Rotational Motion

$ \begin{aligned} &\text { Moment of inertia of solid cylinder }\\ &\begin{aligned} I & =\frac{M R^2}{2}=\frac{2.5 \times(0.1)^2}{2} \\ & =1.25 \times 10^{-2} \mathrm{~kg}-\mathrm{m}^2 \\ & =0.0125 \mathrm{~kg}-\mathrm{m}^2 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} \omega & =\omega_0+\alpha t \\ \Rightarrow \quad \alpha & =\frac{\omega-\omega_0}{t}=\frac{21-6}{1.5} \\ & =10 \mathrm{rad} / \mathrm{s}^2 \\ \text { Since, } L & =I \omega \\ \Rightarrow \quad \frac{d L}{d t} & =I \frac{d \omega}{d t}=I \alpha \\ & =100 \times 10^{-3} \times 10=1 \mathrm{Nm} \end{aligned} $

Step 1: Find the radius and moment of inertia.

The diameter of the disc is 0.8 m, so the radius $R$ is half of that: $R = 0.4$ m.

The mass $M$ is 4 kg. The moment of inertia $I$ for a disc about its center is given by $I = \frac{MR^2}{2}$.

So, $I = \frac{4 \times (0.4)^2}{2}$

$= \frac{4 \times 0.16}{2} = \frac{0.64}{2} = 0.32 \mathrm{~kg} \cdot \mathrm{~m}^2$

Step 2: Use the torque formula to get angular acceleration.

We know the formula: $\tau = I \alpha$, where $\tau$ is torque, $I$ is moment of inertia, and $\alpha$ is angular acceleration.

So, $\alpha = \frac{\tau}{I}$.

Plug in the values: $\alpha = \frac{2.56}{0.32} = 8~\mathrm{rad}/\mathrm{s}^2$

Let:

• M be the mass of both the solid sphere and the solid cylinder.

• R be the radius of both the solid sphere and the solid cylinder.

1. Moment of inertia of a solid sphere about its diameter ($I_{sphere}$):

   The formula for the moment of inertia of a solid sphere of mass M and radius R about any of its diameters is given by:

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

2. Moment of inertia of a solid cylinder about its axis ($I_{cylinder}$):

   The formula for the moment of inertia of a solid cylinder of mass M and radius R about its central longitudinal axis is given by:

   $I_{cylinder} = \frac{1}{2} MR^2$

Now, we need to find the ratio $I_{sphere} : I_{cylinder}$.

Ratio $= \frac{I_{sphere}}{I_{cylinder}} = \frac{\frac{2}{5} MR^2}{\frac{1}{2} MR^2}$

We can cancel out the common terms $MR^2$ from the numerator and the denominator:

Ratio $= \frac{\frac{2}{5}}{\frac{1}{2}}$

To simplify the fraction, we multiply the numerator by the reciprocal of the denominator:

Ratio $= \frac{2}{5} \times \frac{2}{1}$

Ratio $= \frac{4}{5}$

Therefore, the ratio of the moment of inertia of the solid sphere about its diameter and the moment of inertia of the solid cylinder about its axis is $4:5$.

As sphere is in pure rolling, $v=R \omega$

Rotational KE of sphere $=\frac{1}{2} I \omega^2$

$ =\frac{1}{2} \cdot \frac{2}{5} m R^2 \omega^2=\frac{1}{5} m v^2 $

Kinetic energy of translation $=\frac{1}{2} m v^2$

Total $\mathrm{KE}=\left(\frac{1}{2}+\frac{1}{5}\right) m v^2=\frac{7}{10} m v^2$

Required ratio $=\frac{1}{5} \times \frac{10}{7}=\frac{2}{7}$

Moment of inertia of disc $B$ about given axis,

$ I_B=\frac{1}{2} M r^2 $

Moment of inertia of disc $A$ about given axis using parallel axes theorem is,

$ I_A=\frac{1}{2} M r^2+M \cdot(2 r)^2=\frac{9}{2} M r^2 $

$\therefore M \cdot I$ of system about given axis is,

$ I=I_A+I_B=\left(\frac{9}{2}+\frac{1}{2}\right) M r^2=5 M r^2 $

$ \begin{aligned} &\text { Rotational kinetic energy }\\ &\begin{aligned} K_r & =\frac{1}{2} I \omega^2 \\ & =\frac{1}{2} \times\left(\frac{1}{2} M R^2\right) \omega^2 \\ & =\frac{1}{2} \times \frac{1}{2} \times 20 \times 1^2 \times 2^2 \\ & =20 \mathrm{~J} \end{aligned} \end{aligned} $

Given:

A thin uniform rod of length $ L $, and we need the radius of gyration $ k $ about an axis passing through its centre and perpendicular to its length.

Formulae:

Moment of inertia (M.I.) about the given axis for a uniform rod is:

$ I = \frac{1}{12} M L^2 $

The radius of gyration $ k $ is related to the moment of inertia by

$ I = M k^2 $

Substituting:

$ M k^2 = \frac{1}{12} M L^2 $

$ k^2 = \frac{L^2}{12} $

$ k = \frac{L}{\sqrt{12}} $

✅ Final Answer:

$ \boxed{\text{Option A: } \frac{L}{\sqrt{12}}} $

$K_{\text {disc }}=6 \mathrm{~J}$

$ \begin{aligned} & \therefore \quad \frac{1}{2} I \omega^2+\frac{1}{2} m v^2=6 \\ & \Rightarrow \quad \frac{1}{2} \times \frac{1}{2} m R^2 \cdot\left(\frac{v}{R}\right)^2+\frac{1}{2} m v^2=6 \\ & \Rightarrow \quad \frac{3}{4} m v^2=6 \Rightarrow m v^2=8 \mathrm{~J} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { Thus, } K_{\text {Ring }}=\frac{1}{2} I \omega^2+\frac{1}{2} m v^2 \\ & \quad=\frac{1}{2} \times(m R)^2\left(\frac{v}{R}\right)^2+\frac{1}{2} m v^2 \\ & \quad=\frac{1}{2} m v^2+\frac{1}{2} m v^2 \\ & \quad=m v^2=8 \mathrm{~J} \end{aligned} $

$a_{\text {rolling }}=\frac{g \sin \theta}{1+\frac{I}{M R^2}}=\frac{g \sin \theta}{1+\frac{\frac{2}{5} M R^2}{M R^2}}$

$ \begin{aligned} & 3.5=\frac{5}{7}(g \sin \theta) \\ \Rightarrow \quad & g \sin \theta=4.9 \end{aligned} $

Thus, acceleration of the sphere when it slides down without rolling is $g \sin \theta$

$ =4.9 \mathrm{~m} / \mathrm{s}^2 $

Rotational kinetic energy of disc

$ \begin{aligned} K_r & =\frac{1}{2} I \omega^2 \\ & =\frac{1}{2} \times \frac{1}{2} M R^2 \times\left(\frac{v}{R}\right)^2 \\ & =\frac{1}{4} m v^2 \end{aligned} $

Total kinetic energy

$ \begin{aligned} & =\frac{1}{2} I \omega^2+\frac{1}{2} m v^2 \\ & =\frac{1}{2} \times \frac{1}{2} M R^2 \times\left(\frac{v}{R}\right)^2+\frac{1}{2} m v^2 \\ & =\frac{1}{4} m v^2+\frac{1}{2} m v^2 \\ & =\frac{3}{4} m v^2=3\left(K_r\right) \end{aligned} $

The moment of inertia of a rod about an axis perpendicular to the rod and passing through one end is $\frac{M L^2}{3}$. Two rods are in the $X$ and $Y$-axis and they both contribute to the moment of inertia about the $Z$-axis. The rod along

the $Z$-axis does not contribute to the moment of inertia about the $Z$-axis. Therefore the total moment of inertia of the system is

$ \frac{M L^2}{3}+\frac{M L^2}{3}=\frac{2 M L^2}{3} $

Radius of gyration of ring

$ \begin{aligned} K & =\sqrt{\frac{I}{M}}=\sqrt{\frac{M R^2}{M}} \\ K & =R \\ \Rightarrow 10 \sqrt{2} & =R \\ \Rightarrow \quad R & =10 \sqrt{2} \mathrm{~cm} \end{aligned} $

Radius of gyration about diameter

$ K_d=\sqrt{\frac{I_d}{M}}=\sqrt{\frac{M R^2}{\frac{2}{M}}} $

$ =\frac{R}{\sqrt{2}}=\frac{10 \sqrt{2}}{\sqrt{2}}=10 \mathrm{~cm} $

$ \begin{aligned} \theta & =\omega_0 t+\frac{1}{2} \alpha t^2 \\ & =0 \times t+\frac{1}{2} \times \pi \times 6^2 \\ & =18 \pi \end{aligned} $

$\therefore$ Number of rotations made by the wheel,

$ n=\frac{\theta}{2 \pi}=\frac{18 \pi}{2 \pi}=9 $

Let $I_{\rm solid}=\tfrac25MR^2$ and $I_{\rm shell}=\tfrac23MR^2$. Since

$I_{\rm solid}=20\;\Rightarrow\;\frac25\,MR^2=20\quad\Longrightarrow\quad MR^2=\frac{5}{2}\times20=50,$

we get

$I_{\rm shell}=\frac23\,MR^2=\frac23\times50= \;33.3\;\mathrm{kg\cdot m^2}.$

Answer: Option C (33.3 kg·m²).

For the ring rolling down the inclined plane, the moment of inertia $(I)$,

$ I_{\text {ring }}=m r^2 $

For solid sphere, $I_{\text {sphere }}=\frac{2}{5} m r^2$ For solid cylinder, $I_{\text {cylinder }}=\frac{1}{2} m r^2$

Since, the solid sphere has the smallest moment of inertia, it will have the greatest translational kinetic energy when it reaches the bottom of the incline, thus solid sphere will reach the bottom of the inclined plane with maximum velocity.

$ \text { Given, } $

Coeffficient of linear expansion,

$ \alpha=20 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1} $

Temperature change, $\Delta T=50^{\circ} \mathrm{C}$

Initial angular speed $=\omega_0$

Using conservation of angular momentum,

$ \frac{2}{5} m r^2 \omega_0=\frac{2}{5} m[r(1+\alpha \Delta T)]^2 \omega^{\prime} $

$\therefore$ Radius of the sphere after the temperature increase $\left[r^{\prime}=r(1+\alpha \Delta T)]\right.$

$ \begin{aligned} \omega^{\prime} & =\frac{\omega_0}{(1+\alpha \Delta T)^2}=\omega_0[1-2 \alpha \Delta t] \\ & =\omega_0\left[1-2 \times 20 \times 10^{-5} \times 50\right] \\ & =\omega_0[1-0.02]=0.98 \omega_0 \end{aligned} $

The angular speed will be $0.98 \omega_0$.

Let:

$ m $ be the mass

$ r $ be the radius

$ v $ is the linear velocity

$ \omega $ is the angular velocity

The condition for rolling without slipping gives us $ v = r\omega $.

The total kinetic energy (KE) for any rolling object is the sum of its rotational and translational kinetic energies:

$ \mathrm{KE}_{\text{total}} = \mathrm{KE}_{\text{rotational}} + \mathrm{KE}_{\text{translational}} $

The formula for total kinetic energy is:

$ \mathrm{KE}_{\text{total}} = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2 $

where $ I $ is the moment of inertia.

For the ring:

$ I = mr^2 $

For the disc:

$ I = \frac{mr^2}{2} $

Now, using these values, we can calculate the kinetic energies:

Kinetic Energy of the Disc

$ \mathrm{KE}_{\text{disc}} = \frac{1}{2} \cdot \frac{mr^2}{2} \cdot \left(\frac{v^2}{r^2}\right) + \frac{1}{2} mv^2 = \frac{3}{4} mv^2 \quad \quad \quad \quad \quad \quad \quad \quad \left( \text{i} \right) $

Kinetic Energy of the Ring

$ \mathrm{KE}_{\text{ring}} = \frac{1}{2} \cdot mr^2 \cdot \left(\frac{v^2}{r^2}\right) + \frac{1}{2} mv^2 = mv^2 \quad \quad \quad \quad \left( \text{ii} \right) $

Ratio of Kinetic Energy

From equations (i) and (ii):

$ \frac{\mathrm{KE}_{\text{ring}}}{\mathrm{KE}_{\text{disc}}} = \frac{mv^2}{\frac{3}{4} mv^2} = \frac{4}{3} $

Therefore, the ratio of the kinetic energy of the ring to the disc is 1.33.

The total length of a rod is $L$. So, when the rod is bent at the middle, then each part have same length $\left(\frac{L}{2}\right)$ and mass $\left(\frac{M}{2}\right)$.

Moment of inertia of each part about axis passing through its one end.

$ \begin{aligned} &=\frac{1}{3}\left(\frac{M}{L}\right)\left(\frac{1}{2}\right)^2\\ &\text { Thus, Net moment of inertia }\\ &\begin{aligned} I & =\frac{1}{3}\left[\frac{M}{2}\right]\left[\frac{L}{2}\right]^2+\frac{1}{3}\left[\frac{M}{2}\right]\left[\frac{L}{2}\right]^2 \\ & =\frac{1}{3} \frac{M L^2}{8}+\frac{1}{3} \frac{M L^2}{8} \\ \Rightarrow I & =\frac{M L^2}{12} \end{aligned} \end{aligned} $

By the conservation of energy

Loss in potential energy of rod = Gain of rotational kinetic energy

$ \begin{aligned} & m g \times \frac{2 L}{2} \sin \alpha=\frac{1}{2} I \omega^2 \\ & m g \times \frac{2 L}{2} \sin \alpha=\frac{1}{2} \times \frac{m(2 L)^2}{3} \times \omega^2 \\ & \text { Here, } \omega=\sqrt{\frac{3 g \sin \alpha}{2 L}} \end{aligned} $

$ \text { According to work energy theorem, } $

Total energy at point $A=$ Total energy at point $B$

$ M g h=\frac{1}{2} M v^2+\frac{1}{2} I \omega^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

(Rotational energy)

For cylinder, $I=\frac{1}{2} M R^2$ and

Linear velocity, $v=R \omega \Rightarrow \omega=\frac{v}{R}$

Substitute it in Eq. (i)

$ \begin{aligned} & \quad M g h=\frac{1}{2} M v^2+\frac{1}{2} \times \frac{1}{2} \times M R^2 \times \frac{v^2}{R^2} \\ & \Rightarrow \quad g h=\frac{3}{4} v^2 \Rightarrow v=\sqrt{\frac{4}{3} g h} \end{aligned} $

$ \text { Let the axis of rotation be } B C \text {. } $

So, moment of inertia is given by,

$ \begin{aligned} & I=I_A+I_B+I_C \\ & =m_A R^2+m_B(0)+m_C(0) \end{aligned} $

As both masses lie on axis of rotation, separation distance is zero.

$ I=m_A R^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Using Pythagoras theorem,

$ \begin{aligned} l^2 & =\frac{l^2}{4}+R^2 \Rightarrow R^2=\frac{3}{4} l^2 \\ I & \left.=m_A \frac{3}{4} l^2 \quad \text { [From Eq. (i) }\right] \\ \Rightarrow \quad I & =\frac{3}{4} m l^2 \quad\left(\text { as } m_A=m\right) \end{aligned} $

Given:

Mass of solid cylinder, $ m_s = 3.2 \, \text{kg} $

Mass of hollow cylinder, $ m_h = 1.6 \, \text{kg} $

Acceleration of the solid cylinder, $ a_s = 4 \, \text{m/s}^2 $

Moment of Inertia

For a solid cylinder:

$ I_s = \frac{1}{2} m R^2 $

For a hollow cylinder:

$ I_h = m R^2 $

Equation of Motion for Rolling Objects

The equation for motion is given by:

$ m g \sin \theta = m a + I \frac{a}{R^2} $

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

Calculations

Solid Cylinder

For the solid cylinder:

$ g \sin \theta = 4 \left( 1 + \frac{m R^2}{2m R^2} \right) $

$ g \sin \theta = 6 $

Hollow Cylinder

For the hollow cylinder:

$ g \sin \theta = a_h \left( 1 + \frac{m R^2}{m R^2} \right) $

$ 6 = a_h (1 + 1) $

$ a_h = 3 \, \text{m/s}^2 $

Therefore, the acceleration of the hollow cylinder is $ 3 \, \text{m/s}^2 $.

Given:

Mass of the sphere, $ m = 50 \, \text{kg} $

Radius of the sphere, $ r = 20 \, \text{cm} = 0.2 \, \text{m} $

Angular velocity, $ \omega = 420 \times \frac{2\pi}{60} = 14\pi \, \text{rad/s} $

The moment of inertia $ I $ of a solid sphere rotating about its diameter is given by:

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

Calculating the moment of inertia:

$ \begin{align*} I &= \frac{2}{5} \times 50 \times (0.2)^2 \\ &= \frac{2}{5} \times 50 \times 0.04 \\ &= 20 \times 0.04 \\ &= 0.8 \, \text{kg} \cdot \text{m}^2 \end{align*} $

The angular momentum $ L $ of the rotating sphere is calculated as:

$ L = I \cdot \omega $

Substitute the known values:

$ \begin{align*} L &= 0.8 \times 14\pi \\ &= 0.8 \times 14 \times \frac{22}{7} \\ &= 44 \times 0.8 \\ &= 35.2 \, \text{kg} \cdot \text{m}^2/\text{s} \\ &= 35.2 \, \text{J} \cdot \text{s} \end{align*} $

Therefore, the angular momentum of the sphere is $ 35.2 \, \text{J} \cdot \text{s} $.

Given, $ \boldsymbol{R}_{A} < \boldsymbol{R}_{B} < \boldsymbol{R}_{C} $

Frictional force opposes the body from sliding

Here, $f=m g \quad \ldots \text { (i) } $

Normal reaction provided necessary centripetal force.

$N=m \omega^2 R \quad$ (where $R$ is radius)

$ f=\mu N=\mu m \omega^2 R \quad \ldots \text { (ii) } $

Substitute the value of $f$ in Eq. (i),

$ \begin{aligned} m g & =\mu m \omega^2 R \\ \omega & =\frac{\sqrt{g}}{\mu R} \quad \ldots \text { (iii) } \end{aligned} $

From here, $\omega \propto \frac{1}{R}$

$\therefore$ The correct sequence of angular velocity is

$ \omega_C<\omega_B<\omega_A $

To find the moment of inertia of a solid sphere with a mass of 20 kg and a diameter of 20 cm about a tangent to the sphere, we follow these steps:

First, find the moment of inertia of the sphere about its center of mass. The formula for the moment of inertia ($MI$) of a solid sphere about its center is given by:

$ MI = \frac{2}{5} M R^2 $

where:

$M$ is the mass of the sphere (20 kg),

$R$ is the radius of the sphere.

Given the diameter is 20 cm, the radius $R$ is:

$ R = \frac{20}{2} \, \text{cm} = 10 \, \text{cm} = 0.10 \, \text{m} $

Substituting the values into the formula:

$ MI = \frac{2}{5} \times 20 \times (0.10)^2 = 0.08 \, \text{kg} \, \text{m}^2 $

Next, we use the theorem of parallel axes. This theorem states that the moment of inertia of a body about any axis parallel to an axis through its center of mass is the sum of the moment of inertia about the center of mass and the product of the mass and the square of the distance between the two axes.

For a tangent to the sphere, the distance is $R$ (which is the radius), so:

$ \text{MI about tangent} = \frac{2}{5} M R^2 + M R^2 = \frac{7}{5} M R^2 $

Substitute the values:

$ =\frac{7}{5} \times 20 \times (0.10)^2 = 0.28 \, \text{kg} \, \text{m}^2 $

Therefore, the moment of inertia of the solid sphere about a tangent is $0.28 \, \text{kg} \, \text{m}^2$.

The problem discusses a solid cylinder rolling down an inclined plane without slipping. Here's a step-by-step breakdown to find the total kinetic energy of the cylinder:

Translation Kinetic Energy

The translation kinetic energy ($ K_{\text{trans}} $) is given by:

$ K_{\text{trans}} = \frac{1}{2} m v^2 $

Given that:

$ K_{\text{trans}} = 140 \, \text{J} $

We know the relationship between linear velocity ($ v $) and angular velocity ($ \omega $) for rolling without slipping is:

$ v = \omega R $

This gives:

$ 140 = \frac{1}{2} m (\omega R)^2 $

Rearranging, we find:

$ 280 = m R^2 \omega^2 \, \ldots \text{(i)} $

Rotational Kinetic Energy

The rotational kinetic energy ($ K_{\text{rot}} $) can be expressed as:

$ K_{\text{rot}} = \frac{1}{2} I \omega^2 $

For a solid cylinder, the moment of inertia ($ I $) is:

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

Substitute this into the expression for rotational kinetic energy:

$ K_{\text{rot}} = \frac{1}{2} \times \frac{m R^2}{2} \times \omega^2 $

$ K_{\text{rot}} = \frac{280}{4} \quad \text{[Using Eq. (i)]} $

$ K_{\text{rot}} = 70 \, \text{J} $

Total Kinetic Energy

The total kinetic energy of the cylinder is the sum of its translation and rotational kinetic energies:

$ K = K_{\text{trans}} + K_{\text{rot}} $

Thus:

$ K = 140 + 70 = 210 \, \text{J} $

Therefore, the total kinetic energy of the cylinder is 210 J.

Given:

Mass of both cylinders = $ M $

Radius of both cylinders = $ R $

The moment of inertia for a solid cylinder and a hollow cylinder about their axes are as follows:

For the solid cylinder:

$ I_1 = \frac{M R^2}{2} $

For the hollow cylinder:

$ I_2 = M R^2 $

From these formulas, it is clear that:

$ I_1 < I_2 $

A solid cylinder with radius $ R $ is at rest at a height $ h $ on an inclined plane. As it rolls down and reaches the ground, its velocity can be determined as follows:

Let $ v $ be the velocity of the solid cylinder when it reaches the ground.

By applying the conservation of energy principle:

Total energy at point $ A = $ Total energy at point $ B $

$ \begin{aligned} & \text{At point $ A $, the total energy is: } mgh + 0 \quad (\text{Potential Energy at height $ h $}) \\ & \text{At point $ B $, the total energy is: } \frac{1}{2} I \omega^2 + \frac{1}{2} mv^2 + 0 \quad (\text{Rotational and Kinetic Energy at ground level}) \end{aligned} $

Substituting the moment of inertia $ I $ for a solid cylinder $(I = \frac{1}{2} MR^2)$ and the angular velocity $(\omega = \frac{v}{R})$:

$ \begin{aligned} mgh &= \frac{1}{2} \left(\frac{1}{2} MR^2\right) \left(\frac{v}{R}\right)^2 + \frac{1}{2} mv^2 \\ gh &= \frac{v^2}{4} + \frac{v^2}{2} \\ gh &= \frac{3v^2}{4} \\ v^2 &= \frac{4gh}{3} \\ v &= \sqrt{\frac{4gh}{3}} \end{aligned} $

Let $m$ be the mass of the solid sphere of radius $R$, then moment of inertia about its diameter

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

$\begin{aligned} \text { Here, } m & =\text { volume } \times \text { density } \\ & =\frac{4}{3} \pi R^3 \rho \\ \because \quad I & =\frac{2}{5} \times \frac{4}{3} \pi R^3 \rho \cdot R^2 \\ I & =\frac{8}{15} \pi R^5 \rho \quad \text{... (i)} \end{aligned}$

where, $\rho$ is density of sphere.

When outer half of sphere is removed, then its new volume,

$\begin{aligned} V^{\prime} & =\frac{4}{3} \pi\left(\frac{R}{2}\right)^3 \\ & =\frac{4}{3} \pi \frac{R^3}{8}=\frac{\pi R^3}{6} \end{aligned}$

$\because$ New mass of remaining part of sphere,

$M^{\prime}=V^{\prime} \times \rho=\frac{\pi R^3 \rho}{6}$

$\because$ Moment of inertia of newly formed sphere,

$\begin{aligned} I^{\prime} & =\frac{2}{5} M^{\prime}\left(\frac{R}{2}\right)^2 \\ & =\frac{2}{5} \frac{\pi R^3 \rho}{6} \times \frac{R^2}{4}=\frac{\pi R^5 \rho}{60} \\ \Rightarrow \quad I^{\prime} & =\frac{1}{32}\left[\frac{8}{15} \pi R^5 \rho\right]=\frac{1}{32} \times I \quad \text{[From Eq. (i)]}\\ \Rightarrow \quad I^{\prime} & =\frac{I}{32} \end{aligned}$

According to question, a disc of $r=\frac{R}{3}$ is removed from the bigger disc of radius $R$.

The mass of reduced disc is $\frac{M}{9}$ because mass is directly proportional to the area for uniform disc.

Moment of inertia of total disc, $I_{\text {total }}=\frac{M R^2}{2}$

Moment of inertia of removed disc $I_{\text {removed }}=I_{\mathrm{CM}}+\frac{M}{9} \times\left(\frac{R}{3}\right)^2$

$\begin{aligned} I_{\text {removed }} & =\frac{M}{9} \frac{\left(\frac{R}{3}\right)^2}{2}+\frac{M R^2}{81} \\ & =\frac{M R^2}{81}\left(\frac{1}{2}+1\right)=\frac{M R^2}{81} \times \frac{3}{2}=\frac{M R^2}{54} \end{aligned}$

The moment of inertia of remaining disc $=$

$\begin{aligned} & I_{\text {total }}-I_{\text {removed }} \\ & =\frac{M R^2}{2}-\frac{M R^2}{54}=\frac{M R^2(27-1)}{54} \\ & =\frac{26}{54} M R^2=\frac{13}{27} M R^2 \end{aligned}$

Given, mass of solid sphere $=M$

Radius of solid sphere $=R$

Angular frequency, $\omega=600 \mathrm{rpm}=20 \pi \mathrm{~rad} / \mathrm{s}$

As we know that,

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

$\begin{aligned} & =\frac{1}{2} \times \frac{2}{5} M R^2(20 \pi)^2 \\ & =\frac{M R^2}{5}\left(400 \pi^2\right)=80 \pi^2 M R^2\end{aligned}$

We know the moment of inertia is a measure of how hard it is to spin an object. For wheel $A$, it is $5~\mathrm{kg}-\mathrm{m}^2$. For wheel $B$, it is $20~\mathrm{kg}-\mathrm{m}^2$.

Wheel $A$ is spinning at $10$ revolutions per second ($10~\mathrm{rps}$), and wheel $B$ is not moving at all at the start ($0~\mathrm{rps}$).

Let’s call the speed that both wheels spin at after they are connected $\omega$.

When the wheels are joined, the total rotational momentum (angular momentum) stays the same, because there is no friction or any loss.

The total rotational momentum before the clutch is connected is:
$I_1 \omega_1 + I_2 \omega_2$

After they are joined, the total moment of inertia is $I_1 + I_2$ and both wheels move at the new speed $\omega$:

$ I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega $

Now substitute the numbers:
$ 5 \times 10 + 20 \times 0 = (5+20) \omega $

$ 50 = 25\omega $

$ \omega = \frac{50}{25} = 2~\mathrm{rps} $

This means after the clutch is connected, both wheels will rotate together at $2~\mathrm{rps}$.

Moment of inertia of sphere and hollow cylinder are,

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

and $I_{cylinder}=MR^2$

Angle of inclination, $\theta=30\Upsilon$

Let, acceleration of body = a

Friction force = $f$

Mass of body = $m$

Radius of body = R

$mg\sin\theta-f=ma$ ..... (i)

Torque, $\tau=Rf=I\alpha$

$\Rightarrow f=I\alpha/R$ ..... (ii)

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

$m g \sin \theta-\frac{I \alpha}{R}=m a$

Angular acceleration, $\alpha=a / R$

$\begin{array}{ll} \Rightarrow & m g \sin \theta-\frac{I a}{R^2}=m a \\ \Rightarrow & m g \sin \theta=a\left(m+\frac{I}{R^2}\right) \\ \Rightarrow & a=\frac{m g \sin \theta}{m+\frac{I}{R^2}}=\frac{g \sin \theta}{1+\frac{I}{m R}} \end{array}$

$\because$ Distance covered by body, $s=\frac{1}{2} a t^2$

$\begin{aligned} & \because \quad s_{\text {sphere }}=s_{\text {cylinder }} \\ & \frac{1}{2} a_{\text {sphere }} \times t^2=\frac{1}{2} \times s_{\text {cylinder }} \times t^2 \\ & \Rightarrow \quad a_{\text {sphere }}=a_{\text {cylinder }} \\ & \Rightarrow \quad \frac{g \sin 30 \Upsilon}{1+\frac{\frac{2}{5} m R^2}{m R^2}}=\frac{g \sin \theta}{1+\frac{m R^2}{m R^2}} \\ & \Rightarrow \quad \frac{1 / 2}{7 / 5}=\frac{\sin \theta}{2} \Rightarrow \sin \theta=\frac{5}{7} \\ & \Rightarrow \quad \theta=\sin ^{-1}\left(\frac{5}{7}\right)=45.6 \Upsilon \approx 45 \Upsilon \\ & \end{aligned}$

Given,

Diameter of sphere = 2a

Radius of sphere = a

Mass of sphere = m

Side of square = b

$\because$ Moment of inertia of sphere, $I=\frac{2}{5} m a^2$

From parallel axis theorem, $I_p=I_{\mathrm{CM}}+m R^2$

So, net moment of inertia on sphere

$\begin{aligned} 1 & =2 \times\left(\frac{2}{5} m a^2+m b^2\right)+2 \times\left(\frac{2}{5} m a^2\right) \\ & =\frac{8 m a^2}{5}+2 m b^2 \end{aligned}$

Given, Energy, $E=684 \mathrm{~J}$

Initial angular frequency, $\omega_i=180 \mathrm{~rpm}=6 \pi$

Final angular frequency, $\omega_f=360 \mathrm{~rpm}=12 \pi$

Let, $I$ be the moment of inertia.

As we know that,

$\begin{aligned} E & =\frac{1}{2} I\left(\omega_f^2-\omega_i^2\right) \\ 684 & =\frac{1}{2} I\left[(12 \pi)^2-(6 \pi)^2\right] \\ \Rightarrow \quad I & =\frac{684 \times 2}{\left(144 \pi^2-36 \pi^2\right)}=\frac{684 \times 2}{1065.9}=1.28 \mathrm{~kg}-\mathrm{m}^2 \end{aligned}$

Given, mass of sphere $=m$

Spring constant $=k$

Angle of inclined $=\theta$

By using law of conservation of energy,

$\begin{aligned} \frac{1}{2} k x^2 & =m g x \sin \theta \\ \Rightarrow \quad x^2 & =\frac{2 m g x \sin \theta}{k} \\ x & =\frac{2 m g \sin \theta}{k} \end{aligned}$

Given, mass of girl = M

Radius of marry-go-round = R

Rotational inertia of girl = I

Mass of rock = m

Speed of rock with respect to ground = v

Let linear speed of girl be v.

By using law of conservation of angular momentum,

Initial angular momentum $\left(L_i\right)=$ Final angular momentum $L_f$

$\begin{array}{llrl} \Rightarrow & 0 =m v R-\left(I+M R^2\right) \frac{v^{\prime}}{R} \\\\ \Rightarrow & m v R =\left(I+M R^2\right) \frac{v^{\prime}}{R} \\\\ \Rightarrow & m v R^2 =\left(I+M R^2\right) v^{\prime} \\\\ & \therefore \quad v^{\prime}=\frac{m v R^2}{I+M R^2} \end{array}$

Given, mass $M$ and radius $R$ of wheels are same

As we know that,

Moment of inertia (I) for following bodies are

(i) $\operatorname{Ring}=M R^2$

(ii) Angular disc $=\frac{M}{2}\left(R^2-r^2\right)$

(iii) Solid disc $=M R^2 / 2$

(iv) Cylindrical disc $=M R^2 / 2$

Hence, moment of inertia of ring among all four is maximum.