Rotational Motion

A bar, having a mass of $ M = 1 $ kg and a length of $ L = 0.2 $ m, is hinged at a point P. A particle with a mass of $ m = 0.1 $ kg moves in a direction perpendicular to the bar at a velocity of $ u = 5 $ m/s. This particle collides elastically with the bar at a point that is $ \frac{L}{2} $ (0.1 m) away from the pivot point P. As a result of this collision, the bar begins to rotate with an angular velocity $ \omega $, and the particle recoils with a new velocity $ v $.

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.

To determine the value of $n$, let's analyze the problem by considering the moment of inertia for each ring about an axis passing through its center. For a ring, the moment of inertia about an axis through its center can be expressed as:

$I = mR^2$

Where:

• $I$ is the moment of inertia
• $m$ is the mass of the ring
• $R$ is the radius of the ring

Given that both rings are made of the same material, they will have the same density. Let's denote the moment of inertia of the first ring (radius $R$) as $I_1$ and the moment of inertia of the second ring (radius $nR$) as $I_2$. According to the problem:

$\frac{I_1}{I_2} = \frac{1}{64}$

For the first ring (radius $R$):

$I_1 = m_1 R^2$

For the second ring (radius $nR$):

$I_2 = m_2 (nR)^2 = m_2 n^2 R^2$

Assuming both rings are made of the same material and have the same mass per unit length, the mass of the second ring with radius $nR$ will be $n$ times the mass of the first ring (since circumference scales linearly with radius):

$m_2 = n m_1$

Then we can write:

$I_2 = n m_1 (nR)^2 = n m_1 n^2 R^2 = n^3 m_1 R^2$

Using the ratio of the moments of inertia:

$\frac{I_1}{I_2} = \frac{m_1 R^2}{n^3 m_1 R^2} = \frac{1}{n^3} = \frac{1}{64}$

The above equality simplifies to:

$n^3 = 64$

$n = \sqrt[3]{64} = 4$

Thus, the value of $n$ is:

Option C

$\begin{aligned} \frac{x}{\sin \theta} & =\frac{2 r}{\sin 90^{\circ}} \Rightarrow x=2 r \sin \theta \\ \therefore \quad \frac{d x}{d t} & =2 r \cos \theta \frac{d \theta}{d t} \\ \frac{d \theta}{d t} & =\frac{d x / d t}{2 r \cos \theta}=\frac{2 v_0}{2 r \cos 60^{\circ}}=\frac{2 v_0}{r} \end{aligned}$

$a = {{g\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}}$

$v = \sqrt {{{2Sg\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}}} $

$ \Rightarrow {{{v_c}} \over {{v_{ss}}}}\sqrt {{{1 + {{K_{ss}^2} \over {{R^2}}}} \over {1 + {{K_c^2} \over {{R^2}}}}}} = \sqrt {{{1 + {2 \over 5}} \over {1 + {1 \over 2}}}} $

$ \Rightarrow \sqrt {{{{7 \over 5}} \over {{3 \over 2}}}} = \sqrt {{{14} \over {15}}} $

${\overrightarrow L _0} = {\overrightarrow L _{of\,cm}} + {\overrightarrow L _{about\,cm}}$

$ \Rightarrow {a \over 3}{R^2}\omega = mvR + {2 \over 3}m{R^2}\omega = {5 \over 3}m{R^2}\omega $

$ \Rightarrow a = 5$

$\alpha = {{d\omega } \over {dt}} = 6{t^2} - 2t$

$\int_0^\omega {d\omega = \int_0^t {(6{t^2} - 2t)dt} } $

so $\omega = 2{t^3} - {t^2} + 10$

and ${{d\theta } \over {dt}} = 2{t^3} - {t^2} + 10$

so $\int_4^\theta {d\theta = \int_0^t {(2{t^3} - {t^2} + 10)dt} } $

$\theta = {{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4$

Taking torque from B

${F_w} \times 5 = {3 \over 2}mg$

$ \Rightarrow {F_w} = {3 \over {10}} \times 10 \times 10$

$ = 30\,N$

$N = mg = 100\,N$

and ${f_r} = {F_w} = 30\,N$

so ${F_f} = \sqrt {{N^2} + f_r^2} = \sqrt {10900} = 10\sqrt {109} \,N$

so ${{{F_w}} \over {{F_f}}} = {3 \over {\sqrt {109} }}$

(A) Moment of inertia of solid sphere of radius R about a tangent $ = {2 \over 5}M{R^2} + M{R^2} = {7 \over 5}M{R^2}$

$\Rightarrow$ A $-$ (II)

(B) Moment of inertia of hollow sphere of radius R about a tangent $ = {2 \over 3}M{R^2} + M{R^2} = {5 \over 3}M{R^2}$

$\Rightarrow$ B $-$ (I)

(C) Moment of inertia of circular ring of radius (R) about its diameter = ${{\left( {M{R^2}} \right)} \over 2}$

$\Rightarrow$ C $-$ (IV)

(D) Moment of inertia of circular disc of radius (R) about any diameter

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

$\Rightarrow$ D $-$ (III)

$m{\omega ^2}({l_0} + x) = kx$

$ \Rightarrow m{\omega ^2}{l_0} = (k - m{\omega ^2}) \times x$

$ \Rightarrow x = {{m{\omega ^2}{l_0}} \over {(k - m{\omega ^2})}}$

$K{E_R} = {1 \over 2}l{w^2}$

$ = {1 \over 2} \times {2 \over 5} \times {\omega ^2} \times (m{R^2})$

$K{E_{total}} = {1 \over 2} \times {7 \over 5} \times m{R^2} \times {\omega ^2}$

$\therefore$ ${{K{E_R}} \over {K{E_{total}}}} = {2 \over 7}$

${I_1}{\omega _1} = {I_2}{\omega _2}$

$M{R^2}{\omega _1} = (M{R^2} + 2m{R^2}){\omega _2}$

${\omega _2} = \left( {{M \over {M + 2m}}} \right){\omega _1}$

${\omega _2} = 2\left( {{M \over {M + 2m}}} \right)$

$\overrightarrow \tau = \overrightarrow r \times \overrightarrow F $

$ = (2\widehat i + \widehat j + 2\widehat k) \times (3\widehat i + 4\widehat j - 2\widehat k)$

$ = - 10\widehat i + 10\widehat j + 5\widehat k$

(I)

${v_{net}} = \sqrt 2 \omega R = \sqrt 2 $

I $\to$ S

(II) ${v_A}(t = 0.13) = {{5\pi } \over {\sqrt 2 }}\cos 45\widehat i + \left[ {{{5\pi } \over {\sqrt 2 }} \times {1 \over {\sqrt 2 }} - g \times 0.1} \right]\widehat j$

$ = {{5\pi } \over 2}\widehat i + \left( {{{5\pi } \over 2} - 1} \right)\widehat j$

${v_B}(t = 0.1\,\sec ) = {{ - 5\pi } \over 2}\widehat i + \left( {{{5\pi } \over 2}} \right)\widehat j$

After $t = 0.1$, relative velocities should not change.

${v_{rel}}(t = 0.1\,\sec ) = \left| {5\pi \widehat i - \widehat j} \right|$

$ = \sqrt {25{\pi ^2} + 1} $

II $\to$ T

(III) $x = {x_A} - {x_B}$

$ = {x_0}\sin t - {x_0}\sin \left( {t + {\pi \over 2}} \right)$

$ = \sqrt 2 {x_0}\sin \left( {t - {\pi \over 4}} \right)$

${v_{rel}} = {{dx} \over {dt}} = \sqrt 2 {x_0}\cos \left( {t - {\pi \over 4}} \right)$

$ = \sqrt 2 \cos \left( {{\pi \over 3} - {\pi \over 4}} \right)$

$ = \sqrt 2 \times {{\sqrt 3 + 1} \over {2\sqrt 2 }} = {{\sqrt 3 + 1} \over 2}$

III $\to$ P

(IV) ${v_{rel}} = \sqrt {{3^2} + {1^2}} = \sqrt {10} $

IV $\to$ R

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} $

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 $=$

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

Area mass density, $\sigma = {M \over {16{R^2}}}$ ($\because$ Area = $4R \times 4R = 16{R^2}$)

Mass of each hole ${m_1} = \sigma \pi {R^2} = {M \over {16{R^2}}}\pi {R^2} = {{\pi M} \over {16}}$

Distance between centre of plate and centre of hole

$x = {{\sqrt {{{(2R)}^2} + {{(2R)}^2}} } \over 2} = {{2\sqrt 2 R} \over 2}$

Moment of inertia of one hole at about Z-axis

${I_1} = {1 \over 2}{m_1}{R^2} + {m_1}{x^2} = {{5\pi } \over {32}}M{R^2}$

Moment of inertia of whole plate about Z-axis

$I = {{M{{(4R)}^2}} \over 6} = {8 \over 3}M{R^2}$

Required moment of inertia

${I_0} = I - 4{I_1},$ $ = \left[ {{8 \over 3} - 4\left( {{{5\pi } \over {32}}} \right)} \right]M{R^2} = \left[ {{8 \over 3} - {{5\pi } \over 8}} \right]M{R^2}$

Given, R = 5 cm and M = 1 kg

So, ${I_0} = \left[ {{8 \over 3} - {{5\pi } \over 8}} \right]1 \times 25 \times {10^{ - 4}} = 0.0017$ kg-m²