A bead is slide along the parabola without friction. The wire is accelerated parallel to the x axis with acceleration ' $a$ '. Find the new equilibrium position.

$N \cos \theta=m g-(1)$

$N \sin \theta=m a-(2)$

$\tan \theta=\frac{a}{g}$

$y=k x^2$, $\frac{d y}{d x}=2 k x={\tan \theta}$

$x= \frac{a}{2kg}$

A particle is going around the circle with angular velocity $\omega=5 \mathrm{rad} / \mathrm{sec}$, and the radius of the circle is $r=4 {m}$. To calculate, the speed of the foot of the perpendicular. When OP sweeps $\theta=30^{\circ}$.

$O P^{\prime}=x=r \cos \theta$

$\frac{d x}{d t}=-r \sin \theta(\frac{d \theta}{d t})=-r \omega \sin \theta$

$|v\vec{p^{\prime}}|=(-4 \times 5 \times \sin 30)=10 \mathrm{r} / \mathrm{s}$

Angular velocity of $P$ about $Q$

The required angle $\angle MQP$ $=\frac{\theta}{2}=\frac{5}{2}=2.5 \mathrm{rad} / \mathrm{sec}$

To calculated the $\mathrm{CM}$ of the remaining portion.

$X_{C M}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}$

$\sigma^{\prime}$ : mass per unit area of the disk

$x_1 =\pi\left(\frac{l}{2}\right)^2 \sigma ; x_1=R / 2$

$m_2 =\pi\left[R^2-\left(\frac{R}{2}\right)^2\right] \sigma =\frac{3 \pi R^2}{4}$

$X_{C M}=\frac{(\pi \frac{R^2}{4} \sigma \frac{R}{2}+\pi(\frac{3 \pi R^2}{4} \sigma) x}{\pi R^2 \rho}=0$

$\frac{R}{8}+\frac{3 x}{4}=0 \Rightarrow x=-\frac{R}{6}$

Problem involving moment of inertia, $\omega$, linear velocity, rotational kinetic energy, orbital angular momentum, etc.

Given a symmetric body is rotating about axis.

Rotational kinetic energy: $K E_{\text {Rot }}=K E=\frac{1}{2} I \omega^2-(1)$

$l=I \omega$

$\frac{K E}{l}=\frac{\omega}{2} \Rightarrow \omega=\frac{2 K E}{l}$

$I=\frac{l}{\omega}=\frac{l^2}{2 K E}$

$\omega$ is (A)$\frac{{2KE}}{l}$, (B)$\frac{K E}{l}$, (C)$\frac{K E}{2 l}$, (D)$\frac{K E}{\sqrt{2 l}}$

$\omega=\frac{K E}{l}$

Three rods each of length L, joined to form an equilateral triangle.

mass of each rod : m

length of each rod : l

$\angle$ DQO $= {30^{\circ}}$

$\frac{I}{z} = \frac{I}{{z^{\prime}}} + md^2$

$\tan 30^{\circ}$$=\frac{d}{l/ 2}=\frac{1}{\sqrt{3}} \Rightarrow d=\frac{l}{2 \sqrt{3}}$

$I_z =\frac{m l^2}{12} + m(\frac{l}{2 \sqrt{3}})^2=\frac{m l^2}{b}$

$I_{\text {system }}=3 \times \frac{3 L^2}{6}=\frac{m L^2}{2}$

Two solid sphers have the same mass, made of materials of different densities. Which will have larger MI about an axis passing through the center.

$I_1=\frac{2}{5} m r_1^2, I_2=\frac{2}{5} m r_2^2$

$\frac{I_1}{I_2}=\frac{r_1^2}{r_2^2}$

$m_1=m=\frac{4}{3} \pi r_1^3 s_1 \Rightarrow r_1^3=\frac{3 m}{4 \pi s_1}$

$r_1^2=\left(\frac{8 m}{4 a^3}\right)^{2 / 3}$

$m_2=m=\frac{4}{3} \pi r_2^3 s_2 \Rightarrow r_2^3=\frac{3 m}{4 \pi s_2}$

$r_2^2=\left(\frac{8 m}{4 a^3}\right)^{2 / 3}$

$\frac{I_1}{I_2} \propto (\frac{s_2}{s_1})^{2/3} \Rightarrow I \propto$ $\frac{1}{\rho^{2/3}}$

To find the point of application of the force.

Taking moments about c:

$\tau_c=0$

$\tau_D=30 \times 3=90 {L}$

$90=10 \times x \Rightarrow x=9 {m}$

In fact, one can put $\times N$ at $B$, then

$90 =x(10)$

$\Rightarrow$ X =9 N

Thin spherical shell, the shell rolls without slipping. To find the acceleration of the shell.

Translational motion: F + f = ma ....(1)

Rotational motion

$F R-f R=I \alpha=I(\frac{a}{R})$ ....(2)

$F-f=\frac{I a}{R^2}(3)$

(l)+(2)

$2 F=(m+\frac{I}{R^2}) a$

$=(m+\frac{2 /3 m R^2}{R^2}) a \Rightarrow a=\frac{6 R}{5 m}$