momentum $\mathbf{L}$ about origin of co-ordinates is

$$ \begin{aligned} & \mathbf{L}=\mathbf{r} \times \mathbf{p} \\ & =\left(\mathbf{r} _{0}+\mathbf{v} _{0} \mathrm{t}\right) \times \mathbf{m} \mathbf{v} _{0} \\ & =\mathbf{r} _{0} \times \mathrm{m} \mathbf{v} _{0} \quad\left[\because \mathbf{v} _{0} \times \mathbf{v} _{0} \equiv 0\right] \end{aligned} $$

Solution :

Fig. 12 shows straight line along which particle moves. The line makes an angle of $30^{\circ}$ with $x$-axis.

$$ \begin{aligned} & \mathrm{p}=\text { linear momentum of particle } \\ & =\mathrm{mv}=0.1 \times 2=0.2 \mathrm{kgms}^{-1} \\ & \ell _{0}=\text { magnitude of angular momentum of } \mathrm{P} \\ & \quad \quad \text { about } \mathrm{O} \\ & =\mathrm{p} \times \mathrm{OA}=\mathrm{p} \times \mathrm{OC} \sin 30^{0} \\ & =0.2 \times 3 \times 0.5=0.3 \mathrm{kgm}^{2} \mathrm{~s}^{-1} \end{aligned} $$

Fig. 12

Solution :

Fig. 13 shows the conical pendulum. $\mathrm{O} _{1}$ is point of suspension and $\mathrm{O} _{1} \mathrm{P}=\ell=$ length of pendulum.

$\mathrm{OP}=\mathrm{r}=\ell \sin \theta=$ radius of circular path of bob

$\mathrm{v}=$ linear speed of bob $=\mathrm{r} \omega$

$\mathrm{L} _{0}=$ magnitude of angular momentum of bob about 0.

$$ \begin{aligned} & =\mathrm{mvr}=\mathrm{mr}^{2} \omega \\ & =\mathrm{m} \ell^{2} \omega \sin ^{2} \theta \end{aligned} $$

Fig. 13

Solution :

$\mathrm{L} _{\mathrm{P}}=$ Angular momentum of $\mathrm{P}$ about 0

$=2 \mathrm{mv} _{0} \cdot$; clockwise

$\mathrm{L} _{\mathrm{Q}}=$ Angular momentum of $\mathrm{Q}$ about 0

$=8 \mathrm{mv} _{0} . \mathrm{y}$; anticlockwise

$\mathrm{L}=$ The net angular momentum $=\mathrm{L} _{\mathrm{Q}}-\mathrm{L} _{\mathrm{P}}$

Fig. 14

$=2 \mathrm{mv} _{0}[4 \mathrm{y}-\mathrm{a}]$; anticlockwise

$=4 \mathrm{mv} _{0} \mathrm{a} \quad$ (Given)

$$ \therefore \quad 4 y-a=2 a \quad \text { or } \quad y=\frac{3 a}{4} $$

Solution :

Fig. 15

Fig. 16 shows free-body diagram of sphere. Since sphere is in equilibrium there is neither a motion of translation nor rotation. For translation equilibrium

Fig. 16

$$ \begin{equation*} M g \sin \theta=f+T \tag{i} \end{equation*} $$

For rotational equilibrium; the net torque about $\mathrm{O}$ is zero. Therefore

$$ \begin{align*} & \mathrm{TR}=\mathrm{f} . \mathrm{R} \\ & \text { or } \quad \mathrm{T}=\mathrm{f} \tag{ii} \end{align*} $$

From Eqns. (i) and (ii) we have

$$ \mathrm{T}=\frac{\mathrm{Mg} \sin \theta}{2} $$

Solution :

For axis-1

$\mathrm{I} _{\mathrm{A}}=\mathrm{I} _{\mathrm{B}}=0$

Fig. 20

$I _{C}=I _{D}=I _{E}=I _{H}=m a^{2}$

$I _{G}=I _{F}=m(\sqrt{2 a})^{2}$

$\therefore \quad \mathrm{I} _{1}=4 \mathrm{ma}^{2}+2 \mathrm{~m}(\sqrt{2} \mathrm{a})^{2}=8 \mathrm{ma}^{2} \hspace{40mm} . . . . . . . (i)$

For axis-2

Through midpoint of side $\mathrm{AB}$ and $\mathrm{CD}$

$\mathrm{I} _{\mathrm{A}}=\mathrm{I} _{\mathrm{B}}=\mathrm{I} _{\mathrm{C}}=\mathrm{I} _{\mathrm{D}}=\mathrm{m}\left(\frac{\mathrm{a}}{2}\right)^{2}$

The perpendicular distance of points E, F, G and $\mathrm{H}$ from axis $=\sqrt{(\mathrm{a})^{2}+\left(\frac{\mathrm{a}}{2}\right)^{2}}=\frac{\sqrt{5}}{2} \mathrm{a}$

Hence,

$$ \begin{gather*} \mathrm{I} _{2}=4\left[\mathrm{~m}(\mathrm{a} / 2)^{2}\right]+4\left[\mathrm{~m}\left(\sqrt{\frac{5}{2} \mathrm{a}}\right)^{2}\right] \\ =6 \mathrm{ma}^{2} \tag{ii} \end{gather*} $$

$\therefore \quad \frac{\mathrm{I} _{1}}{\mathrm{I} _{2}}=\frac{4}{3}$

Solution :

$\mathrm{I}=$ M.I of ring about given axis $=\mathrm{MR}^{2}$

Let $\mathrm{r}$ be radius of disc. The M.I, of disc, $\mathrm{I} _{1}$, about a tangent in its own plane is

$$ \mathrm{I} _{1}=\frac{\mathrm{Mr}^{2}}{4}+\mathrm{Mr}^{2}=\frac{5 \mathrm{Mr}^{2}}{4} $$

[Theorem of parallel axis]

Given $\frac{I}{2}=I _{1}$. Therefore

$$ \frac{\mathrm{MR}^{2}}{2}=\frac{5}{4} \mathrm{Mr}^{2} $$

or $\quad r=\sqrt{\frac{2}{5}} \mathrm{R}$

Solution :

$\mathrm{I} _{1}=\frac{1}{12}(2 \mathrm{M})(2 \mathrm{~L})^{2}=\frac{2}{3} \mathrm{ML}^{2} \hspace{40mm} . . . . . . .(i)$

In Fig. 21(b); the axis $\mathrm{YY}$ ’ is perpendicular to part $\mathrm{OB}$ and passes through its end $\mathrm{O}$

$\therefore \quad \mathrm{I}^{\prime}=\frac{1}{3}(\mathrm{M})(\mathrm{L})^{2}=\frac{\mathrm{ML}^{2}}{3}$

The part OA; makes an angle $\theta=60^{\circ}$ with axis YOY’ passing through end O. Its M.I,

$\mathrm{I}^{\prime \prime}=\frac{1}{3} \mathrm{ML}^{2} \sin ^{2} 60=\frac{\mathrm{ML}^{2}}{4}$

$\mathrm{I} _{2}=$ M.I of arrangement of both rods about given axis.

$=\mathrm{I}^{\prime}+\mathrm{I}^{\prime \prime}=\frac{\mathrm{ML}^{2}}{3}+\frac{\mathrm{ML}^{2}}{4}=\frac{7}{12} \mathrm{ML}^{2} \hspace{40mm}. . . . . . . (ii)$

$\therefore \frac{\mathrm{I} _{1}}{\mathrm{I} _{2}}=\frac{\frac{2}{3} \mathrm{ML}^{2}}{\frac{7}{4} \mathrm{ML}^{2}}=\frac{8}{21}$

Solution :

$\mathrm{I} _{x}=$ M.I of system about $x$-axis

$$ \begin{aligned} & =\mathrm{I} _{\mathrm{OA}}+\mathrm{I} _{\mathrm{OB}}+\mathrm{I} _{\mathrm{OC}} \\ & =0+\frac{1}{3}\left(\lambda _{2} \mathrm{~L}\right) \mathrm{L}^{2}+\frac{1}{3}\left(\lambda _{3} \mathrm{~L}\right) \mathrm{L}^{2} \\ & =\frac{\left(\lambda _{2}+\lambda _{3}\right)}{3} \mathrm{~L}^{3} \end{aligned} $$

Similarly,

$$ \begin{aligned} & \mathrm{I} _{\mathrm{y}}=\left(\frac{\lambda _{3}+\lambda _{1}}{3}\right) \mathrm{L}^{3} \text { and } \mathrm{I} _{\mathrm{z}}=\left(\frac{\lambda _{1}+\lambda _{2}}{3}\right) \mathrm{L}^{3} \\ & \therefore \quad \mathrm{I} _{x}: \mathrm{I} _{\mathrm{y}}: \mathrm{I} _{\mathrm{z}}::\left(\lambda _{2}+\lambda _{3}\right):\left(\lambda _{3}+\lambda _{1}\right):\left(\lambda _{1}+\lambda _{2}\right) \end{aligned} $$

Solution :

$\mathrm{I} _{1}=\mathrm{I} _{\mathrm{AB}}+\mathrm{I} _{\mathrm{AC}}+\mathrm{I} _{\mathrm{BC}}$

$\mathrm{I} _{\mathrm{AB}}=0 ; \mathrm{I} _{\mathrm{AC}}=\mathrm{I} _{\mathrm{BC}}=\frac{1}{3} \mathrm{ML}^{2} \sin ^{2} 60=\frac{\mathrm{ML}^{2}}{4}$

$\therefore \quad \mathrm{I} _{1}=\frac{\mathrm{ML}^{2}}{2} \hspace{40mm}. . . . . . . . .(1)$

Similarly,

$\mathrm{I} _{2}=\frac{1}{3} \mathrm{ML}^{2}+\frac{1}{3} \mathrm{ML}^{2}+\mathrm{I} _{\mathrm{BC}}$

From right angled $\triangle \mathrm{ABD}$ in fig. 23(b); we have

$$ \begin{align*} & \frac{\mathrm{AD}}{\mathrm{AB}}=\cos 30 \quad \therefore \mathrm{AD}=\frac{\sqrt{3}}{2} \mathrm{~L} \\ & \therefore \quad \mathrm{I} _{\mathrm{BC}}=\frac{1}{12} \mathrm{ML}^{2}+\mathrm{M}\left(\frac{\sqrt{3}}{2} \mathrm{~L}\right)^{2} \quad \text { [Clf theorem of } | \text { axis] } \\ & \quad=\frac{5}{6} \mathrm{ML}^{2} \tag{ii} \\ & \mathrm{I} _{2}=\frac{2}{3} \mathrm{ML}^{2}+\frac{5}{6} \mathrm{ML}^{2}=\frac{3}{2} \mathrm{ML}^{2} \\ & \therefore \quad \frac{\mathrm{I} _{1}}{\mathrm{I} _{2}}=\frac{1}{3} \end{align*} $$

Solution :

$I _{A}, I _{B}, I _{C}, I _{D}, I _{E}$ denote M.I of, spheres A, B, C, D and E respectively about $\mathrm{zz}^{\prime}$ as axis. $\mathrm{zz}^{\prime}$ as tangent to sphere $\mathrm{B}$ and $\mathrm{D}$. Therefore,

$$ \mathrm{I} _{\mathrm{B}}=\mathrm{I} _{\mathrm{D}}=\frac{2}{5} \mathrm{MR}^{2}+\mathrm{MR}^{2} $$

Using theorem of parallel axis

$$ \mathrm{I} _{\mathrm{A}}=\frac{2}{5} \mathrm{MR}^{2}+\mathrm{M}(3 \mathrm{R})^{2} $$

and $\quad \mathrm{I} _{\mathrm{C}}=\mathrm{I} _{\mathrm{E}}=\frac{2}{5} \mathrm{MR}^{2}+\mathrm{M}(4 \mathrm{R})^{2}$

$$ \begin{aligned} \therefore \quad \mathrm{I} & =\left(\frac{2}{5} \mathrm{MR}^{2}+9 \mathrm{MR}^{2}\right)+2\left(\frac{2}{5} \mathrm{MR}^{2}+\mathrm{MR}^{2}\right)+2\left(\frac{2}{5} \mathrm{MR}^{2}+16 \mathrm{MR}^{2}\right) \\ & =45 \mathrm{MR}^{2} \end{aligned} $$

Solution :

In Fig. $25 \mathrm{ABCD}$ is block. $\mathrm{AB}=\mathrm{a} ; \mathrm{AD}=\mathrm{h}$. The free body diagram is as shown. Obviously

$$ \mathrm{f} _{\mu}=\mu \mathrm{N}=\mu \mathrm{mg} \cos \theta $$

If $\mathrm{mg} \sin \theta \leq \mathrm{f} _{\ell t} ;$ block does not slide down the inclined plane. It will topic over if torque due to weight about end $\mathrm{A}$ is more than torque due to force of friction. i.e.

$$ (\mathrm{mg} \sin \theta) \frac{\mathrm{h}}{2}>(\mathrm{mg} \cos \theta)\left(\frac{\mathrm{a}}{2}\right) $$

Fig. 25

i.e. $\quad \tan \theta>\frac{\mathrm{a}}{\mathrm{h}}$

Therefore block does not slide before toppling if

$$ \begin{align*} & \tan \theta>\frac{\mathrm{a}}{\mathrm{h}}<\mu \\ & \therefore \mu>\frac{\mathrm{a}}{\mathrm{h}} \tag{1} \end{align*} $$

The block will slide down before toppling if

$\mathrm{mg} \sin \theta>f _{t \mathrm{t}}$

$\mathrm{mg} \sin \theta>\mu \mathrm{mg} \cos \theta$

or $\quad \tan \theta>\mu$

In other words block slides before toppling if

$$ \mu<\frac{\mathrm{a}}{\mathrm{h}} $$

Solution :

The M.I of disc about its own axis

$$ \mathrm{I}=\frac{\mathrm{MR}^{2}}{2} $$

The instantaneous angular acceleration; $\alpha$, is

$$ \begin{align*} & \alpha=\frac{\tau}{\mathrm{I}}=\frac{2(\alpha \theta+\beta)}{\mathrm{MR}^{2}} \tag{i} \\ & \alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=\frac{\mathrm{d} \omega}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{dt}}=\omega \frac{\mathrm{d} \omega}{\mathrm{d} \theta} \tag{ii} \end{align*} $$

From Eqns. (i) and (ii); we have

$$ \omega \mathrm{d} \omega=\frac{2}{\mathrm{MR}^{2}}(\alpha \theta+\beta) \mathrm{d} \theta $$

Integrate,

$$ \int _{0}^{\omega} \omega \mathrm{d} \omega=\frac{2}{\mathrm{MR}^{2}} \int _{0}^{\theta}(\alpha \theta+\beta) \mathrm{d} \theta $$

or $\quad \frac{\omega^{2}}{2}=\frac{2}{\mathrm{MR}^{2}}\left[\frac{\alpha \theta^{2}}{2}+\beta \theta\right]$

$\therefore \quad \omega=\left[\frac{2}{\mathrm{MR}^{2}}\left(\alpha \theta^{2}+2 \beta \theta\right)\right]^{1 / 2}$

Solution :

Since strings $A B$ and $C D$ are of a fixed length; the cylinder can only rotate about its own axis i.e. it has pure rotational motion. The torque, $\tau$; of external applied force.

Fig. 26

$$ \tau=\mathrm{FR} $$

The angular acceleration; $\alpha$, is

$$ \alpha=\frac{\tau}{\mathrm{I}}=\frac{\mathrm{FR}}{\left(\frac{\mathrm{MR}^{2}}{2}\right)}=\frac{2 \mathrm{~F}}{\mathrm{MR}} $$

The motion of cylinder is uniformly accelerated. Let $\theta$ be angle described in time t. Obviously

$$ \begin{gathered} \theta=\frac{1}{2} \alpha \mathrm{t}^{2}=\frac{\mathrm{Ft}^{2}}{\mathrm{MR}} \\ \ell=\text { The length of string unwound }=\mathrm{R} \theta=\left(\frac{\mathrm{F}}{\mathrm{M}}\right) \mathrm{t}^{2} \end{gathered} $$

Solution :

Regarding cylinder and dust as a system; there is no external force and therefore no external torque, the angular momentum is conserved.

$\mathrm{L} _{1}=$ The initial angular momentum

$$ \begin{equation*} =\left(\frac{1}{2} \mathrm{MR}^{2}\right) \omega _{1} \tag{i} \end{equation*} $$

Let $\mathrm{I} _{2}$ be the M.I of cylinder + dust collected about axis of rotation. Then

$$ \mathrm{I} _{2}=\frac{\mathrm{MR}^{2}}{2}+\frac{1}{2}\left(\frac{\mathrm{M}}{4}\right)\left[(2 \mathrm{R})^{2}+\mathrm{R}^{2}\right]^{*}=\frac{9}{8} \mathrm{MR}^{2} $$

Let $\omega _{2}$ be the final angular speed of system. Then

$\mathrm{L} _{2}=$ The final angular momentum

$$ \begin{equation*} =\mathrm{I} _{2} \omega _{2}=\left(\frac{9}{8} \mathrm{MR}^{2}\right) \omega _{2} \tag{ii} \end{equation*} $$

From law of conservation of angular momentum

$$ \begin{aligned} & \mathrm{L} _{1}=\mathrm{L} _{2} \quad \text { or } \quad \frac{\mathrm{MR}^{2} \omega _{1}}{2}=\frac{9 \mathrm{MR}^{2} \omega _{2}}{8} \\ \therefore \quad \omega _{2} & =\frac{4}{9} \omega _{1} \end{aligned} $$

Solution :

In Fig. 28; OB is instantaneous position of rod when it has rotated through angle $\theta$. The rod is moving in a vertical plane in gravity of earth, there is a loss in gravitational potential energy. C is C.G of rod initially. $C _{1}$ is instantaneous position of C.G as shown in Fig. 28.

$\mathrm{h}=$ The vertical distance by which $\mathrm{C} . \mathrm{G}$ goes down $=\mathrm{CC} _{2}=\mathrm{OC}-\mathrm{OC} _{2}$

$$ =\frac{\mathrm{L}}{2}(1-\cos \theta) $$

Fig. 28

The loss in gravitational P.E $=\operatorname{Mgh}=\frac{\mathrm{MgL}}{2}(1-\cos \theta) \hspace{20mm}. . . . . . . . .(i)$

Let $\omega$ be the instantaneous angular speed of rod.

The gain in rotational $K . E=\frac{I}{2}\left[\omega^{2}-\omega _{0}^{2}\right]$

$$ \begin{equation*} =\frac{1}{2}\left(\frac{\mathrm{ML}^{2}}{3}\right)\left[\omega^{2}-\left(\frac{\mathrm{v} _{0}}{\mathrm{~L}}\right)^{2}\right] \tag{ii} \end{equation*} $$

From law of conservation of energy

$$ \begin{align*} & \frac{\mathrm{MgL}}{2}(1-\cos \theta)=\frac{\mathrm{ML}^{2}}{6}\left[\omega^{2}-\frac{\mathrm{v} _{0}^{2}}{\mathrm{~L}^{2}}\right] \\ & \therefore \quad \omega^{2}=\frac{3 \mathrm{~g}}{\mathrm{~L}}(1-\cos \theta)+\frac{\mathrm{v} _{0}^{2}}{\mathrm{~L}^{2}} \tag{iii} \end{align*} $$

Given $\theta=60^{\circ}$. Therefore

$$ \omega^{2}=\frac{3 g}{2 \mathrm{~L}}+\frac{\mathrm{v} _{0}^{2}}{\mathrm{~L}^{2}}=\left(\frac{2 \mathrm{v} _{0}^{2}+3 \mathrm{gL}}{2 \mathrm{~L}^{2}}\right) $$

or $\quad \omega=\frac{1}{\mathrm{~L}}\left[\frac{2 \mathrm{v} _{0}^{2}+3 \mathrm{gL}}{2}\right]^{1 / 2}$

Solution :

Fig. 29

When $\mathrm{m}$ descends down by a distance of $1 \mathrm{~m}$ along the plane; the vertical distance, $\mathrm{h}$, is

$$ \mathrm{h}=1 \times \cos 53^{\circ}=0.6 \mathrm{~m} $$

The loss in gravitational P.E of $\mathrm{m}=\mathrm{mgh}$

$$ \begin{equation*} =0.2 \times 10 \times 0.6=1.2 \mathrm{~J} \tag{i} \end{equation*} $$

The work done against force of friction between block and inclined plane $=\left(\mu \mathrm{mg} \cos 37^{\circ}\right) \times 1$

$$ \begin{equation*} =0.1 \times 0.2 \times 10 \times 0.8 \times 1 \simeq 0.16 \mathrm{~J} \tag{ii} \end{equation*} $$

The gain in $\mathrm{K} . \mathrm{E}=\frac{1}{2} \times 0.2 \times \mathrm{v}^{2}+\frac{1}{2}(0.8) \omega^{2}$

$\mathrm{v}=$ linear speed of mass M; $\omega=$ angular speed of pulley. Since there is no slipping $\mathrm{v}=\mathrm{R} \omega$. Therefore

The gain in K.E $=0.1 \mathrm{v}^{2}+0.4\left(\frac{\mathrm{v}}{0.4}\right)^{2}=2.6 \mathrm{v}^{2}$

From law conservation of energy

$$ \begin{aligned} 1.2-0.16 & =2.6 \mathrm{v}^{2} \\ \therefore \quad \mathrm{v} & =\sqrt{\frac{1.04}{2.6}} \simeq 0.63 \mathrm{~ms}^{-1} \end{aligned} $$

Answer: $\left.\left[\left(\frac{2}{3}+\frac{2}{\sqrt{3}}\right) \hat{\mathrm{i}}-\frac{2}{3} \hat{\mathrm{j}}\right]\right]$

Fig. 31

Answer: (i) $\left(\mathrm{F} _{1}-\mathrm{F} _{2}\right) \mathrm{R}$, anticlockwise (ii) $\frac{2\left(\mathrm{~F} _{1}-\mathrm{F} _{2}\right)}{\mathrm{MR}}$ ]

Fig. 33

Correct answer: (2)

Solution:

Let $\mathrm{m} _{1}$ be mass moved from B to A as shown in Fig. 1(b). The co-ordinates of C.M is at origin of co-ordinates i.e. $\mathrm{X}=0$

$\therefore 0=\frac{\left(2 \mathrm{~m}+\mathrm{m} _{1}\right)(-\mathrm{a})+\left(4 \mathrm{~m}-\mathrm{m} _{1}\right) \times 2 \mathrm{a}}{6 \mathrm{~m}}$

or $\left(2 m+m _{1}\right)=2\left(4 m-m _{1}\right)$

$\therefore \mathrm{m} _{1}=2 \mathrm{~m}$

$\frac{\mathrm{m} _{1}}{4 \mathrm{~m}} \times 100=$ Percentage of mass of B transferred $=50 \%$

Correct answer: (1)

Solution:

Let $\mathbf{r} _{3}=x \hat{\dot{i}}+y \hat{j}+z \hat{k}$. Then $\mathbf{R}=\frac{m \mathbf{r} _{1}+m \mathbf{r} _{2}+2 m \mathbf{r} _{3}}{m+m+2 m}$

or $\quad \hat{i}+\hat{j}=(\hat{i}+2 \hat{j}-\hat{k})+(4 \hat{i}-\hat{j}+2 \hat{k})+2(x \hat{i}+y \hat{j}+z \hat{k}) / 4$

or $\quad 4(\hat{\mathrm{i}}+\hat{\mathrm{j}})=(5+2 x) \hat{\mathrm{i}}+(1+2 \mathrm{y}) \hat{\mathrm{j}}+(1+2 \mathrm{z}) \hat{\mathrm{k}}$

Equating coefficient of $\hat{i}, \hat{j}$ and $\hat{k}$ we have

$4=5+2 x \hspace{40mm}. . . . . . . . .(i)$

$4=1+2 \mathrm{y} \hspace{40mm}. . . . . . . . .(ii)$

$0=1+2 \mathrm{z} \hspace{40mm}. . . . . . . . .(iii)$

Therefore $x=-\frac{1}{2} ; \mathrm{y}=\frac{3}{2} ; \mathrm{z}=-\frac{1}{2}$. Obviously

$\mathbf{r} _{3}=(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathrm{k}}) / 2$

Correct answer: (4)

Solution:

For the system shown co-ordinates of $\mathrm{A}\left(-\frac{\ell}{2} ;-\frac{\ell}{2 \sqrt{3}}\right) ; \mathrm{B}\left(\frac{\ell}{2} ;-\frac{\ell}{2 \sqrt{3}}\right) \mathrm{C}\left(0 ; \frac{\ell}{\sqrt{3}}\right)$ and $0(0,0)$. Let $\mathrm{X}$ and $\mathrm{Y}$ be the co-ordinates of C.M. Then

$$ \begin{aligned} & \mathrm{X}=\frac{\mathrm{m}\left(-\frac{\ell}{2}\right)+\mathrm{m}\left(\frac{\ell}{2}\right)+\mathrm{m}(0) ;+2 \mathrm{~m}(0)}{5 \mathrm{~m}}=0 \\ & \mathrm{Y}=\frac{\mathrm{m}\left(-\frac{\ell}{2 \sqrt{3}}\right)+\mathrm{m}\left(-\frac{\ell}{2 \sqrt{3}}\right)+2 \mathrm{~m}\left(\frac{\ell}{\sqrt{3}}\right)}{5 \mathrm{~m}}=\frac{\ell}{5 \sqrt{3}} \end{aligned} $$

Correct answer: (3)

Solution:

Choose $x-y$ axis as shown in Fig. 3. Co-ordinates of different points are:

$0(0,0) ; \mathrm{A}(\ell, 0) \cdot \mathrm{B}\left(\frac{\ell}{2}, \frac{\sqrt{3}}{2} \ell\right)$

$\mathrm{C}\left(\frac{\ell}{2} ; 0\right) ; \mathrm{D}\left(\frac{\ell}{4}, \frac{\sqrt{3}}{4} \ell\right) ; \mathrm{E}\left(\frac{3 \ell}{4}, \frac{\sqrt{3}}{4} \ell\right)$

(X, Y) are co-ordinates of C.M.

$X=\frac{m \times 0+m(\ell)+m\left(\frac{\ell}{2}\right)+2 m\left(\frac{\ell}{2}\right)+2 m\left(\frac{\ell}{4}\right)+2 m\left(\frac{3 \ell}{4}\right)}{m+m+m+3(2 m)}$

Fig. 3

$$ =\frac{9 \mathrm{~m} \ell}{2 \times 9 \mathrm{~m}}=\frac{\ell}{2} $$

$$ \begin{aligned} Y= & \frac{\mathrm{m} \times 0+\mathrm{m} \times 0+\mathrm{m}\left(\frac{\sqrt{3}}{2} \ell\right)+2 \mathrm{~m} \times 0+2 \mathrm{~m}\left(\frac{\sqrt{3}}{4} \ell\right)+2 \mathrm{~m}\left(\frac{\sqrt{3}}{4} \ell\right)}{9 \mathrm{~m}} \\ & =\frac{3 \sqrt{3} \mathrm{~m} \ell}{2 \times 9 \mathrm{~m}}=\frac{\ell}{2 \sqrt{3}} \end{aligned} $$

Correct answer: (3)

Solution:

Fig. 4

Fig. 4 shows a uniform circular disc of radius R. Consider the disc as made up of concentric semi-circular rings. The radius of rings varies from 0 to $R$. By symmetry the center of mass lies on y-axis. Consider an elementary ring of radius $r$ thickness dr as shown in Fig. 4.

$\mathrm{dm}=$ Mass of elementary ring considered $=\pi(\mathrm{rdr}) \sigma ; \quad \sigma=$ mass $/$ area $=$ constant

$Y=$ Coordinate of $C . M=\frac{\int _{0}^{R} r d m}{\int _{0}^{R} d m}$

$=\frac{\int _{0}^{\mathrm{R}} \pi \sigma r^{2} d r}{\int _{0}^{\mathrm{R}} \pi \sigma r d r}=\frac{\frac{\mathrm{R}^{3}}{3}}{\frac{\mathrm{R}^{2}}{2}}=\frac{2 \mathrm{R}}{3}$

Correct answer: (3)

Solution:

Let $\mathrm{m}$ be mass of complete disc of radius R.The mass $\mathrm{m} _{1}$ of the part removed is $\frac{\mathrm{m}}{16}$. To obtain C.M of remaining disc we consider mass $\mathrm{m}$ to be placed at $\mathrm{O}$ and $-\left(\frac{\mathrm{m}}{16}\right)$ at $\mathrm{O} _{1}$ as shown in Fig. 6. Distance $\mathrm{OO} _{1}=\frac{3 \mathrm{R}}{4}$. The co-ordinates of $\mathrm{O} _{1}$ are $\left(\frac{3 \mathrm{R}}{4} \times \frac{1}{\sqrt{2}} ; \frac{3 \mathrm{R}}{4} \times \frac{1}{\sqrt{2}}\right)$

The co-ordinates (X, Y) of C.M are:

$X=\frac{m \times 0+\left(-\frac{m}{16}\right) \frac{3 R}{4 \sqrt{2}}}{m+\left(-\frac{m}{16}\right)}=-\left(\frac{R}{20 \sqrt{2}}\right)$

Similarly $\mathrm{Y}=-\left(\frac{\mathrm{R}}{20 \sqrt{2}}\right)$

Fig. 6

Correct answer:(2)

Solution:

Let $\mathrm{m} _{1}=$ Mass of complete disc of radius $\mathrm{R}=\pi \mathrm{R}^{2} \sigma$

$\mathrm{m} _{2}=$ Mass of disc of radius $\mathrm{r}$, removed $=\pi \mathrm{R}^{2} \sigma$

$m _{3}=$ Mass of disc of radius $r$, placed with center at

$\mathrm{O} _{2}=\pi \mathrm{r}^{2}(2 \sigma)=2 \pi \mathrm{r}^{2} \sigma$

The given arrangement is equivalent to what is shown in Fig. 8. Note mass at $\mathrm{O} _{1}$ has negative sign because it has been removed. Co-ordinates of $\mathrm{O}(0$, $0) ; \mathrm{O} _{1}(\mathrm{a}, 0)$ and $\mathrm{O} _{2}(0, \mathrm{~b})$

Fig. 8

Let $(\mathrm{X}, \mathrm{Y})$ be the co-ordinates of $\mathrm{C} . \mathrm{M}$ of the arrangement. Then

$$ \begin{aligned} X= & \frac{m(0)+\left(-m _{1}\right) a+m _{2}(0)}{m+\left(-m _{1}\right)+m _{2}} \\ & =-\left[\frac{m _{1} a}{\left(m-m _{1}+m _{2}\right)}\right]=-\left[\frac{\pi r^{2} \sigma a}{\pi\left[R^{2}-r^{2}+2 r^{2}\right] \sigma}\right] \\ & =-\left(\frac{r^{2} a}{R^{2}+r^{2}}\right) \end{aligned} $$

Similarly,

$$ \begin{aligned} & \mathrm{Y}= \frac{\mathrm{m}(0)+\left(-\mathrm{m} _{1}\right)(0)+\mathrm{m} _{2} \times \mathrm{b}}{\mathrm{m}+\left(-\mathrm{m} _{1}\right)+\mathrm{m} _{2}} \\ &=\frac{2 \pi \mathrm{r}^{2} \sigma \cdot \mathrm{b}}{\pi\left[\mathrm{R}^{2}-\mathrm{r}^{2}+2 \mathrm{r}^{2}\right] \sigma}=\frac{2 \mathrm{r}^{2} \mathrm{~b}}{\left(\mathrm{R}^{2}+\mathrm{r}^{2}\right)} \end{aligned} $$

Correct answer: (1)

Solution:

Particle A is moving freely under gravity; the C.M of A moves downwards along the vertical line path of A. Since A breaks up to two fragments on its own i.e. their is no external force; the equation of motion of C.M of $B$ and $C$ is same as of particle $A$. This means there is no change in path of C.M of fragments A and B.

Correct answer: (4)

Solution:

For uniform rod OA, the center of mass $X _{0}=\frac{L}{2}$. For non-uniform rod, the position X, of C.M is

$$ \begin{array}{r} \mathrm{X}=\frac{\int _{0}^{\mathrm{L}}(x \lambda \mathrm{d} x)}{\int _{0}^{\mathrm{L}} \lambda \mathrm{d} x}=\frac{\lambda _{0} \int _{0}^{\mathrm{L}}(\mathrm{L}-x) x \mathrm{~d} x}{\lambda _{0} \int _{0}^{\mathrm{L}}(\mathrm{L}-x) \mathrm{d} x} \\ =\frac{\left(\frac{\mathrm{L}^{3}}{2}-\frac{\mathrm{L}^{3}}{3}\right)}{\left(\mathrm{L}^{2}-\frac{\mathrm{L}^{2}}{2}\right)}=\frac{\mathrm{L}^{3}}{6} \times \frac{2}{\mathrm{~L}^{2}}=\frac{\mathrm{L}}{3} \end{array} $$

The change in position of $C \cdot M=\frac{L}{2}-\frac{L}{3}=\frac{L}{6}$

Correct answer: (4)

Solution:

The initial position $\mathrm{X} _{0}$ of the C.M. of system (i. e. $t=0$ ) is

$\mathrm{X} _{0}=\frac{\mathrm{m}(0)+2 \mathrm{~m}(3 \mathrm{a})}{\mathrm{m}+2 \mathrm{~m}}=2 \mathrm{a}$

$\mathrm{v} _{0}=$ The initial speed of. $\mathrm{C.M}$ of system

$$ =\frac{\mathrm{m} \times 0-2 \mathrm{mv} _{0}}{\mathrm{~m}+2 \mathrm{~m}}=-\left(\frac{2 \mathrm{v} _{0}}{3}\right) $$

When B starts moving towards A; there is no external force. The speed of center of mass remains same at all times. The instantaneous position, X of C.M is

$$ X=X _{0}+v _{0} t=2 a-\left(\frac{2 v _{0}}{3}\right) \mathrm{t} $$

Let $\mathrm{X}=\mathrm{a}$ at $\mathrm{t}=\mathrm{t} _{1}$. Then $\mathrm{a}=2 \mathrm{a}-\left(\frac{2 \mathrm{v} _{0}}{3}\right) \mathrm{t} _{1}$

$\therefore \mathrm{t} _{1}=\frac{3 \mathrm{a}}{2 \mathrm{v} _{0}}$

Correct answer: (3)

Solution:

Let $a _{1}$ and $a _{2}$ be the acceleration of mass $m _{1}$ and $m _{2}$ respectively. Then $a _{1}=0$ (There is no external force applied on $m _{1}$ ), and

$$ \mathrm{a} _{2}=\frac{\mathrm{F}}{\mathrm{m} _{2}} $$

Let $a _{c}$ be acceleration of C.M. Then

$$ \mathrm{a} _{\mathrm{c}}=\frac{\mathrm{m} _{1} \mathrm{a} _{1}+\mathrm{m} _{2} \mathrm{a} _{2}}{\mathrm{~m} _{1}+\mathrm{m} _{2}}=\frac{\mathrm{F}}{\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right)}=\mathrm{constant} $$

$\mathrm{S}=$ Distance traveled by C.M in time $\mathrm{t}$

$$ =\frac{1}{2} \mathrm{a} _{\mathrm{c}} \mathrm{t}^{2}=\frac{1}{2}\left(\frac{\mathrm{Ft}^{2}}{\mathrm{~m} _{1}+\mathrm{m} _{2}}\right) $$

$\mathrm{S} _{1}=$ Instantaneous distance between $\mathrm{m} _{1}$ and $\mathrm{m} _{2}$ $=\mathrm{D}+$ distance traveled by $\mathrm{m} _{2}$ in time $\mathrm{t}$

$=\mathrm{D}+\frac{1}{2} \mathrm{a} _{2} \mathrm{t}^{2}=\mathrm{D}+\frac{1}{2}\left(\frac{\mathrm{Ft}^{2}}{\mathrm{~m} _{2}}\right)$

Correct answer: (4)

Solution:

The entire mass $4 \mathrm{~m}$ of rod can be considered as concentrated at its center i.e. $x _{1}=\mathrm{L}$. The instantaneous position of $\mathrm{P}=x _{2}=\mathrm{v} _{0} \mathrm{t}$. The instantaneous position of C.M X is

$$ \mathrm{X}=\frac{4 \mathrm{~m} \times \mathrm{L}+\mathrm{mv} _{0} \mathrm{t}}{4 \mathrm{~m}+\mathrm{m}}=0.8 \mathrm{~L}+0.2 \mathrm{v} _{0} \mathrm{t} $$

$\mathrm{X}$ vs $\mathrm{t}$ graph is a straight line. Also $\mathrm{X}(\mathrm{t}=0)=0.8 \mathrm{~L}$

Let $P$ reach end A of rod at $t=t _{1}=\frac{2 L}{v _{0}}$. Then

$$ \mathrm{X}\left(\mathrm{t}=\mathrm{t} _{1}\right)=0.8 \mathrm{~L}+0.4 \mathrm{~L}=1.2 \mathrm{~L} $$

Correct answer: (2)

Solution:

The mass $\mathrm{M}$ of rod is considered as concentrated at its midpoint $\mathrm{O} _{1} ; x _{1}=\frac{\mathrm{L}}{2}$. Let $\mathrm{P}$ be the instantaneous position of particle on rod. The instantaneous position co-ordinate of $\mathrm{P}$

$$ x _{2}=\left(\mathrm{L}-\frac{1}{2} \mathrm{at}{ }^{2}\right) $$

$\mathrm{X}=$ The instantaneous position co-ordinate of C.M

$$ \begin{aligned} & =\frac{\mathrm{m}\left(x _{1}\right)+\mathrm{m}\left(x _{2}\right)}{\mathrm{M}+\mathrm{m}}=\frac{\mathrm{M}\left(\frac{\mathrm{L}}{2}\right)}{(\mathrm{M}+\mathrm{m})}+\left(\frac{\mathrm{m}}{\mathrm{M}+\mathrm{m}}\right)\left(\mathrm{L}-\frac{1}{2} a \mathrm{t}^{2}\right) \\ & =\left[\frac{\mathrm{M}+2 \mathrm{~m}}{2(\mathrm{M}+\mathrm{m})}\right] \mathrm{L}-\left[\frac{\mathrm{ma}}{2(\mathrm{M}+\mathrm{m})}\right] \mathrm{t}^{2} \end{aligned} $$

Correct answer: (3)

Solution:

As the ball moves on planck there is no external force. The center of mass of ‘planck-ball’ system is initially at rest, it remains at rest as ball moves on planck. Let ball travel a distance L (= the length of planck) is time t. Then $\mathrm{v} _{0}=\frac{\mathrm{L}}{\mathrm{t}}$. In this time the planck moves by a distance

$$ \ell=\frac{\mathrm{mL}}{(\mathrm{M}+\mathrm{m})} $$

$\mathrm{v}=$ The speed acquired by planck $=\frac{\ell}{\mathrm{t}}=\frac{\mathrm{m}\left(\frac{\mathrm{L}}{\mathrm{t}}\right)}{\mathrm{M}+\mathrm{m}}$

$$ =\frac{\mathrm{mv} _{0}}{(\mathrm{M}+\mathrm{m})} $$

Given $\mathrm{M}=10 \mathrm{~kg} ; \mathrm{m}=0.5 \mathrm{~kg} ; \mathrm{v}=0.2 \mathrm{~ms}^{-1}$, therefore

$$ \begin{aligned} \mathrm{v} _{0}= & \frac{0.2(10+0.5)}{0.5} \mathrm{~ms}^{-1} \\ & =4.2 \mathrm{~ms}^{-1} \end{aligned} $$

Correct answer: (1)

Solution:

The initial velocity $\mathbf{V} _{0}$ of center of mass A and B is

$$ \begin{equation*} \mathbf{V} _{0}=\frac{\mathrm{m} _{1} \mathbf{u} _{1}+\mathrm{m} _{2} \mathbf{u} _{2}}{\mathrm{~m} _{1}+\mathrm{m} _{2}}=\mathrm{constant} \tag{i} \end{equation*} $$

Let $\mathbf{v} _{1 c}$ and $\mathbf{v} _{2 c}$ be velocity of A and B in C.M frame of reference. Then

$$ \mathbf{v} _{1 \mathrm{c}}=\mathbf{u} _{1}-\mathbf{V} _{0} ; \mathbf{v} _{2 \mathrm{c}}=\mathbf{u} _{2}-\mathbf{V} _{0} $$

$\mathbf{P} _{\mathrm{c}}=$ The total linear momentum of A and B in C.M frame of reference.

$$ \begin{align*} & =\mathbf{m} _{1} \mathbf{v} _{1 \mathrm{c}}+\mathbf{m} _{2} \mathbf{v} _{2 \mathrm{c}}=\mathrm{m} _{1}\left(\mathbf{u} _{1}-\mathbf{V} _{0}\right)+\mathrm{m} _{2}\left(\mathbf{u} _{2}-\mathbf{V} _{0}\right) \\ & =\left(\mathrm{m} _{1} \mathbf{u} _{1}+\mathrm{m} _{2} \mathbf{u} _{1}\right)-\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right) \mathbf{V} _{0} \tag{ii} \end{align*} $$

From Eqns. (i) and (ii) we have,

$$ \mathbf{P} _{\mathrm{c}}=\text { zero } $$

Correct answer: (1)

Solution:

Fig. 12 shows position of masses $\mathrm{m} _{1}$ and $\mathrm{m} _{2}$ at $\mathrm{t}=0$.

$\mathrm{X} _{0}=$ Initial co-ordinate of C.M

$$ =\frac{\mathrm{m} _{1} \times 0+\mathrm{m} _{2}(2 \mathrm{D})}{\mathrm{m} _{1}+\mathrm{m} _{2}}=\frac{2 \mathrm{~m} _{2} \mathrm{D}}{\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right)} $$

Fig. 12

$\mathrm{V} _{0}=$ The initial velocity of $\mathrm{C} . \mathrm{M}$

$$ =\frac{\mathrm{m} _{1} \times 0+\mathrm{m} _{2} \times 0}{\mathrm{~m} _{1}+\mathrm{m} _{2}}=0 $$

The equation of motion of C.M of system is

$$ \left(\mathrm{m} _{1}+\mathrm{m} _{2}\right) \frac{\mathrm{d}^{2} \mathrm{X}}{\mathrm{dt}}=0 $$

Note there is no external force. Masses move towards one another due to their mutual force of interaction, which is an internal force. Therefore

$$ \frac{\mathrm{d}^{2} \mathrm{X}}{\mathrm{dt}^{2}}=\frac{\mathrm{dV}}{\mathrm{dt}}=0 $$

$\therefore \mathrm{V}=\mathrm{V} _{0}=0$

The center of mass is initially at rest and it remains at rest throughout. Therefore distance traveled by center of mass is zero.

Correct answer: (2)

Solution:

Let $\mathrm{h}$ be the perpendicular distance of line along which particle moves from $\mathrm{O}$. The angular momentum $\mathrm{L} _{0}$ is

$$ \mathrm{L} _{0}=\mathrm{mv} _{0} \mathrm{~h} $$

$\mathrm{L} _{0}$ does not depend on $\mathrm{r}$ or $\theta$ ( $\mathrm{r}$ and $\theta$ both change in such a manner that $\mathrm{h}=\mathrm{r} \sin \theta=$ constant.)

Alternative Solution :

Since there is no external force i.e. $\mathbf{F} _{\text {ext }}=0$; there is no external torque, therefore $\mathbf{L}$ is constant.

Correct answer: (3)

Solution:

Since instantaneous velocity is always perpendicular to instantaneous acceleration, the particle moves in a circle with center $\mathrm{O}$. Let $\mathrm{R}$ be radius of circular path. The magnitude of acceleration is

$$ \beta=\frac{\alpha^{2}}{R} \quad \text { or } \quad R=\frac{\alpha^{2}}{\beta} $$

The magnitude of linear momentum of particle $=\mathrm{m} \alpha$. The magnitude of angular momentum; $\ell$; about $\mathrm{O}$ is

$$ \ell=p \times R=(m \alpha) \frac{\alpha^{2}}{\beta}=\frac{m \alpha^{3}}{\beta} $$

Correct answer: (4)

Solution:

$\mathbf{L}=$ Total angular momentum of the two particle system about point $\mathrm{O}=\mathbf{L} _{\mathrm{P}}+\mathbf{L} _{\mathrm{Q}}$

$\mathbf{L} _{\mathrm{P}}=$ Angular momentum of $\mathrm{P}$ about $\mathrm{O}=\mathrm{mv} _{0}a$; clockwise

$\mathbf{L} _{\mathrm{Q}}=$ Angular momentum of $\mathrm{Q}$ about $\mathrm{O}=4 \mathrm{mv} _{0}$; anticlockwise

$$ \begin{aligned} \therefore \quad|\mathbf{L}| & =\left|\mathbf{L} _{\mathrm{Q}}\right|-\left|\mathbf{L} _{\mathrm{P}}\right| \\ & =\mathrm{mv} _{0}[4 \mathrm{~b}-\mathrm{a}] \end{aligned} $$

The direction of $\mathbf{L}$ is anticlockwise

Correct answer: (1)

Solution:

For “disc-mass system”; there is no external force, therefore no external torque. The angular momentum of system is conserved.

$\mathrm{L} _{\mathrm{i}}=\mathrm{L} _{\text {DISC }}+\mathrm{L} _{\text {mass }}=0+\mathrm{mvR}$

$\mathrm{L} _{\mathrm{f}}=\mathrm{I} \omega _{0}$

$\mathrm{I}=\mathrm{I} _{\text {DISC }}+\mathrm{I} _{\text {mass }}=\frac{1}{2}(2 \mathrm{~m}) \mathrm{R}^{2}+\mathrm{mR}^{2}=2 \mathrm{mR}^{2}$

$\therefore \quad \mathrm{mvR}=2 \mathrm{mR}^{2} \omega _{0} \quad$ or $\quad \omega _{0}=\frac{\mathrm{v}}{2 \mathrm{R}}$

The mass m moves on disc, somehow on its own.e. there is no external force. Angular momentum is again conserved.

$$ \begin{aligned} & \mathrm{I} \omega _{0}=\mathrm{I}^{\prime} \omega^{\prime} ; \quad \omega^{\prime}=\text { angular speed of system as mass } \mathrm{m} \text { reach } \mathrm{O} . \\ & \mathrm{I}^{\prime}=\mathrm{I} _{\mathrm{DISC}}=\mathrm{mR}^{2} \\ & \therefore \quad \omega^{\prime}=\left(\frac{2 \mathrm{mR}^{2}}{\mathrm{mR}^{2}}\right) \omega _{0}=2 \omega _{0}=\frac{\mathrm{v}}{\mathrm{R}} \end{aligned} $$

Correct answer: (4)

Solution:

Fig. 16 show the parabolic path of particle when it is in position $P$. The velocity of particle is $\mathrm{v} _{0}$ as shown in Fig. 16. OQ is perpendicular from origin $\mathrm{O}$ on the line of action of linear momentum of particle at $\mathrm{P}$.

$\ell=$ angular momentum of particle at $\mathrm{P}$ about $\mathrm{O}=\mathrm{mv} _{0} \times \mathrm{OQ}$

$=\mathrm{mv} _{0} \mathrm{R} \sin \theta$

Fig. 16

$\text { where } \mathrm{OP}=\mathrm{R}=\frac{\mathrm{v}_0^2 \sin 2 \theta}{\mathrm{g}}=\text { horizontal range of } \mathrm{P} \text {. }$

$$ \therefore \quad \ell=\frac{\mathrm{mv} _{0}^{3} \sin 2 \theta \sin \theta}{\mathrm{g}} $$

$$ =\frac{2 \mathrm{mv} _{0}^{3} \sin ^{2} \theta \cos \theta}{\mathrm{g}} $$

Correct answer: (4)

Solution:

There is no friction between $\mathrm{P}$ and surface in contact. When $\mathrm{P}$ and $\mathrm{Q}$ collide there is no transfer of angular momentum from P to Q. Sphere P only transfers its linear momentum to Q. Hence after collision P has same angular speed as before collision. Therefore $\omega _{1}=\omega$.

Correct answer: (2)

Solution:

Fig. 17

The moment of inertia $(J)$ of the system about axis through C.M $(\mathrm{O})$ and perpendicular to $A B$ is

$$ \begin{aligned} & \mathrm{I}=\mathrm{m} _{1} x _{1}^{2}+\mathrm{m} _{1} x _{2}^{2} \\ & x _{1}=\frac{\mathrm{m} _{2} \ell}{\mathrm{m} _{1}+\mathrm{m} _{2}} ; x _{2}=\frac{\mathrm{m} _{1} \ell}{\mathrm{m} _{1}+\mathrm{m} _{2}} \\ & \therefore \quad \mathrm{I}=\mathrm{m} _{1}\left(\frac{\mathrm{m} _{2} \ell}{\mathrm{m} _{1}+\mathrm{m} _{2}}\right)^{2}+\mathrm{m} _{2}\left(\frac{\mathrm{m} \ell}{\mathrm{m} _{1}+\mathrm{m} _{2}}\right)^{2} \\ & =\frac{\mathrm{m} _{1} \mathrm{~m} _{2} \ell^{2}}{\left(\mathrm{~m} _{1}+\mathrm{m} _{2}\right)} \end{aligned} $$

The angular speed $(\omega)$ of the system is $\omega=2 \pi \mathrm{n}$

The angular momentum L; of the system about axis of rotation;

$$ \begin{aligned} & \mathrm{L}=\mathrm{I} \omega \\ & =2 \pi\left(\frac{\mathrm{m} _{1} \mathrm{~m} _{2}}{\mathrm{~m} _{1}+\mathrm{m} _{2}}\right) \ell^{2} \mathrm{n} \end{aligned} $$

Correct answer: (3)

Solution:

Let $\omega$ be the instantaneous angular speed of stone when instantaneous radius of circle is $r$. The angular momentum $\ell$, of stone about center of circle.

$$ \begin{equation*} \ell=\mathrm{mr}^{2} \omega=\text { constant } \tag{i} \end{equation*} $$

The instantaneous tension $\mathrm{T}$ in string provides necessary centripetal force. Therefore,

$$ \begin{aligned} & \mathrm{T}=\mathrm{mr} \omega^{2} \\ & =\mathrm{mr}\left(\frac{\ell}{\mathrm{mr}^{2}}\right)^{2} \quad[\mathrm{Clf}=\mathrm{n}(\mathrm{i})] \\ & =\frac{\ell^{2}}{\mathrm{mr}^{3}} \quad \text { Since } \ell \text { and m are constant, therefore } \\ & \mathrm{T} \propto \mathrm{r}^{-3} \end{aligned} $$

Correct answer: (2)

Solution:

Consider “rod-particle system”. Since there is no external force angular momentum of system is conserved.

$\mathrm{L} _{\mathrm{i}}=$ Initial angular momentum of system about $\mathrm{O}=\mathrm{mv} _{0}(\mathrm{~L} / 2)$

$\mathrm{L} _{\mathrm{f}}=$ Final angular momentum of system about $\mathrm{O}=\mathrm{I} \omega$.

$\mathrm{I}=$ M.I of rod particle system about axis through midpoint $\mathrm{O}$ and perpendicular to length of rod

$$ \begin{aligned} & =\frac{1}{12}(2 \mathrm{~m})(2 \ell)^{2}+\mathrm{m}\left(\frac{\ell}{2}\right)^{2} \\ & =\frac{2}{3} \mathrm{~m} \ell^{2}+\frac{\mathrm{m} \ell^{2}}{4}=\frac{11}{12} \mathrm{~m} \ell^{2} \end{aligned} $$

From law of conservation of angular momentum

$$ \begin{aligned} & \mathrm{L} _{\mathrm{i}}=\mathrm{L} _{\mathrm{f}} \quad \text { or } \quad \frac{\mathrm{mv} _{0} \mathrm{~L}}{2}=\frac{11}{12} \mathrm{~m} \ell^{2} \omega \\ & \therefore \quad \mathrm{v} _{0}=\frac{11}{6} \ell \omega \end{aligned} $$

Correct answer: (2)

Solution:

Consider disc-particle system; there is no external force and therefore no external torque. The angular momentum of systemabout $\mathrm{O}$ is conserved.

$\mathrm{L} _{\mathrm{i}}=$ The total initial angular momentum $=\mathrm{mvR}$

$\mathrm{L} _{\mathrm{f}}=$ The total final angular momentum $=\left(\mathrm{I} _{\text {disc }}+\mathrm{I} _{\text {particle }}\right) \omega$

$$ =\left[\frac{1}{2}(2 m) R^{2}+m R^{2}\right] \omega=2 m R^{2} \omega $$

Now, $\mathrm{L} _{\mathrm{i}}=\mathrm{L} _{\mathrm{f}}$; therefore

$$ \begin{aligned} & \mathrm{mvR}=2 \mathrm{mR}^{2} \omega \\ & \omega=\frac{\mathrm{v}}{2 \mathrm{R}} \end{aligned} $$

Correct answer: (4)

Solution:

As bead slides along chord PQ there is no external force and therefore no external torque. The angular momentum of system remains constant. However the M.I (I) of the system varies as bead moves. The M.I decreases becomes minimum when bead is at middle of chord PQ and then increases. M.I with bead at end $\mathrm{P}$ and $\mathrm{Q}$ is same. Therefore $\omega$ first increases acquires a maximum value when $\mathrm{I}$ is minimum and then starts decreasing. Final value of $\omega=\omega _{0}$ when bead reaches end Q. The variation of I is non-linear as shown in(4).

Correct answer: (3)

Solution:

As beads slide down sides PQ and PR there is no external force (sides of $\triangle \mathrm{PQR}$ are smooth) and therefore no external torque. The total angular momentum of system is, therefore, conserved.

As beads slide down the beads lose gravitational potential, but their is a gain in K.E. The total energy is, therefore, conserved.

Correct answer: (3)

Solution:

The instantaneous force $\mathbf{F}$ on particle is $\mathbf{F}=\mathrm{q}=\mathrm{qE} _{0} \hat{\mathrm{i}}$. The instantaneous position vector $\mathbf{r}$ of particle is

$$ \mathbf{r}=a \hat{i}+b \hat{j} $$

The instantaneous torque $\tau=\mathbf{r} \times \mathbf{F}$

$$ \begin{aligned} & =(a \hat{i}+b \hat{j}) \times q E _{0} \hat{i} \\ & =-q E _{0} b(\hat{k}) \end{aligned} $$

Correct answer: (1)

Solution:

Since the gravitational force is still a radial force i.e. is directed along the line joining $\mathrm{m}$ and $\mathrm{M}$; torque of gravitational force on $\mathrm{m}$ due to $\mathrm{M}$, about position of $\mathrm{M}$, is zero. The perpendicular distance of line of action of $\mathbf{F}$ from position of $M$ is zero.

Correct answer: (3)

Solution:

$\mathbf{r}=$ The instantaneous position vector of particle $=a \hat{i}+b \hat{j}$

The instantaneous torque of forces on particle about origin of co-ordinates is

$$ \begin{aligned} \tau=\mathbf{r} & \times\left(\mathbf{F} _{1}+\mathbf{F} _{2}\right) \\ & =(\mathrm{a} \hat{\mathrm{i}}+\mathrm{b} \hat{\mathrm{j}}) \times\left[2 \mathrm{~F} _{0} \hat{\mathrm{i}}+\mathrm{F} _{0} \hat{\mathrm{j}}\right] \\ & =2 \mathrm{~F} _{0} \mathrm{~b}(\hat{\mathrm{j}} \times \hat{\mathrm{i}})+\mathrm{F} _{0} \mathrm{a}(\hat{\mathrm{i}} \times \hat{\mathrm{j}}) \\ & =\left[\mathrm{F} _{0}(\mathrm{a}-2 \mathrm{~b})\right] \hat{\mathrm{k}} \end{aligned} $$

Correct answer: (3)

Solution:

$\tau _{1}=$ Torque of force $\mathbf{F} _{1}=\mathbf{r} _{1} \times \mathbf{F} _{1}=3 \hat{\mathrm{i}} \times 0.5 \hat{\mathrm{j}}=1.5 \hat{\mathrm{k}}=1.5 \hat{\mathrm{k}} ; \mathrm{N}-\mathrm{m}$

$\tau _{2}=$ Torque of force $\mathbf{F} _{2}=\mathbf{r} _{2} \times \mathbf{F} _{2}=4 \hat{\mathrm{j}} \times(-0.6 \hat{\mathrm{i}})$

$$ =-2.4 \hat{\mathrm{j}} \times \hat{\mathrm{i}}=2.4 \hat{\mathrm{k}} ; \mathrm{N}-\mathrm{m} $$

Then net torque $\tau$ is

$\tau=\tau _{1}+\tau _{2}=3.9 \mathrm{~N}-\mathrm{m} ; \hat{\mathrm{k}}$

Correct answer:(1)

Solution:

The magnitude of the external applied torque $=\mathrm{F} \times \mathrm{L}$

whether rod is rotating about axis through end $\mathrm{O}$ or $\mathrm{A}$.

The M.I of rod about axis through end $\mathrm{O}$ and $\mathrm{A}$ is not same. $\mathrm{I} _{\mathrm{O}}>\mathrm{I} _{\mathrm{A}}$. This is due to variation of $\mathrm{I}$ with distribution of mass. Therefore angular speed or angular acceleration will be different when axis passes through end $\mathrm{O}$ or $\mathrm{A}$

Correct answer: (4)

Solution:

Due to flick imparted to rod the rod rotates in a vertical plane (X-Y plane) about axis through end O. Let $\theta$ be the instantaneous angle rod makes with vertical as shown in Fig. 24. Then

$\mathbf{w}=$ The weight of rod $=-(\mathrm{mg}) \hat{\mathrm{j}}$

$\mathbf{r}=$ The instantaneous position vector of C.G.

$$ =\left(\frac{L}{2} \sin \theta\right) \hat{\mathrm{i}}+\left(\frac{\mathrm{L}}{2} \cos \theta\right) \hat{\mathrm{j}} $$

The instantaneous torque about $\mathrm{O}$ due to weight of rod is

$$ \begin{aligned} & \tau=\mathbf{r} \times \mathbf{F} \\ & =\left[\left(\frac{L}{2} \sin \theta\right) \hat{i}+\left(\frac{L}{2} \cos \theta\right) \hat{j}\right] \times(-m g) \hat{j} \\ & =-\left(\frac{m g L}{2} \sin \theta\right) \hat{k} \end{aligned} $$

Fig. 24

The magnitude of torque changes with time $t$ because $\theta$ varies with time. However direction of $\tau$ is same at all times.

Correct answer: (3)

Solution:

The forces acting on cylinder are shown in Fig. 26. Taking moment of forces about point A at equilibrium.

$\operatorname{Mg}(\mathrm{AB})=\mathrm{F} . \mathrm{CD}$

$$ =\mathrm{F}(\mathrm{CO}+\mathrm{OD}) $$

or $\quad \operatorname{Mgr} \sin \theta=\mathrm{F}(\mathrm{r}+\mathrm{r} \sin \theta)$

where $\mathrm{OA}=\mathbf{r}=$ The radius of cylinder. Therefore

$$ F=\frac{M g \sin \theta}{(1+\sin \theta)} $$

Fig. 26

Correct answer: (4)

Solution:

$I=$ M.I of “ring bead system” about given axis $=I _{\text {ring }}+I _{P}+I _{Q}$

$$ \begin{aligned} & =\mathrm{MR}^{2}+\mathrm{mR}^{2}+\mathrm{mR}^{2} \\ & =(\mathrm{M}+2 \mathrm{~m}) \mathrm{R}^{2} \end{aligned} $$

When beads move to position $\mathrm{P}^{\prime}$ and $\mathrm{Q}^{\prime}$ their perpendicular distance from axis of rotation remain same. Hence I’ $=$ I

Correct answer: (2)

Solution:

$\mathrm{I}=$ The initial M.I of ring-bead system

$$ =\mathrm{I} _{\mathrm{ring}}+\mathrm{I} _{\mathrm{P}}+\mathrm{I} _{\mathrm{Q}}=\frac{1}{2}(2 \mathrm{~m}) \mathrm{R}^{2}=\mathrm{mR}^{2} $$

$I _{P}$ and $I _{Q}$ is zero because beads are lying on the axis of rotation. Their perpendicular distance from axis of rotation is zero.

$I^{\prime}=$ The final M.I of ring bead system

$$ \begin{aligned} & =\frac{1}{2}(2 m) R^{2}+m(R \sin \alpha)^{2}+m(R \sin \beta)^{2} \\ & =m R^{2}+m R^{2}\left(\sin ^{2} \alpha+\sin ^{2} \beta\right) \end{aligned} $$

$\therefore \Delta \mathrm{I}=$ The change in M.I of the system

$$ =I^{\prime}-I=m R^{2}\left(\sin ^{2} \alpha+\sin ^{2} \beta\right) $$

Correct answer: (1)

Solution:

$\mathrm{M}=$ Total mass of circular loop $=$ Mass of wire $=2 \lambda \mathrm{L}$

Let $r$ be radius of each turn of circular loop formed. Then

$$ 2(2 \pi r)=2 L \quad \text { or } \quad r=\frac{L}{2 \pi} $$

The M.I of one turn about any tangent in its own plane.

$$ \begin{aligned} & =\frac{1}{2}\left(\frac{\mathrm{M}}{2}\right) \mathrm{r}^{2}+\left(\frac{\mathrm{M}}{2}\right) \mathrm{r}^{2} \\ & =\frac{3 \mathrm{Mr}^{2}}{4}=\frac{3 \mathrm{M}}{4}\left(\frac{\mathrm{L}}{2 \pi}\right)^{2}=\frac{3 \mathrm{ML}^{2}}{16 \pi^{2}} \end{aligned} $$

$I=$ The total M.I of the two turn loop

$$ =2\left[\frac{3 \mathrm{ML}^{2}}{16 \pi^{2}}\right]=\frac{3(2 \mathrm{~L} \lambda) \mathrm{L}^{2}}{8 \pi^{2}}=\frac{3 \lambda \mathrm{L}^{3}}{4 \pi^{2}} $$

Correct answer: (4)

Solution:

In Fig. $29 \mathrm{AB}, \mathrm{BC}$ and $\mathrm{AC}$ are the three rods. $\mathrm{O}$ is centroid of the triangle.

$x=\mathrm{OD}=\mathrm{OE}=\mathrm{OF}$

$=$ Perpendicular distance of each rod from axis of rotation.

From right angled triangle OAD

$\frac{x}{(\ell / 2)}=\tan 30^{\circ} \quad \therefore \quad x=\frac{\ell}{2 \sqrt{3}}$

Fig. 29

Let $\mathrm{I} _{1}$ be M.I of any rod about axis through $\mathrm{O}$ and perpendicular to plane of triangle. Using theorem of parallel axis.

$\mathrm{I} _{1}=\frac{\mathrm{m} \ell^{2}}{12}+\mathrm{m}(x)^{2}=\frac{\mathrm{m} \ell^{2}}{12}+\mathrm{m}\left(\frac{\ell}{2 \sqrt{3}}\right)^{2}=\frac{\mathrm{m} \ell^{2}}{6}$

$\therefore \mathrm{I}=$ The total M.I of the system $=3 \mathrm{I} _{1}=\frac{\mathrm{m} \ell^{2}}{2}$

Given $\mathrm{I}=\mathrm{M} \ell^{2}$. Therefore,

$$ \mathrm{M}=\frac{\mathrm{m}}{2} $$

Correct answer: (4)

Solution:

Let $\mathrm{M}$ be the total mass of rod. Then

$$ \begin{equation*} \mathrm{M}=\int _{0}^{\mathrm{L}} \lambda \mathrm{d} x=\lambda _{0} \int _{0}^{\mathrm{L}} x \mathrm{~d} x=\frac{\lambda _{0} \mathrm{~L}^{2}}{2} \tag{i} \end{equation*} $$

Let $x$ be the position co-ordinate of C.M, C of rod. Then

$\mathrm{X}=\frac{\int _{0}^{\mathrm{L}} x \mathrm{dm}}{\int _{0}^{\mathrm{L}} \mathrm{dm}}=\frac{\lambda _{0} \int _{0}^{\mathrm{L}} x^{2} \mathrm{~d} x}{\lambda _{0} \int _{0}^{\mathrm{L}} x \mathrm{~d} x}=\frac{2}{3} \mathrm{~L}$

YOY’ is axis through C.M perpendicular to length of rod. PQ is an elementary mass, dm, as shown in Fig. 31 $\mathrm{dI}=(\mathrm{dm})[\mathrm{X}-x]^{2}=\lambda _{0}\left[\frac{2 \mathrm{~L}}{3}-x\right]^{2} x \mathrm{~d} x$

$\therefore \quad \mathrm{I}=\lambda _{0} \int _{0}^{\mathrm{L}}\left(\frac{2 \mathrm{~L}}{3}-x\right)^{2} x \mathrm{~d} x=\frac{\lambda _{0} \mathrm{~L}^{4}}{36}=\frac{\mathrm{ML}^{2}}{18}$

Correct answer: (3)

Solution:

$I=$ M.I of rod about given axis is

$$ =\frac{\mathrm{ML}^{2}}{12} \sin ^{2} \theta $$

I is $\theta$ graph is as shown in (3)

Correct answer: (1)

Solution:

Choose $\mathrm{X}^{\prime}$ and $\mathrm{Y}^{\prime}$ as shown in Fig. 33. Consider an elementary mass $\delta \mathrm{m}$ at $\mathrm{pt} \mathrm{P}$. Let co-ordinates of $\mathrm{P}$ be $(x, \mathrm{y})$ in $\mathrm{X}-\mathrm{Y}$ axis and be $\left(x^{\prime}, \mathrm{y}^{\prime}\right)$ in $\mathrm{X}^{\prime}-\mathrm{Y}^{\prime}$ axis. Obviously $\mathrm{OA}=x, \mathrm{OB}=\mathrm{y}, \mathrm{OA}^{\prime}=x^{\prime} ; \mathrm{OB}^{\prime}=\mathrm{y}^{\prime}$.

From Fig. 33 we have

$\mathrm{OA}=\mathrm{OA} _{1}-\mathrm{A} _{2} \mathrm{~A}^{\prime}$ $=\mathrm{OA}^{\prime} \cos \theta-\mathrm{PA}^{\prime} \sin \theta$

or $\quad x=x^{\prime} \cos \theta-y^{\prime} \sin \theta$

Fig. 33

Similarly $\mathrm{y}=x^{\prime} \sin \theta+\mathrm{y}^{\prime} \cos \theta$

By definition,

$\mathrm{I}=$ M.I about $\mathrm{COX}^{\prime}$ as axis $=\sum(\delta \mathrm{m}) \mathrm{y}^{2}$

$$ \begin{align*} & =\sum \delta \mathrm{m}\left(x^{\prime} \sin \theta+y^{\prime} \cos \theta\right)^{2} \\ & =\sum\left(\delta m x^{\prime 2}\right) \sin ^{2} \theta+\sum\left(\delta m y^{\prime 2}\right) \cos ^{2} \theta+2 \sum \delta m x^{\prime} \cdot y^{\prime} \sin \theta \cos \theta \tag{i} \end{align*} $$

By definition,

$$ \mathrm{I}^{\prime}=\sum \delta \mathrm{m}\left(x^{\prime}\right)^{2} ; \mathrm{I}^{\prime \prime}=\sum \delta \mathrm{m}\left(\mathrm{y}^{\prime}\right)^{2} $$

and I’ equals II" as the square is completely symmetrical about any two mutually perpendicular lines through center $\mathrm{O}$.

Also $\sum \delta \mathrm{m}\left(x^{\prime} \cdot \mathrm{y}^{\prime}\right)=0$. The body balances about its center of gravity $\mathrm{O}$. Therefore

$$ \sum(\delta m) g \cdot y^{\prime}=0 $$

We can rewrite Eqn. (i) as

$$ \mathrm{I}=\mathrm{I}^{\prime}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\mathrm{I}^{\prime} $$

Correct answer: (3)

Solution:

The symmetry axis of sphere is any diameter of sphere. Therefore

$$ \begin{equation*} \mathrm{I}=\frac{2}{5} \mathrm{MR}^{2} \tag{i} \end{equation*} $$

Let $\mathrm{r}$ be radius of disc into which sphere is reshaped. Mass of disc is same as that of sphere i.e. M. The M.I, I; of disc about given axis is

$$ \begin{equation*} \mathrm{I}^{\prime}=\frac{\mathrm{Mr}^{2}}{2}+\mathrm{Mr}^{2}=\frac{3}{2} \mathrm{Mr}^{2} \tag{ii} \end{equation*} $$

Given I’ $=\frac{3}{5}$ I. Therefore

$$ \frac{3}{5}\left(\frac{3}{2} \mathrm{Mr}^{2}\right)=\frac{2}{5} \mathrm{MR}^{2} $$

or

$$ \mathrm{r}=\frac{2}{3} \mathrm{R} $$

Correct answer: (3)

Solution:

$I=$ M.I of calender about an axis through its midpoint and perpendicular to the length of cylinder.

$$ =\mathrm{M}\left(\frac{\mathrm{R}^{2}}{4}+\frac{\mathrm{L}^{2}}{12}\right)^{*}=\mathrm{M}\left[\frac{\mathrm{R}^{2}}{4}+\frac{(\sqrt{3} \mathrm{R})^{2}}{12}\right]=\frac{\mathrm{MR}^{2}}{2} $$

Let $\mathrm{r}$ be radius of shaved cylinder. Then

mass of shaved cylinder $=\frac{1}{2}$ mass of given cylinder

$\pi \mathrm{r}^{2} \mathrm{~L} \rho=\frac{1}{2}\left[\pi \mathrm{R}^{2} \mathrm{~L} \rho\right] \quad \therefore \mathrm{r}^{2}=\frac{\mathrm{R}^{2}}{2}$

I’ $=$ M.I of “shaved” cylinder about given axis.

$$ =\frac{\mathrm{M}}{2}\left[\frac{\mathrm{r}^{2}}{4}+\frac{\mathrm{L}^{2}}{12}\right]=\frac{\mathrm{M}}{2}\left[\frac{\mathrm{R}^{2}}{8}+\frac{(\sqrt{3} \mathrm{R})^{2}}{12}\right]=\frac{3}{16} \mathrm{MR}^{2} $$

$\Delta \mathrm{I}=$ Change in $\mathrm{M} . \mathrm{I}=\mathrm{I}^{\prime}-\mathrm{I}=-\left(\frac{5}{16}\right) \mathrm{MR}^{2}$

$\frac{\Delta \mathrm{I}}{\mathrm{I}}=$ Fractional change in M.I $=-\left(\frac{5}{8}\right)$

Correct answer: (3)

Solution:

Let $I _{1}$ be moment of inertia, of the part of radius $R$ about given axis;. Then

$$ \mathrm{I} _{1}=\frac{3}{4}\left(\pi \mathrm{R}^{2} \sigma\right)\left(\frac{\mathrm{R}^{2}}{2}\right)=\frac{3 \pi \mathrm{R}^{4} \sigma}{8} $$

Let $\mathrm{I} _{2}$ be the moment of inertia of the part of radius $2 \mathrm{R}$ about given axis. Then

$$ \mathrm{I} _{2}=\frac{1}{4}\left[\pi(2 \mathrm{R})^{2} \sigma\right]\left(\frac{2 \mathrm{R}^{2}}{2}\right)=2 \pi \mathrm{R}^{4} \sigma $$

$\mathrm{I}=$ M.I of the given lamina; about given axis

$$ =\mathrm{I} _{1}+\mathrm{I} _{2}=\left(\frac{3}{8}+2\right) \pi \mathrm{R}^{4} \sigma=\frac{19 \pi \mathrm{R}^{4} \sigma}{8} $$

Correct answer: (3)

Solution:

Fig. 35 shows the wheel. The mass of rim $=\mathrm{M} _{1}=2 \pi \mathrm{R} \lambda$. Mass of each spoke $=\mathrm{M} _{2}=\mathrm{R} \lambda$. The symmetry axis is axis through center $\mathrm{O}$ and perpendicular to plane of wheel. The M.I., about symmetry axis.

$$ \begin{align*} I= & M _{1} R^{2}+12\left(\frac{1}{3} M _{2} R^{2}\right) \\ = & \lambda R^{3}\left[2 \pi+\frac{4}{3}\right] \tag{i} \end{align*} $$

The M.I of the rod of length L; about given axis.

$$ \begin{equation*} \mathrm{I}=\frac{1}{12}(\lambda \mathrm{L}) \mathrm{L}^{2}=\frac{\lambda \mathrm{L}^{3}}{12} \tag{ii} \end{equation*} $$

Fig. 35

From Eqn. (i) and (ii) we have

$\lambda \mathrm{R}^{3}\left(\frac{6 \pi+4}{3}\right)=\frac{\lambda \mathrm{L}^{3}}{12}$

$\therefore \frac{\mathrm{L}}{\mathrm{R}}=[4(6 \pi+4)]^{1 / 3}=2[3 \pi+2]^{1 / 3}$

Correct answer: (4)

Solution:

Let $\mathrm{h}$ be perpendicular distance of point $\mathrm{P}$ from $\mathrm{XOX}^{\prime}$. Consider diameter $\mathrm{X} _{1} \mathrm{OX} _{1}^{\prime}$ parallel to chord $\mathrm{P} 1$. The perpendicular distance of $\mathrm{P} 1$ from $\mathrm{X} _{1} \mathrm{OX} _{1}^{\prime}$ is $\mathrm{h}^{\prime}$. Therefore

$$ \mathrm{I} _{1}=\frac{1}{2} \mathrm{MR}^{2}+\mathrm{M}\left(\mathrm{h}^{\prime}\right)^{2} $$

Similarly,

$$ \mathrm{I} _{2}=\frac{\mathrm{MR}^{2}}{2}+\mathrm{Mh}^{2} $$

and $I _{3}=\frac{\mathrm{MR}^{2}}{2}+\mathrm{M}\left(\mathrm{h}^{\prime \prime}\right)^{2}$

Fig. 37

Since $\mathrm{h}^{\prime}>\mathrm{h}>\mathrm{h} “$, therefore $\mathrm{I} _{1}>\mathrm{I} _{2}>\mathrm{I} _{3}$.

Correct answer: (2)

Solution:

The M.I; of sector $\mathrm{OAB} ; \mathrm{I} _{1}$, is

$\mathrm{I} _{1}=\left(\frac{\theta _{1}}{2 \pi}\right)\left[\frac{\pi \mathrm{R} _{1}^{2} \sigma _{1}}{2} \mathrm{R} _{1}^{2}\right]=\left(\frac{\theta _{1}}{2 \pi}\right)\left[\frac{\pi \sigma _{1} \mathrm{R} _{1}^{4}}{2}\right]$

Similarly; the M.I of sector OCD; $\mathrm{I} _{2}$; is

$\mathrm{I} _{2}=\left(\frac{\theta _{2}}{2 \pi}\right)\left[\frac{\pi \sigma _{1} \mathrm{R} _{2}^{4}}{2}\right]$

Given $I _{1}=I _{2}$. Therefore

$$ \begin{aligned} & \left(\frac{\theta _{1}}{2 \pi}\right)\left(\frac{\pi \mathrm{R} _{1}^{4} \sigma _{1}}{2}\right)=\left(\frac{\theta _{2}}{2 \pi}\right)\left(\frac{\pi \mathrm{R} _{2}^{4} \sigma _{2}}{2}\right) \\ & \therefore \quad \frac{\sigma _{1}}{\sigma _{2}}=\left(\frac{\theta _{2}}{\theta _{1}}\right)\left(\frac{\mathrm{R} _{2}}{\mathrm{R} _{1}}\right)^{4} \end{aligned} $$

Correct answer: (1)

Solution:

Let $\mathrm{M} _{1}$ be mass of complete disc of radius R. Its M.I; $\mathrm{I} _{1}$, about axis considered is (using theorem of parallel axis).

$I _{1}=\frac{M _{1} R^{2}}{2}+M _{1}\left(\frac{R}{2}\right)^{2}=\frac{3 M _{1} R^{2}}{4} \hspace{40mm} . . . . . .(i)$

Let $M _{2}$ be mass of part of disc cut-out. Obviously $M _{2}=\frac{M _{1}}{4}$. The M.I, $I _{2}$, of this part about given axis is

$\mathrm{I} _{2}=\frac{\mathrm{M} _{2}}{2}\left(\frac{\mathrm{R}}{2}\right)^{2}=\frac{\mathrm{M} _{1} \mathrm{R}^{2}}{32} \hspace{40mm} . . . . . .(ii)$

The M.I, I; of the annular disc;

$\mathrm{I}=\mathrm{I} _{1}-\mathrm{I} _{2}=\frac{23}{32} \mathrm{M} _{1} \mathrm{R}^{2} \hspace{40mm} . . . . . .(iii)$

The mass, $M$, of annular disc is $\left(M _{1}-\frac{M _{1}}{4}\right)=3 \frac{M _{1}}{4} \quad$ or $\quad M _{1}=\frac{4}{3} M$

Therefore,

$$ I=\frac{23}{32}\left(\frac{4}{3} M\right) R^{2}=\frac{23}{24} M R^{2} $$

Correct answer: (1)

Solution:

Fig. 40

Fig. 40 shows hemisphere $\mathrm{X} _{1} \mathrm{OX}^{\prime}{ } _{1}$ is a diameter. The M.I, $\mathrm{I} _{1}$ about XOX’ as axis $\frac{2}{5} \mathrm{MR}^{2}$. In Fig. 40, $\mathrm{G}$ is position of C.G. of hemisphere $\mathrm{X} _{1} \mathrm{GX}^{\prime}{ } _{1}$ is axis through C.G. parallel to diameter XOX’. Let $\mathrm{I} _{\mathrm{g}}$ be M.I about $\mathrm{X} _{1} \mathrm{GX}^{\prime}{ } _{1}$ as axis. From theorem of parallel axis.

$$ \begin{align*} \mathrm{I} _{1} & =\mathrm{I} _{\mathrm{g}}+\mathrm{M}\left(\frac{3 \mathrm{R}}{8}\right)^{2} \\ \text { or } \quad \mathrm{I} _{\mathrm{g}} & =\left(\frac{2}{5}-\frac{9}{64}\right) \mathrm{MR}^{2}=\frac{83}{320} \mathrm{MR}^{2} \tag{i} \end{align*} $$

In Fig. $40 \mathrm{AB}$ is the desired axis about which radius of gyration, $\mathrm{k}$ is to be calculated. Let $\mathrm{I} _{2}$ be M.I about $\mathrm{AB}$ as axis. From Theorem of parallel axes

$$ \begin{aligned} I _{2}= & I g+M\left(\frac{3 R}{4}-\frac{3 R}{8}\right)^{2} \\ & =\left(\frac{83}{320}+\frac{9}{64}\right) MR^{2}=\frac{2}{5} M R^{2} \end{aligned} $$

Also $\mathrm{I} _{2}=\mathrm{Mk}^{2}$. Therefore,

$$ \mathrm{k}=\sqrt{\frac{2}{5}} \mathrm{R} $$

Correct answer: (1)

Solution:

Let $\mathrm{I} _{\mathrm{m}}$ and $\mathrm{J}$ denote the linear and angular impulse imparted to cylinder. By definition.

$$ \begin{equation*} I _{m}=\int F \cdot d t \tag{i} \end{equation*} $$

$$ \begin{equation*} \mathrm{J}=\int \tau . \mathrm{dt}=\int(\mathrm{F} \cdot \mathrm{h}) \mathrm{dt} \tag{ii} \end{equation*} $$

From Eqns (i) and (ii) we have,

$$ \begin{equation*} \mathrm{J}=\left(\mathrm{I} _{\mathrm{m}}\right) \times \mathrm{h} \tag{iii} \end{equation*} $$

Let $v$ be the velocity of center of mass and $\omega$ the angular speed of cylinder, just after impulse has been imparted. Obviously

$\mathrm{I} _{\mathrm{m}}=$ Change in linear momentum

$$ \begin{equation*} =(\mathrm{MV}-0)=\mathrm{MV} \tag{iv} \end{equation*} $$

$\mathrm{J}=$ Change in angular momentum $=\mathrm{I} \omega-0$

$$ \begin{equation*} =\frac{1}{2} \mathrm{MR}^{2} \omega \tag{v} \end{equation*} $$

From Eqn. (iii), (iv) and (v); we have

$$ \frac{1}{2} \mathrm{MR}^{2} \omega=\mathrm{MVh} $$

Also $\mathrm{V}=\mathrm{R} \omega$; there is no slipping. Therefore

$$ \mathrm{h}=\frac{\mathrm{R}}{2} $$

Correct answer: (2)

Solution:

Let $\mathrm{I} _{1} ; \mathrm{J} _{1}$ and $\mathrm{K} _{1}$ denote the moment of inertia; angular momentum and rotational K.E. of rotating disc. Then

$$ \mathrm{I} _{1}=\frac{1}{2} \mathrm{MR}^{2} ; \quad \mathrm{J} _{1}=\mathrm{I} _{1} \omega _{1}=\frac{\mathrm{MR}^{2} \omega _{0}}{2} \text { and } \mathrm{K} _{1}=\frac{\mathrm{J} _{1}^{2}}{2 \mathrm{I} _{1}} $$

The ring has been placed on rotating disc and the system has acquire a steady state angular speed $\omega$. Let $\mathrm{I} _{2} ; \mathrm{J} _{2}$ and $\mathrm{K} _{2}$ be the corresponding quantities for the system. Then

$\mathrm{I} _{2}=\frac{1}{2} \mathrm{MR}^{2}+\left(\frac{\mathrm{M}}{2}\right)\left(\frac{\mathrm{R}}{2}\right)^{2}=\frac{5 \mathrm{MR}^{2}}{8}$

$$ \mathrm{J} _{2}=\mathrm{I} _{2} \omega=\mathrm{J} _{1} \quad \text { (Law of conservation of angular momentum) } $$

and $\quad \mathrm{K} _{2}=\frac{\mathrm{J} _{2}^{2}}{2 \mathrm{I} _{2}}=\frac{\mathrm{J} _{1}^{2}}{2 \mathrm{I} _{2}} \quad\left[\because \mathrm{J} _{1}=\mathrm{J} _{2}\right]$

$\Delta \mathrm{K}=$ Change in rotational K.E. $=\mathrm{K} _{1}-\mathrm{K} _{2}=\frac{\mathrm{J} _{1}^{2}}{2}\left(\frac{1}{\mathrm{I} _{1}}-\frac{1}{\mathrm{I} _{2}}\right)$

$$ \begin{aligned} & =\frac{\mathrm{J} _{1}^{2}}{2}\left[\frac{2}{\mathrm{MR}^{2}}-\frac{8}{5 \mathrm{MR}^{2}}\right]=\frac{\mathrm{J} _{1}^{2}}{5 \mathrm{MR}^{2}} \\ & =\frac{\mathrm{MR}^{2} \omega _{0}^{2}}{20} \end{aligned} $$

Correct answer: (3)

Solution:

The moment of inertia, I, of sphere about diameter as axis is

$$ \mathrm{I}=\frac{2}{5} \mathrm{MR}^{2} $$

Let $\mathrm{L}$ be the side of the cube formed when sphere is recast. Then

$$ M=\frac{4 \pi}{3} R^{3} \rho=L^{3} \rho $$

or $\quad \mathrm{L}=\left(\frac{4 \pi}{3}\right)^{1 / 3} \mathrm{R}$

$I _{1}=$ M.I. of cube about given axis

$$ =M\left[\frac{\mathrm{L}^{2}+\mathrm{L}^{2}}{12}\right]=\frac{\mathrm{ML}^{2}}{6}=\frac{\mathrm{M}}{6}\left(\frac{4 \pi}{3}\right)^{2 / 3} \mathrm{R}^{2} $$

$$ \begin{aligned} & =\frac{1}{6}\left(\frac{4 \pi}{3}\right)^{2 / 3} \times \frac{5}{2} \mathrm{I}=\frac{5}{3}\left[\frac{4 \pi}{3(4) \frac{3}{2}}\right]^{2 / 3} \mathrm{I} \\ & =\frac{5}{3}\left[\frac{\pi}{6}\right]^{2 / 3} \mathrm{I} \end{aligned} $$

Correct answer: (3)

Solution:

Let $\mathrm{I} _{1}$ be M.I of sphere about axis having $\mathrm{k} _{1}=\sqrt{2} \mathrm{R}$. Let $\mathrm{h} _{1}$ be distance of the axis from a parallel axis through C.G. From theorem of parallel axis.

$$ \begin{gather*} \mathrm{M}(\sqrt{2} \mathrm{R})^{2}=\mathrm{Ig}+\mathrm{Mh} _{1}^{2} \\ 2 \mathrm{MR}^{2}=\frac{2}{5} \mathrm{MR}^{2}+\mathrm{Mh} _{1}^{2} \\ \text { or } \mathrm{h} _{1}=\sqrt{\frac{8}{5}} \mathrm{R}=\sqrt{1.6} \mathrm{R} \tag{i} \end{gather*} $$

Let $h _{2}$ be distance of second axis from a parallel axis through C.G., so that $I _{2}=\mathrm{Mk} _{2}^{2}$ and $k _{2}=2 R$. Obviously

$\mathrm{M}(2 \mathrm{R})^{2}=\frac{2}{5} \mathrm{MR}^{2}+\mathrm{Mh} _{2}^{2}$

$\therefore \mathrm{h} _{2}=\sqrt{\frac{18}{5}} \cdot \mathrm{R}=\sqrt{3.6} \mathrm{R}\hspace{40mm} . . . . . .(ii)$

Change in position of axis of rotation.

$$ =\mathrm{h} _{2}-\mathrm{h} _{1}=[\sqrt{3.6}-\sqrt{1.6}] \mathrm{R} $$

Correct answer: (4)

Solution:

Let $\mathrm{I} _{1}$ be M.I of part AO about given axis. Axis passes through one end A of rod; therefore

$$ \mathrm{I} _{1}=\mathrm{m} _{1} \frac{\mathrm{L}^{2}}{3}=\frac{\lambda _{1} \mathrm{~L}^{3}}{3} \quad\left[\because \mathrm{m} _{1}=\lambda _{1} \mathrm{~L}\right] $$

Let $\mathrm{I} _{2}$ be moment of inertia of part $\mathrm{OB}$ of rod. Let $\mathrm{I} _{\mathrm{g}}$ be M.I of part $\mathrm{OB}$ about exist through its midpoint $\mathrm{O} _{1}$ and perpendicular to $\mathrm{OB}$. Then

Fig. 43

$$ \mathrm{I} _{\mathrm{g}}=\mathrm{m} _{2} \frac{\mathrm{L}^{2}}{12}=\frac{\lambda _{2} \mathrm{~L}^{3}}{12} $$

From theorem of parallel axis

$$ \mathrm{I} _{2}=\mathrm{I} _{\mathrm{g}}+\mathrm{m} _{2}\left(\frac{3 \mathrm{~L}}{2}\right)^{2}=\mathrm{m} _{2} \frac{\mathrm{L}^{2}}{12}+\frac{9}{4} \mathrm{~m} _{2} \mathrm{~L}^{2} $$

$$ =\frac{7}{3} \mathrm{~m} _{2} \mathrm{~L}^{2}=\frac{7}{3} \lambda _{2} \mathrm{~L}^{3} $$

$I=$ M.I of complete rod about given axis.

$$ =\mathrm{I} _{1}+\mathrm{I} _{2}=\left(\lambda _{1}+7 \lambda _{2}\right) \frac{\mathrm{L}^{3}}{3} $$

Correct answer: (4)

Solution:

$I _{1}=$ M.I of rod about given axis.

$$ =0.28 \mathrm{M}\left[\frac{(2 \sqrt{3} \mathrm{R})^{2}}{12}\right]=0.28 \mathrm{MR}^{2} $$

$I _{2}=$ M.I of each sphere about given axis.

$$ \begin{aligned} & =\frac{2}{5} \mathrm{MR}^{2}+\mathrm{M}[(1+\sqrt{3}) \mathrm{R}]^{2} \\ & =0.4 \mathrm{MR}^{2}+7.46 \mathrm{MR}^{2}=7.86 \mathrm{MR}^{2} \end{aligned} $$

$I=$ M.I of the arrangement about gives axis.

$$ \begin{aligned} & =\mathrm{I} _{1}+2 \mathrm{~J} _{2}=0.28 \mathrm{MR}^{2}+2 \times 7.86 \mathrm{MR}^{2} \\ & =(0.28+15.72) \mathrm{MR}^{2} \\ & =16 \mathrm{MR}^{2} \end{aligned} $$

Correct answer:(2)

Solution:

When rotating cylinder is gently placed on rough horizontal surface at $t=t _{1}$; it initially slips. But at $t=t _{2}$, slipping stops. There is loss of energy due to friction between horizontal surface and cylinder i.e $\mathrm{K} _{2}<\mathrm{K} _{1}$. At $t>t _{2}$; the cylinder moves with speed it has acquired and its energy remains same as at $t=t _{2}$ i.e $\mathrm{K} _{2}=\mathrm{K} _{3}$.

Correct answer: (4)

Solution:

The wheel has a non-uniformly acceleration motion. Given

$$ \begin{gather*} \alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=\mathrm{a}-\mathrm{bt} \\ \therefore \quad \omega=\int _{0}^{\omega} \mathrm{d} \omega=\int _{0}^{\omega}(\mathrm{a}-\mathrm{bt}) \mathrm{dt} \\ \omega=\mathrm{at}-\frac{\mathrm{bt}^{2}}{2} \tag{i} \end{gather*} $$

$\omega=0$ at $\mathrm{t}=\mathrm{t} _{1}$, when wheel is momentarily at rest. Obviously $\mathrm{t} _{1}=\frac{2 \mathrm{a}}{\mathrm{b}}$. Let $\theta$ be the angular displacement of wheel. We have

$$ \omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}=\mathrm{at}-\frac{\mathrm{bt}^{2}}{2} $$

$\therefore \int _{0}^{\theta} \mathrm{d} \theta=\int _{0}^{\mathrm{t}}\left(\mathrm{at}-\frac{\mathrm{bt}^{2}}{2}\right) \mathrm{dt} \quad$ or $\quad \theta=\frac{\mathrm{at}^{2}}{2}-\frac{\mathrm{bt}^{3}}{6}$

$\therefore \theta\left(a t \quad \mathrm{t}=\mathrm{t} _{1}\right)=\frac{\mathrm{at} _{1}^{2}}{2}-\frac{\mathrm{b}}{6} \mathrm{t} _{1}^{3}=\frac{\mathrm{a}}{2}\left(\frac{2 \mathrm{a}}{\mathrm{b}}\right)^{2}-\frac{\mathrm{b}}{6}\left(\frac{2 \mathrm{a}}{\mathrm{b}}\right)^{3}=\frac{2 \mathrm{a}^{3}}{3 \mathrm{~b}^{2}}$

The number of revolutions $=\frac{\theta}{2 \pi}=\frac{\mathrm{a}^{3}}{3 \pi \mathrm{b}^{2}}$

Correct answer: (1)

Solution:

Let $\tau$ be the torque produces by the motor when rotating at $\omega=1500$ r.p.m. $=\frac{1500 \times 2 \pi}{60} \mathrm{rod} / \mathrm{s}$.

We know

$$ \begin{gathered} \tau \omega=\mathrm{P} \quad \text { or } \quad \tau=\frac{\mathrm{P}}{\omega} \\ \therefore \quad \tau=\frac{0.5 \times 10^{3}}{\left(\frac{1500}{60} \times 2 \pi\right)} \mathrm{N}-\mathrm{m} \\ \quad=\frac{10}{\pi} \mathrm{N}-\mathrm{m} \end{gathered} $$

Let $\Delta \mathrm{T}=\mathrm{T} _{1}-\mathrm{T} _{2}$ be the difference in the tension in the two parts of belt. $\mathrm{R}=$ radius of pulley. Clearly $\tau=\left(\mathrm{T} _{1}-\mathrm{T} _{2}\right) \cdot \mathrm{R}=(\Delta \mathrm{T}) \cdot \mathrm{R}$

$\therefore \frac{10}{\pi}=(\Delta \mathrm{T}) \times 5 \times 10^{-2}$

or $\Delta \mathrm{T}=\frac{200}{\pi}$ newton

Correct answer: (4)

Solution:

The torque, $\tau$, on the system is due to weight $\mathrm{Mg}$ of rod and $\mathrm{mg}$ of point mass at A. From Fig. 46

$$ \begin{gathered} \tau=\text { Total torque }=\mathrm{Mg} \times \mathrm{CC} _{1}+\mathrm{Mg} \times \mathrm{AA} _{1} \\ =\frac{\mathrm{MgL}}{2} \sin \theta+\mathrm{mgL} \sin \theta \end{gathered} $$

Fig. 46

$$ =\left(\frac{M}{2}+m\right) g L \sin \theta=\left(\frac{M+2 m}{2}\right) g L \sin \theta $$

$\mathrm{I}=$ The M.I of system about horizontal axis through end O.

$$ =\frac{1}{3} \mathrm{ML}^{2}+\mathrm{mL}^{2}=\left(\frac{\mathrm{M}+3 \mathrm{~m}}{3}\right) \mathrm{L}^{2} $$

The initial acceleration $=\alpha=\frac{\tau}{I}=\left[\frac{3(M+2 m)}{2(M+3 m)}\right]\left[\frac{g \sin \theta}{L}\right]$

Correct answer:(1)

Solution:

Force of friction coming into play is not necessarily equal to the limiting friction.It is $\mathrm{f} ; 0<\mathrm{f} \leq \mathrm{f} _{\mathrm{tt}}$.

Friction does not act as a dissipative force. The loss in translation K.E., due to friction, is exactly equal to gain in rotational K.E., which is due to force of friction.

Correct answer: (2)

Solution:

Fig. 48 shows ladder when $\theta=37^{\circ}$.

Let $(x, 0)$ and $(0, \mathrm{y})$ be the instantaneous co-ordinates of end $\mathrm{B}$ and $\mathrm{A}$ respectively. The length $\mathrm{AB}=\ell$ of rod is $\ell^{2}=x^{2}+y^{2}$

Differentiate w.r.t time $(\mathrm{t})$

$$ 0=2\left[x \frac{\mathrm{d} x}{\mathrm{dt}}+\mathrm{y} \frac{\mathrm{dy}}{\mathrm{dt}}\right] $$

or $\frac{\mathrm{dy}}{\mathrm{dt}}=-\left(\frac{x}{\mathrm{y}}\right) \frac{\mathrm{d} x}{\mathrm{dt}} \hspace{40mm} . . . . . .(i)$

Fig. 48

Obviously $\frac{\mathrm{d} x}{\mathrm{dt}}=\mathrm{v} _{\mathrm{B}}=0.5 \mathrm{~ms}^{-1}$ (Given). Also $\frac{x}{\mathrm{y}}=\cot 37=\frac{0.8}{0.6}=\frac{4}{3}$

Therefore, velocity of end $A ;{ v} _{A}$ is

$$ \mathrm{v} _{\mathrm{A}}=\frac{\mathrm{dy}}{\mathrm{dt}}=-\frac{4}{3} \times \mathrm{v} _{\mathrm{B}}=-\frac{4}{3} \times 0.5=-0.67 \mathrm{~ms}^{-1} $$

Negative sign indicates that end A moves vertically downwards.

Correct answer: (2)

Solution:

Fig. 50 shows instantaneous position $\mathrm{OB}$ of rod when it has turned through an angle $\theta$. The work done, $w$, on the rod by the applied force.

$$ \begin{align*} \mathrm{w}= & \int _{0}^{\theta}(\mathrm{F} \ell \cos \theta) \mathrm{d} \theta \quad[\because \tau=\mathrm{F} \ell \cos \theta] \\ & =\mathrm{F} \ell \sin \theta \tag{i} \end{align*} $$

Let $\omega$ be the angular speed acquired by rod. From work energy theorem

Fig. 50

$\mathrm{K}=$ Rotation K.E. of $\mathrm{rod}=\mathrm{w}=\mathrm{F} \ell \sin \theta$

The instantaneous angular momentum of rod

$$ =\mathrm{J}=\sqrt{2 \mathrm{~J}} \mathrm{~K}=\sqrt{\frac{2 \mathrm{FMl}^{3} \sin \theta}{3}} $$

Given $\theta=30^{\circ}$; therefore

$$ \mathrm{J}=\sqrt{\frac{\mathrm{FML}^{3}}{3}} $$

Correct answer: (2)

Solution:

The linear acceleration or retardation of C.M during ascent and descent have same magnitude. This is only possible if direction of force of friction; f; coming into play is directed up the plane; both during ascent and descent.

Correct answer: (3)

Solution:

Since there is no rotational motion of rod; the net torque on rod about its center O; due to applied forces must be zero. Therefore,

$$ \begin{equation*} \mathrm{F} _{1} \times \frac{\mathrm{L}}{2}=\mathrm{F} _{2} \times \frac{\mathrm{L}}{8} \end{equation*} $$

or $\mathrm{F} _{2}=4 \mathrm{~F} _{1} \hspace{40mm} . . . . . .(i)$

The net force on $\operatorname{rod}=\mathrm{F}=\mathrm{F} _{2}-\mathrm{F} _{1}=3 \mathrm{~F} _{1}$

The linear acceleration, a, of rod is

$$ \begin{equation*} \mathrm{a}=\frac{3 \mathrm{~F} _{1}}{\mathrm{M}} \tag{ii} \end{equation*} $$

Correct answer: (2)

Solution:

Fig. 53(a) and (b) show free body diagram of mass $m$ and pulley respectively. $\mathrm{T}$ is tension in string. Let a be linear acceleration of $\mathrm{m}$. Its equation of motion is

$$ \begin{equation*} \mathrm{ma}=\mathrm{mg}-\mathrm{T} \tag{i} \end{equation*} $$

or $\quad \frac{M a}{2}=\frac{M}{2} g-T$

Fig. 53

$\mathrm{I} \alpha=\tau \quad \text { or } \quad\left(\frac{1}{2} \mathrm{MR}^2\right) \alpha=\mathrm{TR} \hspace{40mm} . . . . . .(ii)$

Let $\alpha$ be the angular acceleration of pulley. Equation of rotational motion of pulley is

Since there is no slipping between pulley and string;

$$ \begin{equation*} \mathrm{a}=\mathrm{R} \alpha \tag{iii} \end{equation*} $$

From Eq ns (i), (ii) and (iii) we have

$$ a=\frac{g}{2} $$

Correct answer: (3)

Solution:

Let a be linear acceleration of mass $\mathrm{m}$ and $\mathrm{T}$ the tension in string. $\mathrm{f}$ is force of friction coming into play between table top and cylinder. Let $\alpha$ be the angular acceleration of cylinder. The linear acceleration of C.M of cylinder also is a. Fig. 55 (a) and (b) show free-body diagram of mass $m$ and cylinder.The equations of motion are:

$$ \begin{align*} & \mathrm{mg}-\mathrm{T}=\mathrm{ma} \tag{i} \\ & \mathrm{T}-\mathrm{f}=4 \mathrm{ma} \tag{ii} \\ & (\mathrm{T}+\mathrm{f}) \mathrm{R}=\mathrm{I} \alpha \tag{iii} \end{align*} $$

Fig. 55

Since there is no slipping;

$$ \begin{equation*} \mathrm{a}=\mathrm{R} \alpha \tag{iv} \end{equation*} $$

From above equations;

$$ \mathrm{a}=\left(\frac{4}{5}\right) \mathrm{g} $$

Correct answer:(4)

Solution:

The slant side, $\ell$; of inclined plane is

$$ \frac{\mathrm{h}}{\ell}=\sin 30 \quad \text { or } \quad \ell=\frac{\mathrm{h}}{\sin 30}=2 \times 0.875 \mathrm{~m}=1.75 \mathrm{~m} $$

Let a be the linear acceleration of center of mass of body. Given

$\mathrm{u}=0 ; \mathrm{S}=\ell=1.75 \mathrm{~m} ; \mathrm{t}=1 \mathrm{~s} ; \mathrm{a}=$ ?

$\therefore 1.75=\frac{1}{2} \times \mathrm{a} \times(1)^{2}$

or $\quad \mathrm{a}=3.5 \mathrm{~ms}^{-2} \hspace{40mm} . . . . . .(i)$

Let I be the M.I of the body about axis through its center of mass. We know

$$ \mathrm{a}=\frac{\mathrm{g} \sin \theta}{\left(1+\frac{\mathrm{I}}{\mathrm{mR}^{2}}\right)} $$

3.$5=\frac{9.8 \times 0.5}{1+\frac{\mathrm{I}}{\mathrm{mR}^{2}}}$

or $\quad \mathrm{I}=\mathrm{mR}^{2}\left[\frac{4.9}{3.5}-1\right]$

$=\frac{2}{5} \mathrm{mR}^{2}$

The body is a solid sphere.

Correct answer: (2)

Solution:

The acceleration of body down smooth inclined plane is $g \sin \theta$. The speed $v _{1}$ is

$$ \begin{equation*} \mathrm{v} _{1}=\sqrt{2 \mathrm{gh}} \tag{i} \end{equation*} $$

Let a be the linear acceleration of C.M of circular disc as it rolls down the inclined plane. Then

$$ \mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\left(\frac{\frac{1}{2} \mathrm{mR}^{2}}{\mathrm{mR}^{2}}\right)}=\frac{2}{3} \mathrm{~g} \sin \theta $$

The speed, $\mathrm{v} _{2}$ as it reaches the bottom of inclined plane is

$$ \mathrm{v} _{2}^{2}=2\left(\frac{2}{3} \mathrm{~g} \sin \theta\right) \ell $$

$$ \begin{aligned} & =\frac{4}{3} \mathrm{gh} \quad[\because \mathrm{h}=\ell \sin \theta] \\ & \therefore \quad \frac{\mathrm{v} _{2}^{2}}{\mathrm{v} _{1}^{2}}=\frac{4}{3 \times 2}=\frac{2}{3} \quad \text { or } \quad \frac{\mathrm{v} _{2}}{\mathrm{v} _{1}}=\sqrt{\frac{2}{3}} \end{aligned} $$

Correct answer: (3)

Solution:

Fig. 56 shows cylinder with vertical string pulled up by a force F. As C.M of cylinder is at rest; $\mathrm{F}=\mathrm{Mg}$. The torque; $\tau$; due to applied force about $\mathrm{O}$ is

$$ \tau=\mathrm{F} . \mathrm{R}=\mathrm{MgR} $$

The M.I of cylinder about its symmetry axis $=\mathrm{I}=\frac{\mathrm{MR}^{2}}{2}$

$\alpha=$ The angular acceleration of cylinder $=\frac{\tau}{\mathrm{I}}=\frac{2 \mathrm{~g}}{\mathrm{R}}$

Fig. 56

Let $\theta$ be the total angle describe by cylinder as it accquires angular speed $\omega _{1}$. Then

$\omega _{1}^{2}-(0)^{2}=2\left(\frac{2 \mathrm{~g}}{\mathrm{R}}\right) \theta$

$\therefore \theta=\frac{\mathrm{R} \omega _{1}^{2}}{4 \mathrm{~g}}$

The work done $=\tau . \theta$

$$ =\frac{\mathrm{MR}^{2} \omega _{1}^{2}}{4} $$

Alternately,

$\mathrm{w}=$ Change in rotational $\mathrm{KE}=\frac{1}{2} \cdot\left(\frac{1}{2} \mathrm{MR}^{2}\right) \omega _{1}^{2}=\frac{1}{4} \mathrm{MR}^{2} \omega _{1}^{2}$

Correct answer: (3)

Solution:

There is an external torque $\tau$, on disc due to friction between horizontal surface and disc. To calculated $\tau$ consider an elementary ring of radius $x$ thickness $\mathrm{d} x$. The mass, $\mathrm{dm}$; of ring considered.

Fig. 57

$$ \begin{aligned} \mathrm{dm}= & (2 \pi x \quad \mathrm{~d} x)\left(\frac{\mathrm{M}}{\pi \mathrm{R}^{2}}\right) \\ & =\left(\frac{2 \mathrm{M}}{\mathrm{R}^{2}}\right) x \mathrm{~d} x \end{aligned} $$

$\mathrm{d} \tau=$ Torque on elementary ring about center $\mathrm{O}$

$$ =\mu(\mathrm{dm}) \mathrm{g} \cdot x=\left(\frac{2 \mu \mathrm{Mg}}{\mathrm{R}^{2}}\right) x^{2} \mathrm{~d} x $$

$\therefore \quad \tau=$ The total torque on disc $=\int _{0}^{R} \mathrm{~d} \tau$

$$ =\frac{2 \mu \mathrm{MgR}}{3} $$

Let $\alpha$ be angular retardation of disc

$\alpha=\frac{\tau}{\mathrm{I}}=\frac{2 \mu \mathrm{Mg} \frac{\mathrm{R}}{3}}{\frac{\mathrm{MR}^{2}}{2}}=\frac{4 \mu \mathrm{g}}{3 \mathrm{R}}$

Let $\omega _{0}$ be initial angular speed of disc. Then

$$ \omega=0 ; \quad \alpha=-\frac{4 \mu \mathrm{g}}{3 \mathrm{R}} ; \quad \mathrm{t}=? $$

$\therefore 0=\omega _{0}-\omega \mathrm{t} \quad$ or $\quad \mathrm{t}=\frac{\omega _{0}}{\alpha}=\frac{3 \omega _{0} \mathrm{R}}{4 \mu \mathrm{g}}$

Given $\omega _{0}=\frac{600}{\pi} \mathrm{rpm}=\frac{600 \times 2 \pi}{\pi \times 60} \mathrm{rad} \quad \mathrm{s}^{-1}=20 \mathrm{rad} \quad \mathrm{s}^{-1}$

$R=200 \mathrm{~mm}=0.2 \mathrm{~m} ; \mu=0.15 ; \mathrm{g}=10 \mathrm{~ms}^{-2}$

$\therefore \mathrm{t}=\frac{3 \times 20 \times 0.2}{4 \times 0.15 \times 10} \mathrm{~s}$

$=2 \mathrm{~s}$

Correct answer: (4)

Solution:

The sphere has energy $\mathrm{K} _{1}$ due to both rotational and translation motion. Obviously

$$ \mathrm{K} _{1}=\frac{1}{2}\left[\frac{2}{3} \times 6 \mathrm{mR}^{2}\right] \cdot \omega^{2}+\frac{1}{2}(6 \mathrm{~m}) \mathrm{v}^{2} $$

Since sphere rolls without slipping.

$$ \begin{aligned} \mathrm{v} & =\mathrm{R} \omega \\ \therefore \quad \mathrm{K} _{1} & =\frac{1}{2}\left[\frac{2}{3} \times 6 \mathrm{mv}^{2}\right]+3 \mathrm{mv}^{2}=5 \mathrm{mv}^{2} \end{aligned} $$

The water inside sphere has only translational motion. Its energy $\mathrm{K} _{2}$ is

$\mathrm{K} _{2}=\frac{1}{2} \mathrm{mv}^{2}=0.5 \mathrm{mv}^{2}$

$\mathrm{K}=$ The total energy of the system.

$$ =\mathrm{K} _{1}+\mathrm{K} _{2}=5.5 \mathrm{mv}^{2} $$

Correct answer: (1)

Solution:

Let $\alpha$ and $\omega$ denote the instantaneous acceleration and angular speed of disc. Given

$$ \alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=-\mathrm{k} \omega $$

where $\mathrm{k}$ is a constant. Therefore

$\int_{-w_0}^{w} \frac{d \omega}{\omega}=-k \int_0^t d t$

$\therefore \omega=\omega _{0} \mathrm{e}^{-k t} \hspace{40mm} . . . . . .(i)$

We know $\omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}$. Therefore

$$ \mathrm{d} \theta=\omega _{0} \mathrm{e}^{-\mathrm{kt}} \mathrm{dt} $$

or $\quad \int _{0}^{\theta} d \theta=\omega _{0} \int _{0}^{t} e^{-k t} d t$

$\therefore \quad \theta=\frac{\omega _{0}}{\mathrm{R}}\left[1-\mathrm{e}^{-\mathrm{kt}}\right]$

$n=$ Number of revolution completed in time $t$.

$$ \begin{equation*} =\frac{\theta}{2 \pi}=\frac{\omega _{0}}{2 \pi \mathrm{k}}\left[1-\mathrm{e}^{-\mathrm{kt}}\right] \tag{ii} \end{equation*} $$

Let $\omega=\frac{\omega _{0}}{2}$ at $\mathrm{t}=\mathrm{t} _{1}$

From Eqn(i) we have

$$ \frac{\omega _{0}}{2}=\omega _{0} \mathrm{e}^{-\mathrm{k} t _{1}} \quad \text { or } \quad \mathrm{e}^{-k t _{1}}=\frac{1}{2} $$

From Eqn (ii) we have

$$ \mathrm{n}=\mathrm{z}=\frac{\omega _{0}}{2 \pi \mathrm{k}}\left[1-\mathrm{e}^{-\mathrm{kt _{1 }}}\right]=\frac{\omega _{0}}{4 \pi \mathrm{k}} $$

Let $\mathrm{N}$ be total number of revolutions made by wheel in coming to rest, i.e. $\mathrm{t} \rightarrow \infty$. From Eqn (i)

$$ \mathrm{N}=\frac{\omega _{0}}{2 \pi \mathrm{k}}=2 \mathrm{z} $$

The additional number of revolution mode $=\mathrm{N}-\mathrm{n}$

$$ =2 \mathrm{z}-\mathrm{z}=\mathrm{z} $$

Correct answer: (4)

Solution:

Regarding ball and sphere as a system; there is no external force and therefore no external torque. The angular momentum of system is conserved. Let $\mathrm{v} _{2}$ be linear speed of ball as it jumps off to table $\mathrm{Q}$ from sphere. Let $\omega$ be angular speed acquired by sphere. Since there is no slipping between $A$ and sphere, $\mathrm{v} _{2}=\mathrm{R} \omega$. Using law of conservation of angular momentum (Take angular momentum about point $\mathrm{O}$ ).

$$ \begin{aligned} & \mathrm{mv} _{1} \mathrm{R}=\frac{2}{5} \mathrm{MR}^{2} \omega+\mathrm{mv} _{2} \mathrm{R} \\ & \mathrm{mv} _{1} \mathrm{R}=\frac{2}{5} \mathrm{MR}^{2}\left(\frac{\mathrm{v} _{2}}{\mathrm{R}}\right)+\mathrm{mv} _{2} \mathrm{R} \\ & \mathrm{mv} _{1}=\left(\frac{2 \mathrm{M}}{5}+\mathrm{m}\right) \mathrm{v} _{2} \\ & \therefore \frac{\mathrm{v} _{2}}{\mathrm{v} _{1}}=\frac{5 \mathrm{~m}}{5 \mathrm{~m}+2 \mathrm{M}} \end{aligned} $$

Correct answer: (2)

Solution:

There is no external force and therefore no external torque. The angular momentum of system is conserved.

$\mathrm{L} _{1}=$ The initial angular momentum $=\mathrm{I} _{1} \omega _{0}$

$\mathrm{I} _{1}=\mathrm{I} _{\mathrm{A}}+\mathrm{I} _{\mathrm{B}}=\frac{1}{3}\left(\frac{\mathrm{M}}{3}\right) \mathrm{L}^{2}+\frac{1}{3} \mathrm{ML}^{2}=\frac{4 \mathrm{ML}^{2}}{9}$

$\therefore \quad \mathrm{L} _{1}=\frac{4 \mathrm{ML}^{2} \omega _{0}}{9} \hspace{40mm} . . . . . .(i)$

Fig. 60 shows the system when rod A is about to emerge out of B. Let $\mathrm{I} _{2}$ be M.I of, arrangement about $\mathrm{YY}^{\prime}$ as axis.

$$ \begin{aligned} \mathrm{I} _{2}= & \frac{1}{3}\left(\frac{\mathrm{M}}{3}\right) \mathrm{L}^{2}+\left[\frac{1}{12} \mathrm{ML}^{2}+\mathrm{M}\left(\frac{3 \mathrm{~L}}{2}\right)^{2}\right] \\ & =\frac{22}{9} \mathrm{ML}^{2} \end{aligned} $$

$\mathrm{L} _{2}=$ The final angular momentum of the system

$$ \begin{equation*} =\mathrm{I} _{2} \omega=\frac{22}{9} \mathrm{ML}^{2} \omega \tag{ii} \end{equation*} $$

From law of conservation of angular momentum $\mathrm{L} _{1}=\mathrm{L} _{2}$. Therefore

$$ \frac{4 \mathrm{ML}^{2} \omega _{0}}{9}=\frac{22}{9} \mathrm{ML}^{2} \omega \quad \text { or } \quad \omega=\frac{2}{11} \omega _{0} $$

Correct answer: (4)

Solution:

Given $\omega _{0}=\frac{\mathrm{v} _{0}}{3 \mathrm{R}}$. Since $\mathrm{v} _{0} \neq \mathrm{R} \omega$; the cylinder slips an horizontal surface. Consider motion about an axis fixed to ground and passing through the instantaneous point of contact of cylinder with ground. There is no net torque about this axis; therefore angular momentum is conserved. Let $\mathrm{v}$ and $\omega$ denote the linear and angular speed when cylinder stops slipping i.e. it starts rolling. Then

$$ \begin{aligned} & \left(\frac{1}{2} \mathrm{MR}^{2}\right) \omega _{0}+\mathrm{Mv} _{0} \mathrm{R}=\left(\frac{1}{2} \mathrm{MR}^{2}\right) \omega+\mathrm{MvR} \\ & \frac{1}{2} \mathrm{MR}^{2}\left(\frac{\mathrm{v} _{0}}{3 \mathrm{R}}\right)+\mathrm{Mv} _{0} \mathrm{R}=\frac{1}{2} \mathrm{MR}^{2}\left(\frac{\mathrm{v}}{\mathrm{R}}\right)+\mathrm{MvR} \end{aligned} $$

$$ \begin{aligned} & \frac{7 \mathrm{Mv} _{0} \mathrm{R}}{6}=\frac{3 \mathrm{MvR}}{2} \\ & \therefore \quad \mathrm{v}=\frac{7}{9} \mathrm{v} _{0} \end{aligned} $$

Correct answer: (1)

Solution:

Let $\mathrm{M}$ be the instantaneous mass of wheel when its instantaneous radius is $\mathrm{r}$. Then

$$ \begin{equation*} \frac{\mathrm{M}}{\mathrm{M} _{0}}=\left(\frac{\mathrm{r}}{\mathrm{R}}\right)^{2} \quad \text { or } \quad \mathrm{M}=\mathrm{M} _{0}\left(\frac{\mathrm{r}}{\mathrm{R}}\right)^{2} \tag{i} \end{equation*} $$

The instantaneous M.I, of wheel about axis of rotation.

$$ =\mathrm{I}=\frac{1}{2} \mathrm{Mr}^{2}=\frac{\mathrm{M} _{0} \mathrm{r}^{4}}{2 \mathrm{R}^{2}} $$

The instantaneous torque $=\tau=$ F.r

The instantaneous acceleration is

$$ \begin{equation*} \alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=\frac{\tau}{\mathrm{I}}=\frac{2 \mathrm{FR}^{2}}{\mathrm{M} _{0} \mathrm{r}^{3}} \tag{ii} \end{equation*} $$

We have $\mathrm{M}=\mathrm{M} _{0}-\alpha \mathrm{t} \quad$ or $\quad \pi \mathrm{r}^{2} \sigma=\pi \mathrm{R}^{2} \sigma-\alpha \mathrm{t}$

$$ \mathrm{r}^{2}=\mathrm{R}^{2}-\left(\frac{\alpha}{\pi \sigma}\right) \mathrm{t} $$

where $\sigma=$ mass per unit area of wheel.

Differentiate w.r.t time $(\mathrm{t})$

$$ 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=-\left(\frac{\alpha}{\pi \sigma}\right) $$

or $\frac{\mathrm{dr}}{\mathrm{dt}}=-\frac{\alpha}{2 \pi \mathrm{r} \sigma}$

From Eqns (ii) and (iii) we have

$$ \begin{aligned} & \frac{\mathrm{d} \omega}{\mathrm{dt}}=\frac{\mathrm{d} \omega}{\mathrm{dr}} \cdot \frac{\mathrm{dr}}{\mathrm{dt}}=-\left(\frac{\alpha}{2 \pi \sigma r}\right) \frac{\mathrm{d} \omega}{\mathrm{dr}} \\ & \therefore \quad \mathrm{d} \omega=-\frac{4 \pi \mathrm{F} \sigma \mathrm{R}^{2}}{\mathrm{M} _{0} \alpha \mathrm{r}^{2}} \mathrm{dr} \end{aligned} $$

Integrating we have

$$ \int _{0}^{\omega} \mathrm{d} \omega=-\frac{4 \pi \mathrm{F} \sigma \mathrm{R}^{2}}{\mathrm{M} _{0} \alpha} \int _{\mathrm{R}}^{\mathrm{R} / 2} \mathrm{r}^{-2} \mathrm{dr} $$

$$ \therefore \quad \omega=\frac{4 \pi \mathrm{F} \sigma \mathrm{R}}{\mathrm{M} _{0} \alpha} $$

Correct answer: (1)

Solution:

Let $v$ and $\omega$ be linear speed of C.M and angular speed of cylinder at O. From law of conservation of energy.

Loss in gravitational P.E. $=$ Gain in K.E.

$$ \operatorname{Mg}(\mathrm{H}-\mathrm{R})=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{2}\left(\frac{1}{2} \mathrm{MR}^{2}\right) \omega^{2} $$

Since there is no slipping $\mathrm{v}=\mathrm{R} \omega$. Therefore

$$ \begin{align*} & \mathrm{Mg}(\mathrm{H}-\mathrm{R})=\frac{1}{2} \mathrm{Mv}^{2}+\frac{1}{4} \mathrm{MR}^{2}\left(\frac{\mathrm{V}}{\mathrm{R}}\right)^{2} \\ & \text { or } \quad \mathrm{v}^{2}=\frac{4 \mathrm{~g}(\mathrm{H}-\mathrm{R})}{3} \tag{i} \end{align*} $$

Since the part OB of track is smooth; as cylinder climbs up it losses translational part of its kinetic energy.

The loss in translation K.E equals gain in gravitational P.E. Therefore

$$ \begin{equation*} \frac{1}{2} \mathrm{Mv}^{2}=\mathrm{Mg}(\mathrm{h}-\mathrm{R}) \end{equation*} $$

or $\quad \mathrm{v}^{2}=2 \mathrm{~g}(\mathrm{~h}-\mathrm{R}) \hspace{40mm} . . . . . .(ii)$

From Eqns (i) and (ii) we have

$\frac{4 \mathrm{~g}(\mathrm{H}-\mathrm{R})}{3}=2 \mathrm{~g}(\mathrm{~h}-\mathrm{R})$

or $\mathrm{h}=\frac{(2 \mathrm{H}+\mathrm{R})}{3}$

Correct answer:(2)

Solution:

The forces acting are shown in Fig. 63. There is no translational motion therefore.

$$ \begin{equation*} \mathrm{N} _{1}+\mathrm{f} _{2}=\mathrm{Mg} \tag{i} \end{equation*} $$

and $\mathrm{N} _{2}=\mu \mathrm{N} _{1} \hspace{40mm} . . . . . .(ii)$

Fig. 63

From Eqns (i) and (ii) we have

$\mathrm{N} _{1}=\frac{\mathrm{Mg}}{\mu^{2}+1}$ and $\mathrm{N} _{2}=\frac{\mu \mathrm{Mg}}{\mu^{2}+1}$

Taking torque of forces about $\mathrm{O}$.

$\tau=\left(\mathrm{f} _{1}+\mathrm{f} _{2}\right) \mathrm{R}=\frac{\mu(\mu+1)}{\left(\mu^{2}+1\right)} \operatorname{MgR}$

$\tau$ is in clockwise direction.

Let $\alpha$ be the angular retardation. Then

$$ \begin{align*} & \left(\frac{1}{2} \mathrm{MR}^{2}\right) \alpha=\tau \\ \therefore \quad & \alpha=\frac{2 \mu \mathrm{g}(\mu+1)}{\left(\mu^{2}+1\right) \mathrm{R}} \tag{iv} \end{align*} $$

Given $\omega _{0}=2 \pi \mathrm{n} ; \omega=0 ; \alpha=-\frac{2 \mu \mathrm{g}(\mu+1)}{\left(\mu^{2}+1\right) \mathrm{R}} ; \theta=$ ?

$\therefore \quad \theta=\frac{\mathrm{n}^{2} \mathrm{R}\left(\mu^{2}+1\right)}{16 \mu \mathrm{g}(\mu+1)}$

The number of revolution $==\frac{\theta}{\pi}=\frac{n^{2}\left(\mu^{2}+1\right) R}{32 \pi^{2} \mu g(\mu+1)}$

Correct answer: (1)

Solution:

The loss is gravitational P.E in moving from position OA to position OB.

$$ =\operatorname{Mg} \frac{\mathrm{L}}{2}[\cos 30-\cos 60]=\frac{\mathrm{MgL}}{4}[\sqrt{3}-1] $$

This energy is converted into rotational K.E. of rod. Let $\omega$ be angular speed of rod in position OB.

Gain in rotational $\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{I} \omega^{2}=\frac{1}{2}\left(\frac{1}{3} \mathrm{ML}^{2}\right) \omega^{2}$

From law of conservation of energy

$\frac{\operatorname{MgL}}{4}[\sqrt{3}-1]=\frac{\mathrm{M}}{6} \mathrm{~L}^{2} \omega^{2}$

or $\quad \omega^{2}=\left[\frac{3(\sqrt{3}-1)}{2}\right]\left(\frac{g}{L}\right)$

$\therefore \omega=\left[\frac{(3 \sqrt{3}-1) \mathrm{g}}{2 \mathrm{~L}}\right]^{1 / 2}$