Rotational Motion

$ \begin{aligned} &\text { } \alpha=\frac{\omega_2-\omega_1}{\Delta t}=\left(\frac{1200-600}{10}\right) \frac{2 \pi}{60}=2 \pi \mathrm{rad} / \mathrm{s}^2\\ &\begin{array}{ll} \Rightarrow & \text { Use equation of motion, } \omega_2^2=\omega_1^2+2 \alpha \theta \\ \Rightarrow & \omega_2=1200 \times \frac{2 \pi}{60}=40 \pi \\ & \omega_1=600 \times \frac{2 \pi}{60}=20 \pi \\ & (40 \pi)^2=(20 \pi)^2+2 \times 2 \pi \theta \\ \Rightarrow & 1200 \pi^2=4 \pi \theta \\ \Rightarrow & \theta=300 \pi \text { radian } \\ \Rightarrow & \text { Number of revolution }=\frac{\theta}{2 \pi}=150 \end{array} \end{aligned} $

Mass of thin wire $=($ Linear mass density $) \times($ Length $)$

$ \Rightarrow \quad M=m L $

$L=2 \pi r$, where $r=$ radius of circular ring $=\frac{L}{2 \pi}$

Using parallel axis theorem,

$ \begin{aligned} & I_{y y^{\prime}}=I_{C M}+M r^2=\frac{M r^2}{2}+M r^2=\frac{3}{2} M r^2 \\ & \Rightarrow I_{y y^{\prime}}=\frac{3}{2} \times(m L) \times\left(\frac{L}{2 \pi}\right)^2=\frac{3 m L^3}{8 \pi^2} \end{aligned} $

For translational equilibrium

$\begin{aligned} & N_1=M g \\ & N_2=f \end{aligned}$

For rotational equilibrium

Torque about $A, M g \frac{L}{2} \cos \theta=N_2 L \sin \theta$

$\begin{aligned} & \frac{M g}{2} \cot \theta=N_2=f \\ & \frac{M g}{2} \cot 30^{\circ}=f \\ & \frac{M g}{2} \sqrt{3}=N_2 \\ & 100 \sqrt{3}=f \end{aligned}$

When considering the Sun as a solid sphere, the formula for the moment of inertia is given by:

$ I = \frac{2}{5} m R^2 $

Here, $ m $ is the mass and $ R $ is the radius of the Sun.

To find the new period of revolution if the Sun expands to twice its current radius, we apply the conservation of angular momentum. The principle states that angular momentum before and after the expansion must be equal:

$ l' \omega' = l \omega $

Substituting the expression for angular momentum $ (I \times \omega) $ for both initial and expanded states, we get:

$ \frac{2}{5} m (2R)^2 \times \frac{2\pi}{T'} = \frac{2}{5} m R^2 \times \frac{2\pi}{T} $

This simplifies to:

$ \Rightarrow 4mR^2 \times \frac{2\pi}{T'} = mR^2 \times \frac{2\pi}{T} $

Cancelling out common terms, we find:

$ \Rightarrow 4 \times \frac{1}{T'} = \frac{1}{T} $

Thus:

$ \Rightarrow T' = 4T = 4 \times 27 = 108 \text{ days} $

Therefore, the new period of revolution would be 108 days if the Sun were to expand to twice its present radius, assuming it remains a sphere of uniform density and there are no external influences.

For larger solid sphere about diameter $Y$-axis,

$I_{\text {whole }}=\frac{2}{5} M(2 R)^2=\frac{8}{5} M R^2$

Density of sphere is uniform

$\begin{aligned} & \Rightarrow \frac{M}{V_{\text {whole }}}=\frac{M_{\text {smaller }}}{V_{\text {smaller }}} \Rightarrow \frac{M}{\frac{4}{3} \pi(2 R)^3}=\frac{M^{\prime}}{\frac{4}{3} \pi R^3} \\ & \Rightarrow M^{\prime}=\frac{M}{8} \end{aligned}$

Using parallel axis theorem for smaller sphere,

$\begin{aligned} & I^{\prime}=I_{\mathrm{cm}}+M^{\prime} R^2=\frac{2}{5} \frac{M R^2}{8}+\frac{M R^2}{8}=\frac{7}{40} M R^2 \\ & \therefore \text { Ratio }=\frac{I_{\text {smaller }}}{I_{\text {remaining }}}=\frac{I^{\prime}}{I_{\text {whole }}-I^{\prime}}=\frac{\frac{7}{40} M R^2}{\left(\frac{8}{5}-\frac{7}{40}\right) M R^2}=\frac{7}{64-7}=\frac{7}{57} \end{aligned}$

$I_{X Y}=I_{C M}+M R^2=\frac{2}{5} M R^2+M R^2=\frac{7}{5} M R^2=\frac{7}{5} \times 5 R^2=7 R^2\quad \text{.... (1)}$

$\begin{aligned} & I_{X Y}=M K^2=5 \times 5^2 \quad \ldots(2) \\ & \therefore 5 \times 5^2=7 \times R^2 \quad \text { [From (1) and (2)] } \\ & \Rightarrow R=\sqrt{\frac{5}{7}} \times 5=\frac{5 x}{\sqrt{7}} \quad \text { (Given) } \\ & \therefore x=\sqrt{5} \end{aligned}$

$ \begin{aligned} & \text { Moment of inertia of rod }=I=\frac{m \ell^2}{12} \\ & \Rightarrow \quad 2400=400 \frac{\ell^2}{12} \\ & \Rightarrow \quad 72=\ell^2 \\ & \Rightarrow \quad \ell=\sqrt{72}=8.48 \mathrm{~cm} \simeq 8.5 \mathrm{~cm} \end{aligned}$

In the case of pure rolling,

The topmost point will have velocity $2 v$ while point $Q$ i.e. lowest point will have zero velocity. Hence point $P$ moves faster than point $Q$.

$\begin{aligned} \tau & =\mathrm{I} \alpha \Rightarrow \alpha=\frac{\tau}{\mathrm{I}}=\frac{100}{300}=\frac{1}{3} ~\mathrm{rad} / \mathrm{sec}^2 \\ \omega_{\mathrm{i}} & =0 \\ \omega_{\mathrm{f}} & =\omega_{\mathrm{i}}+\alpha \mathrm{t} \\ & =0+\frac{1}{3} \times 3 \\ \omega_{\mathrm{f}} & =1 ~\mathrm{rad} / \mathrm{sec}\end{aligned}$

To solve this problem, we need to find the ratio of the radius of gyration for a solid sphere (K₁) to the radius of gyration for a thin hollow sphere (K₂) of the same mass M and radius R.

The moment of inertia (I) of a solid sphere about its own axis is given by :

$I_{solid} = \frac{2}{5}MR^2$

The radius of gyration (K) is related to the moment of inertia (I) and mass (M) by the formula :

$I = MK^2$

So for the solid sphere, we can find K₁ using :

$K^2_{1} = \frac{I_{solid}}{M} = \frac{2}{5}R^2$

$K_{1} = R\sqrt{\frac{2}{5}}$

Now, for a thin hollow sphere, the moment of inertia about its axis is given by :

$I_{hollow} = \frac{2}{3}MR^2$

We can find K₂ using :

$K^2_{2} = \frac{I_{hollow}}{M} = \frac{2}{3}R^2$

$K_{2} = R\sqrt{\frac{2}{3}}$

Now, we need to find the ratio K₁ : K₂ :

$\frac{K_{1}}{K_{2}} = \frac{R\sqrt{\frac{2}{5}}}{R\sqrt{\frac{2}{3}}}$

The R terms cancel out, and we are left with :

$\frac{K_{1}}{K_{2}} = \frac{\sqrt{\frac{2}{5}}}{\sqrt{\frac{2}{3}}}$

Simplify by taking the square root of the fraction :

$\frac{K_{1}}{K_{2}} = \frac{\sqrt{2}\sqrt{3}}{\sqrt{5}\sqrt{2}}$

Now, the square root of 2 terms cancel out, giving us :

$\frac{K_{1}}{K_{2}} = \frac{\sqrt{3}}{\sqrt{5}}$

Along the axis of rotation.

From work - energy theorem

$W = \Delta k$ (change in Kinetic Energy)

In rotation, $KE = {1 \over 2}I{\omega ^2}$

$484 = {1 \over 2}I(\omega _f^2 - \omega _i^2)$

$ \Rightarrow 484 = {1 \over 2}I\left[ {{{\left( {2\pi {{360} \over {60}}} \right)}^2} - {{\left( {2\pi \times {{60} \over {60}}} \right)}^2}} \right]$

$ \Rightarrow 484 = {1 \over 2}I4{\pi ^2}(36 - 1)$

$ \Rightarrow I \simeq 0.7$ kg-m²

Angular acceleration $\alpha = {{{\omega _f} - {\omega _i}} \over t}$

${\omega _f} = 3120 \times {{2\pi } \over {60}}$ rad/s

${\omega _i} =1200 \times {{2\pi } \over {60}}$ rad/s

$ \Rightarrow \alpha = {{(3120 - 1200)} \over {16}} \times {{2\pi } \over {60}} = 4\pi $

${l_1} = {{M{R^2}} \over 2}$

${k_1} = \sqrt {{{{l_1}} \over M}} $

$ = {R \over {\sqrt 2 }}$

${l_2} = {{M{R^2}} \over 4}$

${k_2} = \sqrt {{{{l_2}} \over M}} $

$ = {R \over 2}$

${{{k_1}} \over {{k_2}}} = {{{R \over {\sqrt 2 }}} \over {{R \over 2}}}$

$ = \sqrt 2 :1$

Given, initial velocity of sphere, $u=2.8 \mathrm{~m} / \mathrm{s}$

Acceleration on the inclined plane,

$a=\frac{g \sin \theta}{1+\frac{k^2}{R^2}}$ .... (i)

where, $k$ and $R$ are the radius of gyration and radius of sphere, respectively.

Moment of inertia of sphere,

$\begin{aligned} I & =m k^2=\frac{2}{5} m R^2 \\ \frac{k^2}{R^2} & =\frac{2}{5} \quad \text{.... (ii)} \end{aligned}$

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

$\begin{aligned} a & =\frac{g \sin 30^{\circ}}{1+\frac{2}{5}} \quad\left[\because \theta=30^{\circ}, \text { given }\right] \\ & =\frac{5 g}{14}=\frac{25}{7} \mathrm{~m} / \mathrm{s}^2 \quad\left[\therefore g=10 \mathrm{~m} / \mathrm{s}^2\right] \end{aligned}$

If $s$ be the maximum distance travelled by the sphere, then at maximum distance, $v=0$.

$\therefore$ From equation of motion,

$\begin{aligned} v^2 & =u^2-2 a s \\ 0 & =u^2-2 a s \\ \Rightarrow \quad s & =\frac{u^2}{2 a}=\frac{(2.8)^2 \times 7}{2 \times 25}=1.09 \mathrm{~m} \simeq 1.38 \mathrm{~m} \end{aligned}$

Moment of inertia of the insect disc system,

$\mathrm{MI}=\frac{1}{2} M R^2+m x^2$

where, $m=$ mass of insect

and $\quad x=$ distance of insect from centre.

Clearly, as the insect moves along the diameter of the disc. Moment of inertia first decreases and then increases.

By conservation of angular momentum, angular speed first increases and then decreases.

$\begin{aligned} & \text { } \tau_{\mathrm{ext}}=\frac{d L}{d t}=0 \\ & \Rightarrow \frac{d L}{d t}=0 \Rightarrow L=\text { constant, i.e. } L_{\text {initial }}=L_{\text {final }} \end{aligned}$

Here, mass of ring $M=3 \mathrm{~kg}$.

radius of ring $r=0.6 \mathrm{~m}$

force applied $f=3 \mathrm{~N}$

acceleration $a=0.4 \mathrm{~m} / \mathrm{s}^2$

coefficient of friction between stick and ring $=\frac{F}{10}$

Now, according figure $f=M a$

$\begin{aligned} 3-f_s & =f=M a \\ \text{or} \quad f_s & =3-M a \\ & =3-3 \times 0.4=1.8 \mathrm{~N} \end{aligned}$

Now taking torque about $O$ We have

$f_s R-f_k R=I \cdot \alpha$

$=M R^2 \times \frac{a}{R}$

$\begin{array}{r} \left(\because \alpha=\text { angular acceleration }=\frac{a}{R}\right. \\ \text { and } \left.I=M R^2\right) \end{array}$

$\begin{gathered} \text { or } 1.8 \times 0.6-f_k \times 0.6=3 \times 0.6 \times 0.4 \\ f_k=1.8-1.2=0.6 \mathrm{~N} \end{gathered}$

If $\mu$ is the coefficient of kinetic friction, then

$\begin{aligned} f_k & =\mu \times 3 \\ \Rightarrow \quad \mu & =\frac{f_k}{2}=\frac{0.6}{3}=0.2=\frac{2}{10}=\frac{F}{10} \end{aligned}$

Hence on comparing, we get

$F=2 \mathrm{~N}$

The kinetic energy of a rolling solid sphere

$\begin{aligned} & =k_{\text {trans }}+k_{\text {rot }} \\ & =\frac{1}{2} m v^2+\frac{1}{2} I \omega^2 \\ & =\frac{1}{2} m v^2+\frac{1}{2} \times \frac{2}{5} m R^2 \omega^2 \\ & =\frac{1}{2} m v^2+\frac{1}{5} m v^2 \\ & =7 / 10 ~m v^2 \end{aligned}$

Different solid bodies have different kinetic energy as for different solid bodies, $I$ is different.