Rotational Motion

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{K_{\mathrm{rot}}}{K_{\mathrm{trans}}}=\frac{\frac{1}{2} I \omega^2}{\frac{1}{2} m v^2} \\ & \quad=\frac{\frac{1}{2} \times \frac{2}{5} m r^2 \times\left(\frac{v}{r}\right)^2}{\frac{1}{2} m v^2}=\frac{\frac{1}{5}}{\frac{1}{2}} \\ & \quad=\frac{2}{5} \\ & \therefore K_{\mathrm{rot}}: K_{\text {trans }}=2: 5 \end{aligned} \end{aligned} $

Moment of inertia of thin rod about an axis perpendicular to its length and passing through one end

$ \begin{aligned} & I=\frac{1}{3} M L^2 \\ \text { But } I= & M K^2 \\ \therefore \quad & M K^2=\frac{1}{3} M L^2 \\ \Rightarrow \quad & K=\frac{L}{\sqrt{3}} \Rightarrow \frac{K}{L}=\frac{1}{\sqrt{3}} \\ \Rightarrow \quad & K: L=1: \sqrt{3} \end{aligned} $

Density $\rho$, the length of the wire is

$ L=\frac{m}{\rho} $

Since, the wire is bent into a circular ring, its length is equal to the circumference

$ L=2 \pi R $

Therefore, $R=\frac{L}{(2 \pi)}=\frac{\left(\frac{m}{\rho}\right)}{(2 \pi)}=\frac{m}{(2 \pi \rho)}$

The moment of inertia of a circular ring about its diameter is given by

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

Using the parallel axis theorem, the moment of inertia about an axis parallel to the diameter at a distance $R$ (the radius) from the centre is

$ \begin{aligned} I_T & =I_D+m R^2 \\ & =\frac{1}{2} m R^2+m R^2=\frac{3}{2} m R^2 \\ & =\frac{3}{2} m\left(\frac{m}{2 \pi \rho}\right)^2=\frac{3 m^3}{8 \pi^2 \rho^2} \end{aligned} $

The acceleration (a) of a body rolling down on an inclined plane without slipping

$ a=\frac{g \sin \theta}{1+\frac{I}{M R^2}} $

For solid sphere, $I=\frac{2}{5} M R^2$

$ \therefore \quad a_s=\frac{g \sin \theta}{1+\frac{\frac{2}{5}}{M R^2} M R^2} $

$ \begin{aligned} &\begin{aligned} a_s & =\frac{5}{7} g \sin \theta \\ \text { but } \quad a_s & =3 \mathrm{~m} / \mathrm{s}^2 \\ \therefore \quad 3 & =\frac{5}{7} g \sin \theta \Rightarrow \sin \theta=\frac{21}{5 \mathrm{~g}} \end{aligned}\\ &\text { Thus, for disc, } I=\frac{1}{2} M R^2\\ &\begin{aligned} \therefore \quad a_d & =\frac{g \sin \theta}{\frac{1}{2} M R^2}=\frac{2}{3} g \sin \theta \\ & 1+\frac{M R^2}{M} \\ & =\frac{2}{3} g \times \frac{21}{5 g}=\frac{14}{5}=2.8 \mathrm{~m} / \mathrm{s}^2 \end{aligned} \end{aligned} $

Step 1: Understanding the Setup

We have a uniform meter scale that weighs 100 g. Each end of the scale has a plate that weighs 200 g. The scale is balanced on a support at the 45 cm mark. We put a 300 g weight on the plate at the 0 cm end and put vegetables (unknown weight $x$) on the plate at the 100 cm end.

Step 2: Calculating Clockwise Torque

The things making a clockwise turning effect (torque) around the pivot at 45 cm are:

• The 200 g plate at 100 cm, which is 55 cm from the pivot ($100 - 45 = 55$),
• The vegetable weight ($x$) on this plate, also 55 cm from pivot,
• Part of the scale's own mass to the right of the pivot (100 g is the total mass, half of it is on each side, but for 5 cm past pivot, that's $5$ g at 5 cm, but original calculation uses $100 \times 5$). Here we use $100$ g at a distance of $5$ cm ($50 - 45 = 5$ cm, but the calculation uses 5 cm for the rest mass).

Step 3: Calculating Anticlockwise Torque

The things making an anticlockwise torque are:

• The 200 g plate at 0 cm, 45 cm away from pivot,
• The 300 g weight at 0 cm, 45 cm away from pivot.

Step 4: Setting Torques Equal (Balance)

For balance, clockwise torque = anticlockwise torque.

$ \begin{aligned} 11,500 + 55x &= 22,500 \\ 55x &= 22,500 - 11,500 = 11,000 \\ x &= \frac{11,000}{55} = 200~\text{g} \\ \end{aligned} $

So, the actual mass of the vegetables is $200$ g.

Step 5: Calculate the Error

The balance shows $300$ g (apparent weight), but the real weight is $200$ g (actual weight).

Error $= 300 - 200 = 100~\text{g}$

Moment of inertia of ring about a tangential axis in its plane

$ I=I_{\mathrm{CM}}+M R^2=\frac{M R^2}{2}+M R^2 $

[The moment of inertia of a ring about its diameter $I_{\mathrm{CM}}=\frac{M R^2}{2}$ ]

$ I=\frac{3}{2} M R^2 $

If $K_R$ be the radius of gyration, then

$ M K_R^2=\frac{3}{2} M R^2 \Rightarrow K_R=\sqrt{\frac{3}{2}} R $

Moment of inertia of circular disc about tangential axis in its plane

$ \begin{aligned} I & =I_{C M}+M R^2 \\ & =\frac{M R^2}{4}+M R^2=\frac{5}{4} M R^2 \end{aligned} $

[The moment of inertia of a disc about its diameter is $I_{\mathrm{CM}}=\frac{1}{4} M R^2$ ]

$ \begin{aligned} &\text { If } K_C \text { be the radius of gyration, then }\\ &\begin{aligned} M K_C^2 & =\frac{5}{4} M R^2 \\ K_C & =\sqrt{\frac{5}{4}} R \\ \text { Since, } \frac{K_R}{K_C} & =\frac{\sqrt{12}}{\sqrt{K}} \\ \frac{\sqrt{\frac{3}{2}} R}{\sqrt{\frac{5}{4}} R} & =\frac{\sqrt{12}}{\sqrt{K}} \Rightarrow \frac{3}{2} \times \frac{4}{5}=\frac{12}{K} \\ \Rightarrow \quad K & =10 \end{aligned} \end{aligned} $

Step 1: Find the moment of inertia of the disc.

The moment of inertia $(I)$ for a disc about its center is $I = \frac{1}{2} M R^2$

Here, the mass $M = \frac{10}{\pi^2}\ \mathrm{kg}$, and the radius $R = 2\ \mathrm{m}$.

So, $I = \frac{1}{2} \times \frac{10}{\pi^2} \times (2)^2 = \frac{1}{2} \times \frac{10}{\pi^2} \times 4 = \frac{20}{\pi^2}\ \mathrm{kg\,m}^2$

Step 2: Change revolutions per minute (rev/min) to angular speed (rad/s).

The first speed is $90\ \mathrm{rev}/\mathrm{min}$. To convert to radians per second:

$w_1 = 90 \times \frac{2\pi}{60} = 3\pi\ \mathrm{rad/s}$

The second speed is $120\ \mathrm{rev}/\mathrm{min}$: $w_2 = 120 \times \frac{2\pi}{60} = 4\pi\ \mathrm{rad/s}$

Step 3: Use the rotational kinetic energy formula to find work done.

The work done to increase speed equals the change in rotational kinetic energy: $\text{Work Done} = \Delta KE = \frac{1}{2}I(w_2^2 - w_1^2)$

Plug in the values: $\text{Work} = \frac{1}{2} \times \frac{20}{\pi^2} \left[(4\pi)^2 - (3\pi)^2\right]$

Calculate:

$(4\pi)^2 = 16\pi^2$

$(3\pi)^2 = 9\pi^2$

So, $16\pi^2 - 9\pi^2 = 7\pi^2$

Therefore, $\text{Work} = \frac{1}{2} \times \frac{20}{\pi^2} \times 7\pi^2 = 10 \times 7 = 70\ \mathrm{J}$

When ice from the polar regions melts and flows towards the equator, the mass distribution of the Earth changes. This shift in mass causes in increases in the Earth's moment of inertia. According to the principle of conservation of angular momentum, if the angular momentum ( $L=I \omega$ ) remains constant and the moment of inertia ( $I$ ) increases, the angular velocity $(\omega)$ must decrease. A decrease in angular velocity means the Earth rotates more slowly, leading to an increase in the duration of a day.

Moment of inertia of ring about its center:

$ I_{\mathrm{CM}}=M R^2 $

Using the parallel axis theorem:

The parallel axis theorem says: If you know the moment of inertia of an object about its center ($I_{\mathrm{CM}}$), then the moment of inertia about a parallel axis a distance $R$ away is:

$ I = I_{\mathrm{CM}} + M R^2 $

For a thin circular ring, this axis through the edge is a distance $R$ from the center, so:

$ I = M R^2 + M R^2 = 2 M R^2 $

So, $ M R^2 = \frac{I}{2} \qquad \cdots (i) $

Finding the moment of inertia about the diameter:

The moment of inertia of a ring about its diameter is:

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

Substituting from equation (i):

Replace $M R^2$ with $\frac{I}{2}$: $ I_d = \frac{\frac{I}{2}}{2} = \frac{I}{4} $

As we know,

Moment of inertia about axis of cylinder,

$ I_{\text {long }}=\frac{1}{2} M R^2 $

Moment of inertia about an axis through centre and perpendicular to length.

$ I_{\text {transverse }}=\frac{1}{12} M L^2+\frac{1}{4} M R^2 $

By using, given ratio,

$ \begin{aligned} & \frac{1}{2} M R^2=\frac{1}{n}\left(\frac{1}{12} M L^2+\frac{1}{4} M R^2\right) \\ & \left(\frac{2 n-1}{4}\right) R^2=\frac{1}{12} L^2 \\ & 3(2 n-1)=\frac{L^2}{R^2} \Rightarrow \frac{L}{R}=\sqrt{3(2 n-1)} \end{aligned} $

Let moment of inertia of body $ =I $

The initial kinetic energy $ =\frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2} $

Final potential energy $ =m g h $

Given, $ h=\frac{v^{2}}{g} $

After putting value, potential energy

$ =m \cdot g \cdot \frac{v^{2}}{g}=m v^{2} $

When a body roll up an inclined plane of height $ h $, its kinetic energy converted into potential energy.

So, $ \mathrm{KE}=\mathrm{PE} $

$ \begin{array}{l} \frac{1}{2} I \omega^{2}+\frac{1}{2} m v^{2}=m v^{2} \\\\ \frac{1}{2} I \omega^{2}=\frac{1}{2} m v^{2} \\\\ I=\frac{m v^{2}}{\omega^{2}}=\frac{m(r \omega)^{2}}{\omega^{2}} \\\\ I=m r^{2} \end{array} $

So, body is a ring.

Time taken by body to roll down from a slope without slipping is given by,

$ t=\sqrt{\frac{2 s}{g \sin \theta \beta}} $

where $ \beta=\frac{1}{1+\frac{K^{2}}{R^{2}}} $

From Eq. (i)

$ t \propto \frac{1}{\beta} $

$ t_{1}= $ time to reach bottom (by hollow cylinder)

$ t_{2}= $ time to reach bottom (by solid cylinder)

$ \frac{t_{1}}{t_{2}}=\sqrt{\frac{\beta_{2}}{\beta_{1}}} $

$ \beta_{1} $, for hollow cylinder

$ \beta_{1}=\frac{1}{1+\frac{K^{2}}{R^{2}}}=\frac{1}{1+1}=\frac{1}{2} \quad\left[\text { as } \frac{K^{2}}{R^{2}}=1\right] $

For solid cylinder

$ \beta_{2}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3} $

Substitute all value in Eq. (iv)

$ \begin{aligned} \frac{2}{t_{2}} & =\sqrt{\frac{\frac{2}{3}}{\frac{1}{2}}} \Rightarrow \frac{2}{t_{2}}=\sqrt{\frac{4}{3}} \\\\ t_{2} & =2 \times \sqrt{\frac{3}{4}} \\\\ t_{2} & =1.732 \mathrm{~s} \end{aligned} $

Disc and sphere has same radius and mass.

Now, moment of inertia of sphere at centre (I)

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

Using parallel axis theorem, moment of inertia at point $ X $ of sphere

$ \begin{aligned} I_{X S} & =I+M R^{2} \\\\ & =\frac{2}{5} M R^{2}+M R^{2}=\frac{7 M R^{2}}{5} \end{aligned} $

Moment of inertia of disc at centre perpendicularly

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

Using parallel axis theorem, moment of inertia at point $ X $ of disc.

$ I_{\chi D}=I+M R^{2}=\frac{3 M R^{2}}{2} $

Total moment of inertia at point $ X $

$ \begin{aligned} I_{X S}+I_{X D} & =\frac{3 M R^{2}}{2}+\frac{7 M R^{2}}{5} \\\\ & =\frac{29 M R^{2}}{10} \end{aligned} $

Given, Mass of wire, $=m$

linear mass density $=\rho$

Let length of wire be $l$ and radius ber

$ \rho=\frac{m}{l} $

or

$ m=\rho l \quad \ldots \text { (i) } $

also,

$ \begin{aligned} 2 \pi r & =1 \\\\ \Rightarrow r & =\frac{1}{2 \pi} \quad \ldots \text { (ii) } \end{aligned} $

$\therefore$ The moment of inertia around a circular loop around COM, will be

$ \begin{aligned} I & =\frac{1}{2} m r^2 \\\\ I & \left.=\frac{1}{2}(\rho l)\left(\frac{l}{2 \pi}\right)^2 \text { \{using Eqs. (i) and (ii)\}}\right\} \\\\ & =\frac{1}{2} \rho \frac{l^3}{4 \pi^2} \\\\ & =\frac{1}{8 \pi^2} \rho l^3 \end{aligned} $

$ \begin{aligned} &\text { From parallel axis theorem, }\\\\ &\begin{aligned} I_{\text {diameter }} & =I_{\mathrm{COM}}+M R^2 \\\\ & =\frac{l^3 \rho}{8 \pi^2}+\frac{l^3 \rho}{4 \pi^2} \\\\ & =\frac{3 l^3 \rho}{8 \pi^2} \\\\ \text { or } \quad I_{\text {diameter }} & =\frac{3}{8} \frac{m^3}{\pi^2 \rho^2} \end{aligned} \end{aligned} $

Given, height of plane $=28 \mathrm{~m}$

angle of inclination, $\theta=30^{\circ}$

By conservation of energy,

Potential energy lost by the solid sphere in rolling down the inclined plane

$=$ kinetic energy gain by the sphere

$ \begin{aligned} m g h & =\frac{1}{2} m v^2+\frac{1}{2} I \omega^2 \\\\ & =\frac{1}{2} m v^2+\frac{1}{5} m v^2 \\\\ m g h & =\frac{7}{10} m v^2 \\\\ v^2 & =\frac{10}{7} g h=\sqrt{\frac{10}{7} \times 10 \times 28} \\\\ v & =20 \mathrm{~m} / \mathrm{s} \end{aligned} $

The velocity of the sphere, when it reaches the bottom of the plane is $20 \mathrm{~m} / \mathrm{s}$.

Given,

Initial angular momentum, $L_i=A_0$

Final angular momentum, $L_f=4 A_0$

Time taken $=4$ second.

Change in angular momentum

$ \Delta L=L_f-L_i=4 A_0-A_0=3 A_0 $

Magnitude of torque is given by

$ \tau=\frac{\Delta L}{\Delta T}=\frac{3 A_0}{4} $

Hence, magnitude is $\frac{3 A_0}{4}$.

Given,

Initial angular momentum $=L$

Initial frequency (angular) $=\omega$

Initial kinetic energy $=\mathrm{KE}$

Final angular frequency $=2 \omega$

Final kinetic energy $=\frac{\mathrm{KE}}{2}$

Rotational kinetic energy $\mathrm{KE}=\frac{1}{2} I \omega^2$

$I=$ moment of inertia

Angular momentum

$ \begin{aligned} L & =I \omega \\ L & =\frac{2 \mathrm{KE}}{\omega^2} \cdot \omega \\ \Rightarrow \quad L & =\frac{2 \mathrm{KE}}{\omega} \end{aligned} $

Now, new kinetic energy, $\mathrm{KE}_N=2 \mathrm{KE}$,

$ \omega_N=2 \omega $

New angular momentum $L_N$

$ L_N=\frac{2 \mathrm{KE}_N}{\omega_{\mathrm{N}}}=\frac{2\left(\frac{\mathrm{KE}}{2}\right)}{2 \omega} \Rightarrow L_N=\frac{L}{4} $

Hence, the new angular momentum will be $L / 4$.

Given,

Ratio of radii of two sphere $\left(\frac{r_1}{r_2}\right)=\frac{2}{3}$

Mass of both sphere are same.

Moment of inertia is given for solid sphere by formula,

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

For sphere $1, R=r_1$

$ (M)_1=\frac{2}{5} M_1 r_1^2 $

For sphere $2, R=r_2$

$ (M I)_2=\frac{2}{5} M_2 r_2^2 $

Ratio of moment of inertia is

$ \begin{aligned} \frac{(M)_1}{(M)_2} & =\frac{\frac{2}{5} M_1 r_1^2}{\frac{2}{5} M_2 r_2^2} \\ & =\left(\frac{r_1}{r_2}\right)^2\left[\text { As } M_1=M_2 \text { same mass }\right] \end{aligned} $

Put the value of $r_1 / r_2$

$ \frac{\left(M I_1\right.}{(M I)_2}=\frac{4}{9} $

Hence, the ratio is $4: 9$.

$ \begin{aligned} &\text { Perpendicular distance }(O A)\\ &=\frac{a}{\sqrt{1+1}}=\frac{9}{\sqrt{2}} \mathrm{~m} \end{aligned} $

$ \begin{aligned} & \text { Angular momentum }=m v r \\ & =\frac{m v a}{\sqrt{2}} \mathrm{~kg}-\mathrm{m}^2 / \mathrm{s} \end{aligned} $

Given,

$ \begin{aligned} (K E)_R & =50 \% \text { of }(K E)_T \\ & =\frac{1}{2} \times(K E)_T \\ \Rightarrow \frac{1}{2} I \omega^2 & =\frac{1}{2}\left(\frac{1}{2} m v^2+\frac{1}{2} I \omega^2\right) \\ I \omega^2 & =m v^2 \end{aligned} $

Here, body moves without slipping i.e. $r=\omega R$

$\Rightarrow \quad I\left(\frac{v^2}{R^2}\right)=m v^2$

$ \Rightarrow \quad I=m R^2 $

It is MOI of thin circular ring.

To determine the total kinetic energy of the moon, we consider both its revolution around the Earth and its rotation about its own axis. Here are the given parameters:

The moon revolves around the Earth in an orbit with radius $ R $.

The time period of this revolution is $ T $.

The radius of the moon is $ r $.

The mass of the moon is $ M $.

The total kinetic energy can be divided into two components:

Kinetic energy due to revolution around the Earth:

The formula for rotational kinetic energy is $ \frac{1}{2} I \omega^2 $.

For revolution around the Earth, the moment of inertia is $ I = M R^2 $.

The angular velocity $ \omega $ is $ \frac{2\pi}{T} $.

Substituting these values, we find:

$ \text{Kinetic energy (revolution)} = \frac{1}{2} (M R^2) \left(\frac{2\pi}{T}\right)^2 = \frac{2 M \pi^2 R^2}{T^2} $

Kinetic energy due to rotation about its own axis:

For rotation on its own axis, the moment of inertia is $ I = \frac{2}{5} M r^2 $.

Again, the angular velocity $ \omega $ is $ \frac{2\pi}{T} $.

Substituting these values, we find:

$ \text{Kinetic energy (rotation)} = \frac{1}{2} \left(\frac{2}{5} M r^2\right) \left(\frac{2\pi}{T}\right)^2 = \frac{4 M r^2 \pi^2}{5 T^2} $

Finally, to find the total kinetic energy of the moon, we sum both components:

$ \text{Total kinetic energy} = \frac{2 M \pi^2 R^2}{T^2} + \frac{4 M r^2 \pi^2}{5 T^2} $

The spinning of a Diwali cracker, known as a 'ground chakkar,' demonstrates the principle of conservation of angular momentum.

The formula for angular momentum is given by:

$ I \omega = \text{constant} $

which can be expressed as:

$ I_{\text{Initial}} \omega_{\text{Initial}} = I_{\text{Final}} \omega_{\text{Final}} $

Where:

$I$ is the moment of inertia and $\omega$ is the angular velocity.

Initially, let's denote:

Radius of the chakkar as $R$.

Mass of the chakkar as $M$.

Assume the surface on which the chakkar rotates is smooth.

After some time:

Final radius becomes $R'$.

Final mass is $M'$.

Initial angular speed is $\omega_i$.

Final angular speed is $\omega_f$.

The formula for the moment of inertia for the chakkar is:

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

Substituting this into the conservation equation:

$ M_i \left(\frac{R_i^2}{2}\right) \omega_i = M_f \left(\frac{R_f^2}{2}\right) \omega_f $

Given that:

$M_i > M_f$

$R_i > R_f$

It follows that:

$ \omega_f > \omega_i $

This explains why the chakkar begins with a slower angular speed and then accelerates as it burns, losing mass and thus increasing its angular speed.

To determine how the period of the Earth's rotation would change if its radius became $x$ times its current value, consider the following:

Let the original radius of Earth be $ R $ and its mass be $ M $.

Increase the radius: If the radius is increased to $ xR $, we need to apply the conservation of angular momentum because there are no external torques acting on the Earth:

$ I_1 \omega_1 = I_2 \omega_2 $

Moment of Inertia: For a sphere, the moment of inertia $ I $ is given by:

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

where $ M $ is the Earth's mass and $ R $ is its radius. The angular frequency $ \omega $ is related to the period $ T $ by $ \omega = \frac{2\pi}{T} $.

Substitute into the Conservation Equation: Initially, the moment of inertia and angular frequency are:

$ I_1 = \frac{2}{5} M R^2, \quad \omega_1 = \frac{2\pi}{24 \ \text{hours}} $

New Moment of Inertia and Angular Frequency: With the expanded radius:

$ I_2 = \frac{2}{5} M (xR)^2 = \frac{2}{5} M x^2 R^2 $

New Period Calculation: Substitute these into the conservation equation:

$ \left(\frac{2}{5} M R^2 \right) \frac{2\pi}{24 \ \text{hours}} = \left(\frac{2}{5} M x^2 R^2 \right) \frac{2\pi}{T} $

Simplifying this equation for $ T $ gives:

$ T = 24 x^2 $

Thus, if the radius of the Earth becomes $ x $ times its original value, the new period of rotation will be $ 24 x^2 $ hours.

For a rolling spherical ball, total energy at bottom

$ \begin{aligned} E_K= & \text { KE of rolling } \\ = & (\text { KE of rotation about CM }) \\ & +(\text { KE of translation of CM }) \\ = & \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 \end{aligned} $

(For solid sphere, $I=\frac{2}{5} m r^2$ )

$ \begin{aligned} & =\frac{1}{5} m v^2+\frac{1}{2} m v^2 \\ \therefore \quad E_K & =\frac{7}{10} m v^2 \end{aligned} $

Total energy at topmost point, where ball comes to rest momentarily is

$ U=m g h $

As there is no loss of energy,

$ \begin{aligned} & \quad E_K=U \\ & \text { or } \quad \frac{7}{10} m v^2=m g h \\ & \Rightarrow \quad h=\frac{7 v^2}{10 g}=\frac{7 \times 4^2}{10 \times 10} \\ & \Rightarrow \quad h=1.12 \mathrm{~m} \\ & \text { Also, } \quad \frac{h}{s}=\sin \theta \\ & \quad \begin{aligned}\Rightarrow s & =\frac{h}{\sin \theta}=h \operatorname{cosec} \theta \\ & =h \operatorname{cosec} 30^{\circ} \\ & =1.12 \times 2 \\ & =2.24 \mathrm{~m} \\ & =224 \mathrm{~cm} \approx 225 \mathrm{~cm} \end{aligned} \end{aligned} $

The given situation is shown below.

Mass of each coin, $m=2 \mathrm{~g}$

Mass of 4 coin $=4 \times m$

$ M^{\prime}=4 \times 2 \Rightarrow M^{\prime}=8 \mathrm{~g} $

Let mass of the metre stick be $m^{\prime}$.

The centre of mass lies at its mid-point i.e at the mark of 50 cm .

When the coins are placed 10 cm away from the end $B$, the centre of mass gets shifted by 4 cm from point $G$ towards the coins side.

Now, using the principle of moment about point $P$, we get

$ \begin{aligned} 8 \mathrm{~g}(46-10)-m g(50-46)=0 \\ \Rightarrow 8 \times 36-m \times 4=0 \\ m=\frac{8 \times 36}{4}= & 72 \mathrm{~g} \end{aligned} $

Radius of wheel, $R=0.5 \mathrm{~m}$

Moment of inertia of wheel,

$ I=10 \mathrm{~kg} \cdot \mathrm{~m}^2 $

Initial angular velocity $=2 \pi \times \frac{70}{60} \mathrm{rad} / \mathrm{s}$

$ \Rightarrow \quad \omega_0=\frac{14 \pi}{6} \mathrm{rad} / \mathrm{s} $

$t=5 \mathrm{~s}$, final angular velocity, $\omega=0$

Using equation of rotational motion,

$ \omega=\omega_0+\alpha t $

$ \Rightarrow \alpha=\frac{\omega-\omega_0}{t}=\frac{0-\frac{14 \pi}{6}}{5}=\frac{14 \pi}{30} \mathrm{rad} / \mathrm{s}^2 $

∴ Torque on the wheel,

$ \begin{aligned} \tau & =I \alpha=10 \times \frac{14}{30} \pi=\frac{14 \pi}{3} \mathrm{~N}-\mathrm{m} \\ \Rightarrow \quad \tau & =\frac{14 \pi}{3} \end{aligned} $

But torque on the wheel to stop it,

$ \tau^{\prime}=F \times R=88 \times 0.5=44 \mathrm{~N}-\mathrm{m} $

Coefficient of kinetic friction between wheel and wet cloth,

$ \mu=\frac{\tau}{\tau^{\prime}}=\frac{\frac{14 \pi}{3}}{44}=0.33 $

When the cylinder reaches the ground, then its gravitational potential energy will get converted to translational as well as rotational kinetic energy.

$ \therefore \quad U=K_T+K_R $

where, $K_T=$ Translational kinetic energy

$K_R=$ Rotational kinetic energy

$ \begin{aligned} & \therefore \quad m g h=\frac{1}{2} m v^2+\frac{1}{2} l \omega^2 \\ & m g h=\frac{1}{2} m v^2+\frac{1}{2} \times \frac{m R^2}{2} \times \omega^2 \,\,\,\,\,\,\,\left[\because I_{\mathrm{CM}}=\frac{m R^2}{2}\right]\\ & \Rightarrow \quad g h=\frac{v^2}{2}+\frac{R^2}{4} \frac{v^2}{R^2} \end{aligned} $

( $\because$ In pure Rolling $v=R \omega$ )

$ \begin{array}{ll} \Rightarrow & 10 \times 30=\frac{3 v^2}{4} \Rightarrow v^2=400 \\ \Rightarrow & V=20 \mathrm{~m} / \mathrm{s} \end{array} $

The given situation is as shown below,

Here, $O$ is the point where centre of mass of whole strip exist.

The point $O^{\prime}$ is 100 cm or 1 m away from one end. So, its distance from centre of mass or from point $O$ is 1.5 m .

∴ Moment of inertia about an axis perpendicular to strip at point $O^{\prime}$ can be calculated using parallel axis theorem as

$ \begin{aligned} I^{\prime} & =I_{C M}+M R^2=\frac{M L^2}{12}+M R^2 \\ & =\frac{1 \times 5 \times 5}{12}+1 \times 1.5 \times 1.5 \\ & =2.08+2.25=4.33 \mathrm{~kg} \mathrm{~m}^2 \end{aligned} $

According to the question, given situation is shown below

In $\triangle A B C$,

$ \sin \theta=\frac{B C}{A C} \Rightarrow B C=A C \sin \theta=l \sin \theta $

From law of conservation of mechanical energy,

$ \begin{aligned} m g(l \sin \theta)= & \frac{1}{2} I \omega^2+\frac{1}{2} m v^2 \\ m g(l \sin \theta)= & \frac{1}{2}\left(\frac{1}{2} m r^2\right)\left(\frac{v}{r}\right)^2+\frac{1}{2} m v^2 \\ & {\left[\because I=\frac{1}{2} m r^2 \text { and } \omega=\frac{v}{r}\right] } \end{aligned} $

$ \begin{aligned} m g l \sin \theta & =\left(\frac{1}{4}+\frac{1}{2}\right) m v^2 \\ g l \sin \theta & =\frac{3}{4} v^2 \Rightarrow \quad v^2=\frac{4 g l \sin \theta}{3} \\ \Rightarrow \quad v & =\sqrt{\frac{4 g l \sin \theta}{3}} \\ & =\sqrt{\frac{4 \times 10 \times 0.6 \times \sin 30^{\circ}}{3}} \end{aligned} $

$ \begin{aligned} & \quad[\because l=60 \mathrm{~cm}=0.6 \mathrm{~m}] \\ &=\sqrt{\frac{12}{3}}=\sqrt{4}=2 \mathrm{~m} / \mathrm{s} \end{aligned} $

$ \begin{aligned} &\text { Mass per unit length, }\\ &\frac{d M}{d x}=\lambda|x| \Rightarrow d M=\lambda|x| d x \end{aligned} $

$ \begin{aligned} &\text { Moment of inertia about axis } A A\\ &I_{A A}=\int x^2 d M \end{aligned} $

$ \begin{aligned} &\text { Put the value of } d M\\ &\begin{aligned} I_{A A}= & \int_{-L / 2}^{+L / 2} x^2 \times \lambda|x| d x \\ = & \int_{-L / 2}^{+L / 2} x^2 \times 16|x| d x \\ & \quad\left[\because \lambda=16 \mathrm{~kg} / \mathrm{m}^2 \text { (given) }\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =16 \int_{-L / 2}^{+L / 2} x^2|x| d x \\ & =16\left[\int_{-L / 2}^0 x^2(-x) d x+\int_0^{+L / 2} x^2(+x) d x\right] \\ & \quad\left[\because|x|=\left\{\begin{array}{ll} -x & ; \\ +x & \text { if } x<0 \\ +x & \text { if } x \geq 0 \end{array}\right]\right. \end{aligned} $

$ \begin{aligned} &\text { So, }\\ &\begin{aligned} I_{A A} & =16\left[\int_{-L / 2}^0\left(-x^3\right) d x+\int_0^{+L / 2}\left(x^3\right) d x\right] \\ & =16\left[-\left[\frac{x^4}{4}\right]_{-L / 2}^0+\left[\frac{x^4}{4}\right]_0^{+L / 2}\right] \\ & =4\left[-\left[x^4\right]_{-L / 2}^0+\left[x^4\right]_0^{+L / 2}\right] \\ & =4\left[\begin{array}{l} \left.-\left\{(0)^4-\left(-\frac{L}{2}\right)^4\right\}\right] \\ \left.+\left\{\left(+\frac{L}{2}\right)^4-(0)^4\right\}\right] \end{array}\right. \end{aligned} \end{aligned} $

$ \begin{aligned} & =4\left[\left\{0-\frac{L^4}{16}\right\}+\left\{\frac{L^4}{16}-0\right\}\right] \\ & =4\left[\frac{L^4}{16}+\frac{L^4}{16}\right]=4\left[\frac{L^4}{8}\right]=\frac{L^4}{2} \end{aligned} $

Now by the theorem of parallel axes, the moment of inertia about axis $B B$,

$ \begin{aligned} I_{B B} & =I_{\mathrm{COM}}+\int\left(\frac{L}{2}\right)^2 d M \\ & =I_{\mathrm{COM}}+\frac{L^2}{4} \int d M \end{aligned} $

$ \begin{aligned} &\text { Put the value of } d M \text { and } I_{\text {COM }}\\ &\begin{aligned} I_{B B} & =I_{A A}+\frac{L^2}{4} \int_0^{L / 2} \lambda|x| d x \\ & =\frac{L^4}{2}+\frac{L^2}{4} \lambda \int_0^{L / 2}|x| d x \\ & =\frac{L^4}{2}+\frac{L^2}{4} \times 16 \int_0^{L / 2}(+x) d x \\ & \left.\quad [\because \lambda=16 \mathrm{~kg} / \mathrm{m}^2 \text { (given) }\right] \\ & =\frac{L^4}{2}+4 L^2\left[\frac{x^2}{2}\right]_0^{L / 2} \\ & =\frac{L^4}{2}+2 L^2\left[x^2\right]_0^{L / 2} \\ & =\frac{L^4}{2}+2 L^2\left[\left(\frac{L}{2}\right)^2-(0)^2\right] \\ & =\frac{L^4}{2}+2 L^2\left[\frac{L^2}{4}-0\right] \\ & =\frac{L^4}{2}+\frac{L^4}{2}=L^4=1^4 \\ & =1 \mathrm{~kg}-\mathrm{m}^2 \end{aligned} \end{aligned} $

Given, mass of solid cylinder, $M=10 \mathrm{~kg}$

Radius, $R=40 \mathrm{~cm}=0.4 \mathrm{~m}$

$ L=9 R=9 \times 0.4=3.6 \mathrm{~m} $

Moment of inertia of the solid cylinder about the line passing through its centre and perpendicular to its axis is given as

$ \begin{aligned} I_{\mathrm{COM}} & =M\left[\frac{L^2}{12}+\frac{R^2}{4}\right] \\ & =10\left[\frac{(3.6)^2}{12}+\frac{(0.4)^2}{4}\right] \\ & =10[1.08+0.04]=10 \times 1.12 \\ \Rightarrow \quad I_{\mathrm{COM}} & =11.2 \mathrm{~kg}-\mathrm{m}^2 \end{aligned} $

According to parallel axis theorem, moment of inertia of cylinder about a line passing through its edge and perpendicular to its axis is given as

$ \begin{aligned} I^{\prime} & =I_{\mathrm{COM}}+M\left(\frac{L}{2}\right)^2 \\ & =11.2+10\left(\frac{3.6}{2}\right)^2 \\ & =11.2+10(3.24) \\ & =11.2+324=43.6 \mathrm{~kg}-\mathrm{m}^2 \end{aligned} $

$ \text { We take origin as fixed point. } $

Now, angular momentum about origin $O=$ angular momentum due to linear motion of centre of mass + angular momentum due to rotation of body about its' centre of mass

$ \begin{aligned} & \\ & \begin{aligned}& \\ \therefore L_O=L_{\text {linear motion }} & +L_{\text {rotational motion }} \\ & =m v R+I \omega \\ & =m R^2 \omega+I \omega \quad[\therefore v=R \omega] \\ & =\left(m R^2+I\right) \omega \end{aligned} \end{aligned} $

For solid sphere, $I=\frac{2}{5} m R^2$ and for solid cylinder, $I=\frac{1}{2} m R^2$

$ \begin{aligned} \text { So, }\left(L_{\text {sphere }}\right)_{\text {about ' } O^{\prime}} & =\left(m R^2+\frac{2}{5} m R^2\right) \omega \\ & =\frac{7}{5} m R^2 \omega \\ \text { and }\left(L_{\text {sphere }}\right)_{\text {about ' } O^{\prime}} & =\left(m R^2+\frac{1}{2} m R^2\right) \omega \\ & =\frac{3}{2} m R^2 \omega \end{aligned} $

$ \begin{aligned} & \text { So, ratio of } \frac{L_{\text {sphere }}}{L_{\text {cylinder }}}=\frac{\frac{7}{5} m R^2 \omega}{\frac{3}{2} m R^2 \omega} \\ & \Rightarrow \quad \frac{\frac{L_1}{L_2}}{2}=\frac{14}{15} \end{aligned} $

Let block falls through a height $h$, then loss of PE of block appears in form of KE of block and KE of rotation of pulley.

$ \Rightarrow \quad m g h=\frac{1}{2} m v^2+\frac{1}{2} \frac{M R^2}{2} \cdot \omega^2 $

Here, $\quad m=M=2 \mathrm{~kg}$,

So, $\quad g h=\frac{v^2}{2}+\frac{R^2}{4} \omega^2$

As $v=R \omega, g h=\frac{3}{4} R^2 \omega^2$

Differentiating with respect to time,

$ \begin{gathered} g \cdot \frac{d h}{d t}=\frac{3}{4} R^2 \cdot 2 \omega \cdot \frac{d \omega}{d t} \\ g v=\frac{3}{4} R^2 \times 2 \times \frac{v}{R} \cdot \alpha \\ \Rightarrow g=\frac{6}{4} R \alpha \Rightarrow \frac{4 g}{6}=R \alpha \Rightarrow R \alpha=\frac{2 g}{3} \end{gathered} $

But $R \alpha=a_t=$ tangential acceleration of pulley

= acceleration of block

So, $a_t=R \alpha=\frac{2}{3} g=\frac{20}{3} \mathrm{~m} / \mathrm{s}^2$