## Systems of Particles And Rotational Motion Exercises

### Exercises

**6.1** Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body?

## Show Answer

**Answer**

Geometric centre; No

The centre of mass (C.M.) is a point where the mass of a body is supposed to be concentrated. For the given geometric shapes having a uniform mass density, the C.M. lies at their respective geometric centres.

The centre of mass of a body need not necessarily lie within it. For example, the C.M. of bodies such as a ring, a hollow sphere, etc., lies outside the body.

**6.2** In the $\mathrm{HCl}$ molecule, the separation between the nuclei of the two atoms is about $1.27 \mathring{A}\left(1 \mathring{A}=10^{-10} \mathrm{~m}\right)$. Find the approximate location of the $\mathrm{CM}$ of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

## Show Answer

**Answer**

The given situation can be shown as:

Distance between $H$ and $Cl$ atoms $=1.27 \mathring{A}$

Mass of $H$ atom $=m$

Mass of $Cl$ atom $=35.5 m$

Let the centre of mass of the system lie at a distance $x$ from the $Cl$ atom.

Distance of the centre of mass from the $H$ atom $=(1.27-x)$

Let us assume that the centre of mass of the given molecule lies at the origin. Therefore, we can have:

$ \begin{aligned} & \frac{m(1.27-x)+35.5 m x}{m+35.5 m}=0 \\ & m(1.27-x)+35.5 m x=0 \\ & 1.27-x=-35.5 x \\ & \therefore x=\frac{-1.27}{(35.5-1)}=-0.037 \mathring{A} \end{aligned} $

Here, the negative sign indicates that the centre of mass lies at the left of the molecule. Hence, the centre of mass of the $HCl$ molecule lies $0.037 \mathring{A}$ from the $Cl$ atom.

**6.3** A child sits stationary at one end of a long trolley moving uniformly with a speed $V$ on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?

## Show Answer

**Answer**

No change

The child is running arbitrarily on a trolley moving with velocity $v$. However, the running of the child will produce no effect on the velocity of the centre of mass of the trolley. This is because the force due to the boy’s motion is purely internal. Internal forces produce no effect on the motion of the bodies on which they act. Since no external force is involved in the boy-trolley system, the boy’s motion will produce no change in the velocity of the centre of mass of the trolley.

**6.4** Show that the area of the triangle contained between the vectors $\mathbf{a}$ and $\mathbf{b}$ is one half of the magnitude of $\mathbf{a} \times \mathbf{b}$.

## Show Answer

**Answer**

Consider two vectors $\overrightarrow{{}OK}=|\vec{a}| _{\text{and }} \overrightarrow{{}OM}=|\vec{b}|$, inclined at an angle $\theta$, as shown in the following figure.

In $\triangle OMN$, we can write the relation:

$ \begin{aligned} & \sin \theta=\frac{MN}{OM}=\frac{MN}{|\vec{b}|} \\ & MN=|\vec{b}| \sin \theta \\ & |\vec{a} \times \vec{a}|=|\vec{a}||\vec{b}| \sin \theta \end{aligned} $

$ =OK \cdot MN \times \frac{2}{2} $

$=2 \times$ Area of $\triangle OMK$

$\therefore$ Area of $\triangle OMK=\frac{1}{2}|\vec{a} \times \vec{b}|$

**6.5** Show that $\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})$ is equal in magnitude to the volume of the parallelepiped formed on the three vectors, $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$.

## Show Answer

**Answer**

A parallelepiped with origin $O$ and sides $a, b$, and $c$ is shown in the following figure.

Volume of the given parallelepiped $=a b c$

$\overrightarrow{{}OC}=\vec{a}$

$\overrightarrow{{}OB}=\vec{b}$

$\overrightarrow{{}OC}=\vec{c}$

Let $\hat{\mathbf{n}}$ be a unit vector perpendicular to both $b$ and $c$. Hence, $\hat{\mathbf{n}}$ and $a$ have the same direction.

$ \begin{aligned} & \therefore \vec{b} \times \vec{c}=b c \sin \theta \hat{\mathbf{n}} \\ & =b c \sin 90^{\circ} \hat{\mathbf{n}} \\ & =b c \hat{n} \\ & \vec{a} \cdot(\vec{b} \times \vec{c}) \\ & =a \cdot(b c \hat{\mathbf{n}}) \\ & =a b c \cos \theta \hat{\mathbf{n}} \\ & =a b c \cos 0^{\circ} \\ & =a b c \end{aligned} $

$=$ Volume of the parallelepiped

**6.6** Find the components along the $x, y, z$ axes of the angular momentum 1 of a particle, whose position vector is $\mathbf{r}$ with components $x, y, z$ and momentum is $\mathbf{p}$ with components $p_{\mathrm{x}}, p_{\mathrm{y}}$ and $p_{\mathrm{z}}$. Show that if the particle moves only in the $x-y$ plane the angular momentum has only a $z$-component.

## Show Answer

**Answer**

$l_x=y p_z-z p_y$

$l_y=z p_x-x p_z$

$l_z=x p_y-y p_x$

Linear momentum of the particle, $\vec{p}=p_x \hat{\mathbf{i}}+p_y \hat{\mathbf{j}}+p_z \hat{\mathbf{k}}$

Position vector of the particle, $\vec{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$

Angular momentum, $\vec{l}=\vec{r} \times \vec{p}$

$ =(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \times(p_x \hat{\mathbf{i}}+p_y \hat{\mathbf{j}}+p_z \hat{\mathbf{k}}) $

$= \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ x & y & z \\ p_x & p_y & p_z\end{vmatrix} $

$l_x \hat{\mathbf{i}}+l_y \hat{\mathbf{j}}+l_z \hat{\mathbf{k}}=\hat{\mathbf{i}}(y p_z-z p_y)-\hat{\mathbf{j}}(x p_z-z p_x)+\hat{\mathbf{k}}(x p_y-z p_x)$

Comparing the coefficients of $\hat{\mathbf{i}}, \hat{\mathbf{j}}$, and $\hat{\mathbf{k}}$, we get:

$$ \left. \begin{matrix} l_x=y p_z-z p_y \\ l_y=x p_z-z p_x \tag{i}\\ l_z=x p_y-y p_x \end{matrix} \right\rbrace $$

The particle moves in the $x-y$ plane. Hence, the $z$-component of the position vector and linear momentum vector becomes zero, i.e.,

$z=p_z=0$

Thus, equation $(i)$ reduces to:

$ \left.\begin{matrix} l_x=0 \\ l_y=0 \\ l_z=x p_y-y p_x \end{matrix} \right\rbrace $

Therefore, when the particle is confined to move in the $x-y$ plane, the direction of angular momentum is along the $z$-direction.

**6.7** Two particles, each of mass $m$ and speed $v$, travel in opposite directions along parallel lines separated by a distance $d$. Show that the angular momentum vector of the two particle system is the same whatever be the point about which the angular momentum is taken.

## Show Answer

**Answer**

Let at a certain instant two particles be at points $P$ and $Q$, as shown in the following figure.

Angular momentum of the system about point P:

$$ \begin{align*} \vec{L} _P & =m v \times 0+m v \times d \\ & =m v d \tag{i} \end{align*} $$

Angular momentum of the system about point $Q$ :

$$ \begin{align*} \vec{L} _Q & =m v \times d+m v \times 0 \\ & =m v d \tag{ii} \end{align*} $$

Consider a point $R$, which is at a distance $y$ from point $Q$, i.e.,

$QR=y$

$\therefore PR=d-y$

Angular momentum of the system about point R:

$$ \begin{align*} \vec{L} _R & =m v \times(d-y)+m v \times y \\ & =m v d-m v y+m v y \\ & =m v d \tag{iii} \end{align*} $$

Comparing equations (i), (ii), and (iii), we get:

$ \vec{L} _P= \vec{L} _Q= \vec{L} _R$

We infer from equation (iv) that the angular momentum of a system does not depend on the point about which it is taken.

**6.8** A non-uniform bar of weight $W$ is suspended at rest by two strings of negligible weight as shown in Fig.6.33. The angles made by the strings with the vertical are $36.9^{\circ}$ and $53.1^{\circ}$ respectively. The bar is $2 \mathrm{~m}$ long. Calculate the distance $d$ of the centre of gravity of the bar from its left end.

**Fig. 6.33**

## Show Answer

**Answer**

The free body diagram of the bar is shown in the following figure.

Length of the bar, $l=2 m$

$T_1$ and $T_2$ are the tensions produced in the left and right strings respectively.

At translational equilibrium, we have:

$T_1 \sin 36.9^{\circ}=T_2 \sin 53.1$

$\frac{T_1}{T_2}=\frac{\sin 53.1^{\circ}}{\sin 36.9}$

$=\frac{0.800}{0.600}=\frac{4}{3}$

$\Rightarrow T_1=\frac{4}{3} T_2$

For rotational equilibrium, on taking the torque about the centre of gravity, we have:

$ \begin{aligned} & T_1 \cos 36.9 \times d=T_2 \cos 53.1(2-d) \\ \\ & T_1 \times 0.800 d=T_2 0.600(2-d) \\ \\ & \frac{4}{3} \times T_2 \times 0.800 d=T_2[0.600 \times 2-0.600 d] \\ \\ & 1.067 d+0.6 d=1.2 \\ \\ & \therefore d=\frac{1.2}{1.67} \\ \\ & \quad=0.72 m \end{aligned} $

Hence, the C.G. (centre of gravity) of the given bar lies $0.72 m$ from its left end.

**6.9** A car weighs $1800 \mathrm{~kg}$. The distance between its front and back axles is $1.8 \mathrm{~m}$. Its centre of gravity is $1.05 \mathrm{~m}$ behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.

## Show Answer

**Answer**

Mass of the car, $m=1800 kg$

Distance between the front and back axles, $d=1.8 m$

Distance between the C.G. (centre of gravity) and the back axle $=1.05 m$

The various forces acting on the car are shown in the following figure.

$R_f$ and $R_b$ are the forces exerted by the level ground on the front and back wheels respectively.

At translational equilibrium:

$ \begin{aligned} & R_f+R_b=m g \\ = & 1800 \times 9.8 \\ = & 17640 N \ldots(i) \end{aligned} $

For rotational equilibrium, on taking the torque about the C.G., we have:

$$ \begin{align*} & R_f(1.05)=R_b(1.8-1.05) \\ & R_f \times 1.05=R_b \times 0.75 \\ & \frac{R_f}{R_b}=\frac{0.75}{1.05}=\frac{5}{7} \\ & \frac{R_b}{R_f}=\frac{7}{5} \\ & R_b=1.4 R_f \tag{ii} \end{align*} $$

Solving equations $(i)$ and (ii), we get:

$ \begin{aligned} & 1.4 R_f+R_f=17640 \\ & R_f=\frac{17640}{2.4}=7350 N \end{aligned} $

$\therefore R_b=17640-7350=10290 N$

Therefore, the force exerted on each front wheel $=\frac{7350}{2}=3675 N$, and

The force exerted on each back wheel

$ =\frac{10290}{2}=5145 N $

**6.10** Torques of equal magnitude are applied to a hollow cylinder and a solid sphere, both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry, and the sphere is free to rotate about an axis passing through its centre. Which of the two will acquire a greater angular speed after a given time.

## Show Answer

**Answer**

Let $m$ and $r$ be the respective masses of the hollow cylinder and the solid sphere.

The moment of inertia of the hollow cylinder about its standard axis, $I_1=m r^{2}$

The moment of inertia of the solid sphere about an axis passing through its centre, $I_{II}=\frac{2}{5} m r^{2}$

We have the relation:

$ \tau=I \alpha $

Where,

$\alpha=$ Angular acceleration

$\tau=$ Torque

$I=$ Moment of inertia

For the hollow cylinder, $ _1=I_1 \alpha_1$

For the solid sphere, $\tau_{II}=I_{II} \alpha_{II}$

As an equal torque is applied to both the bodies, $\tau_1=\tau_2$

$\therefore \frac{\alpha_{II}}{\alpha_I}=\frac{I_I}{I_{Il}}=\frac{m r^{2}}{\frac{2}{5} m r^{2}}=\frac{2}{5}$

$$\alpha_{\text{II }}>\alpha_{\text{I }}\tag{i} $$

Now, using the relation:

$\omega=\omega_0+\alpha t$

Where,

$\omega_0=$ Initial angular velocity

$t=$ Time of rotation

$\omega=$ Final angular velocity

For equal $\omega_0$ and $t$, we have:

$\omega \propto \alpha \ldots(ii)$

From equations ( $i$ ) and (ii), we can write:

$\omega_{\text{II }}>\omega _{\text{I }}$

Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.

**6.11** A solid cylinder of mass $20 \mathrm{~kg}$ rotates about its axis with angular speed $100 \mathrm{rad} \mathrm{s}^{-1}$. The radius of the cylinder is $0.25 \mathrm{~m}$. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of angular momentum of the cylinder about its axis?

## Show Answer

**Answer**

Mass of the cylinder, $m=20 kg$

Angular speed, $\omega=100 rad s^{-1}$

Radius of the cylinder, $r=0.25 m$

The moment of inertia of the solid cylinder:

$I=\frac{m r^{2}}{2}$

$=\frac{1}{2} \times 20 \times(0.25)^{2}$

$=0.625 kg m^{2}$

$\therefore$ Kinetic energy $=\frac{1}{2} I \omega^{2}$

$=\frac{1}{2} \times 6.25 \times(100)^{2}=3125 J$

$\therefore$ Angular momentum, $L=I \omega$

$=6.25 \times 100$

$=62.5 Js$

**6.12** (a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of $40 \mathrm{rev} / \mathrm{min}$. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to $2 / 5$ times the initial value? Assume that the turntable rotates without friction.

(b) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

## Show Answer

**Answer**

$100 \text{ rev/min }$

Initial angular velocity, $\omega_1=40 \text{ rev/min }$

Final angular velocity $=\omega_2$

The moment of inertia of the boy with stretched hands $=I_1$

The moment of inertia of the boy with folded hands $=I_2$

The two moments of inertia are related as:

$I_2=\frac{2}{5} I_1$

Since no external force acts on the boy, the angular momentum $L$ is a constant.

Hence, for the two situations, we can write:

$ \begin{aligned} & I_2 \omega_2=I_1 \omega_1 \\ & \omega_2=\frac{I_1}{I_2} \omega_1 \\ & =\frac{I_1}{\frac{2}{5} I_1} \times 40=\frac{5}{2} \times 40 \\ & =100 \text{ rev/min } \end{aligned} $

**(b)** Final K.E. = 2.5 Initial K.E.

Final kinetic rotation, $E_F=\frac{1}{2} I_2 \omega_2^{2}$

Initial kinetic rotation, $E_I=\frac{1}{2} I_1 \omega_1^{2}$

$ \begin{aligned} \frac{E_F}{E_1} & =\frac{\frac{1}{2} I_2 \omega_2^{2}}{\frac{1}{2} I_1 \omega_1^{2}} \\ \\ & =\frac{2}{5} \frac{I_1}{I_1} \frac{(100)^{2}}{(40)^{2}} \\ \\ & =\frac{2}{5} \times \frac{100 \times 100}{40 \times 40} \\ \\ & =\frac{5}{2}=2.5 \\ \\ \therefore E_F & =2.5 E_1 \end{aligned} $

The increase in the rotational kinetic energy is attributed to the internal energy of the boy.

**6.13** A rope of negligible mass is wound round a hollow cylinder of mass $3 \mathrm{~kg}$ and radius $40 \mathrm{~cm}$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 \mathrm{~N}$ ? What is the linear acceleration of the rope? Assume that there is no slipping.

## Show Answer

**Answer**

Mass of the hollow cylinder, $m=3 kg$

Radius of the hollow cylinder, $r=40 cm=0.4 m$

Applied force, $F=30 N$

The moment of inertia of the hollow cylinder about its geometric axis:

$I=m r^{2}$

$=3 \times(0.4)^{2}=0.48 kg m^{2}$

Torque, ${ }^{\tau}=F \times r$

$=30 \times 0.4=12 Nm$

For angular acceleration $\alpha$, torque is also given by the relation:

$\tau=I \alpha$

$\alpha=\frac{\tau}{I}=\frac{12}{0.48}$

$=25 rad s^{-2}$

Linear acceleration $=r \alpha=0.4 \times 25=10 m s^{-2}$

**6.14** To maintain a rotor at a uniform angular speed of $200 \mathrm{rad} \mathrm{s}^{-1}$, an engine needs to transmit a torque of $180 \mathrm{~N} \mathrm{~m}$. What is the power required by the engine ? (Note: uniform angular velocity in the absence of friction implies zero torque. In practice, applied torque is needed to counter frictional torque). Assume that the engine is $100 \%$ efficient.

## Show Answer

**Answer**

Angular speed of the rotor, $\omega=200 \text{ rad / s}$

Torque required, $\tau=180 Nm$

The power of the rotor $(P)$ is related to torque and angular speed by the relation:

$P=\tau \omega$

$=180 \times 200=36 \times 10^{3}$

$=36 kW$

Hence, the power required by the engine is $36 kW$.

**6.15** From a uniform disk of radius $R$, a circular hole of radius $R / 2$ is cut out. The centre of the hole is at $R / 2$ from the centre of the original disc. Locate the centre of gravity of the resulting flat body.

## Show Answer

**Answer**

$R / 6$; from the original centre of the body and opposite to the centre of the cut portion.

Mass per unit area of the original disc $=\sigma$

Radius of the original disc $=R$

Mass of the original disc, $M=\pi R^{2} \sigma$

The disc with the cut portion is shown in the following figure:

Radius of the smaller disc $=\frac{R}{2}$

Mass of the smaller disc, $M^{\prime}=\pi(\frac{R}{2})^{2} \sigma=\frac{1}{4} \pi R^{2} \sigma=\frac{M}{4}$

Let $O$ and $O^{\prime}$ be the respective centres of the original disc and the disc cut off from the original. As per the definition of the centre of mass, the centre of mass of the original disc is supposed to be concentrated at $O$, while that of the smaller disc is supposed to be concentrated at $O^{\prime}$.

It is given that: $OO^{\prime}=\frac{R}{2}$

After the smaller disc has been cut from the original, the remaining portion is considered to be a system of two masses. The two masses are:

$M$ (concentrated at $O$), and

$-M^{\prime}(=\frac{M}{4}) _{\text{concentrated at } O^{\prime}}$

(The negative sign indicates that this portion has been removed from the original disc.)

Let $x$ be the distance through which the centre of mass of the remaining portion shifts from point $O$.

The relation between the centres of masses of two masses is given as:

$ x=\frac{m_1 r_1+m_2 r_2}{m_1+m_2} $

For the given system, we can write:

$ \begin{aligned} x & =\frac{M \times 0-M^{\prime} \times(\frac{R}{2})}{M+(-M^{\prime})} \\ \\ & =\frac{\frac{-M}{4} \times \frac{R}{2}}{M-\frac{M}{4}}=\frac{-M R}{8} \times \frac{4}{3 M}=\frac{-R}{6} \end{aligned} $

(The negative sign indicates that the centre of mass gets shifted toward the left of point O.)

**6.16** A metre stick is balanced on a knife edge at its centre. When two coins, each of mass $5 \mathrm{~g}$ are put one on top of the other at the $12.0 \mathrm{~cm}$ mark, the stick is found to be balanced at $45.0 \mathrm{~cm}$. What is the mass of the metre stick?

## Show Answer

**Answer**

Let $W$ and $W^{\prime}$ be the respective weights of the metre stick and the coin.

The mass of the metre stick is concentrated at its mid-point, i.e., at the $50 ~cm$ mark.

Mass of the meter stick $=m$

Mass of each coin, $m=5 ~g$

When the coins are placed $12 cm$ away from the end $P$, the centre of mass gets shifted by $5 cm$ from point $R$ toward the end $P$. The centre of mass is located at a distance of $45 ~cm$ from point $P$.

The net torque will be conserved for rotational equilibrium about point $R$.

$ \begin{aligned} & 10 \times g(45-12)-m^{\prime} g(50-45)=0 \\ & \therefore m^{\prime}=\frac{10 \times 33}{5}=66 ~g \end{aligned} $

Hence, the mass of the metre stick is $66 g$.

**6.17** The oxygen molecule has a mass of $5.30 \times 10^{-26} \mathrm{~kg}$ and a moment of inertia of $1.94 \times 10^{-46} \mathrm{~kg} \mathrm{~m}^{2}$ about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500 \mathrm{~m} / \mathrm{s}$ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

## Show Answer

**Answer**

Mass of an oxygen molecule, $m=5.30 \times 10^{-26} kg$

Moment of inertia, $I=1.94 \times 10^{-46} kg m^{2}$

Velocity of the oxygen molecule, $v=500 m / s$

The separation between the two atoms of the oxygen molecule $=2 r$

Mass of each oxygen atom $=\frac{m}{2}$

Hence, moment of inertia $I$, is calculated as:

$ \begin{aligned} & (\frac{m}{2}) r^{2}+(\frac{m}{2}) r^{2}=m r^{2} \\ \\ & r=\sqrt{\frac{I}{m}} \\ \\ & \sqrt{\frac{1.94 \times 10^{-46}}{5.36 \times 10^{-26}}}=0.60 \times 10^{-10} m \end{aligned} $

It is given that:

$ \begin{aligned} & KE _{\text{rot }}=\frac{2}{3} KE _{\text{trans }} \\ \\ & \frac{1}{2} I \omega^{2}=\frac{2}{3} \times \frac{1}{2} \times m v^{2} \\ \\ & m r^{2} \omega^{2}=\frac{2}{3} m v^{2} \\ \\ & \omega=\sqrt{\frac{2}{3}} \frac{v}{r} \\ \\ & =\sqrt{\frac{2}{3}} \times \frac{500}{0.6 \times 10^{-10}} \\ \\ & =6.80 \times 10^{12} \text{ rad / s} \end{aligned} $