Rotational Motion

Let's disc rolls to $\phi$ angular displacement about its centre:

As the disc rolls without slipping. we can write,

$ x=r \phi=(R-r) \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Now, Let's apply $\tau=I \alpha$ about contact

$ \begin{aligned} & \text { point } \\ & \tau=m g \sin \theta \cdot r+k x r=\left(\frac{m r^2}{2}+m r^2\right) \alpha^{\prime} \\ & \Rightarrow m g \theta \cdot r+k(R-r) \theta r=\frac{3 m r^2}{2} \alpha^{\prime} \\ & \Rightarrow[m g+k(R-r)] \theta r=\frac{3}{2} m r^{\prime} \alpha^{\prime} \\ & \Rightarrow[m g+k(R-r)] \theta=\frac{3}{2} m r \alpha^{\prime}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

here, $\alpha^{\prime}=\frac{d^2 \phi}{d t^2}$

from (1), $\phi=\left(\frac{R-r}{r}\right) 0$

$ \begin{aligned} & \Rightarrow \quad \alpha^{\prime}=\frac{d^2 \phi}{d t^2}=\left(\frac{R-r}{r}\right) \frac{d^2 \theta}{d t^2} \\ & \Rightarrow \quad \alpha^{\prime \prime}=\left(\frac{R-r}{r}\right) \alpha\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

Now, from (2) and (3),

$ \begin{aligned} & {[m g+k(R-r)] \theta=\frac{3}{2} m r\left(\frac{R-r}{r}\right) \alpha} \\ & \Rightarrow[m g+k(R-r)] \theta=\frac{3}{2} m(R-r) \alpha \end{aligned} $

Now, we know, for SHM,

$ \begin{gathered} \alpha=\omega^2 \theta \\ \Rightarrow \quad\left[\frac{m g+k(R-r)}{\frac{3}{2} m(R-r)}\right] \phi=\omega^2 \phi \\ \Rightarrow \frac{2}{3} \frac{g}{(R-r)}+\frac{2}{3} \frac{k}{m}=\omega^2 \\ \Rightarrow \omega=\sqrt{\frac{2}{3}\left[\frac{g}{(R-r)}+\frac{k}{m}\right]} \end{gathered} $

$ \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} $

$ \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 $

To determine the new angular velocity of the system, we use the principle of conservation of angular momentum. When no external torque acts on a system, its angular momentum remains constant. Let's denote the initial angular momentum and the final angular momentum, respectively, as $L_{\text{initial}}$ and $L_{\text{final}}$.

The initial angular momentum of the system is given by:

$ L_{\text{initial}} = I_{\text{initial}} \cdot \omega $

Here, $I_{\text{initial}}$ is the moment of inertia of the first disc. For a thin circular disc, the moment of inertia about its center is:

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

Thus,

$ L_{\text{initial}} = \left( \frac{1}{2} M R^2 \right) \omega $

When the second disc is placed gently on the first disc, the two discs rotate together with a common angular velocity $\omega'$. The moment of inertia of the second disc is:

$ I_{\text{second}} = \frac{1}{2} \left( \frac{M}{2} \right) R^2 = \frac{1}{4} M R^2 $

The combined moment of inertia of the system after placing the second disc is:

$ I_{\text{final}} = I_{\text{initial}} + I_{\text{second}} = \frac{1}{2} M R^2 + \frac{1}{4} M R^2 = \frac{3}{4} M R^2 $

Thus, the final angular momentum of the system is:

$ L_{\text{final}} = I_{\text{final}} \cdot \omega' = \left( \frac{3}{4} M R^2 \right) \omega' $

By the conservation of angular momentum:

$ L_{\text{initial}} = L_{\text{final}} $

This simplifies to:

$ \left( \frac{1}{2} M R^2 \right) \omega = \left( \frac{3}{4} M R^2 \right) \omega' $

Solving for $\omega'$:

$ \omega' = \frac{\left( \frac{1}{2} M R^2 \right) \omega}{\left( \frac{3}{4} M R^2 \right)} = \frac{\omega}{\frac{3}{2}} = \frac{2}{3} \omega $

So, the new angular velocity of the system is:

$\omega' = \frac{2}{3} \omega$

Thus, the correct answer is:

Option D: $\frac{2}{3} \omega$

For hollow sphere

$\begin{aligned} & I_1=\frac{2}{3} m R^2=\left(\sqrt{\frac{2}{3}} R\right)^2 m \\ & \therefore \quad K_1=\sqrt{\frac{2}{3}} R \end{aligned}$

For solid cylinder

$\begin{aligned} & I_2=\frac{1}{4} m R^2+\frac{m}{3}(4 R)^2 \\ & =\left(\sqrt{\frac{67}{12}} R\right)^2 m \\ & \therefore \quad K_2=\sqrt{\frac{67}{12}} R \\ & \Rightarrow \frac{K_1}{K_2}=\sqrt{\frac{2}{3} \times \frac{12}{67}} \\ & =\sqrt{\frac{8}{67}} \end{aligned}$

$\Rightarrow \quad x=67$

Initial Conditions:

Consider a solid disc of mass $ M $ and radius $ R $ rolling without slipping on a horizontal surface. The center of mass of the disc has an initial speed $ v $.

Kinetic Energy of the Rolling Disc:

For an object that rolls without slipping, its total kinetic energy $ K $ is the sum of its translational kinetic energy and its rotational kinetic energy.

Translational Kinetic Energy:

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

Rotational Kinetic Energy:

The moment of inertia $ I $ of a solid disc about its center is:

$ I = \frac{1}{2} M R^2. $

Since the disc rolls without slipping, the angular velocity $\omega$ is related to the linear speed by:

$ v = \omega R \implies \omega = \frac{v}{R}. $

Therefore, the rotational kinetic energy is:

$ K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v^2}{R^2}\right) = \frac{1}{4} M v^2. $

Total Initial Kinetic Energy:

Summing these gives:

$ K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2. $

Energy Conservation as the Disc Climbs the Incline:

As the disc moves up the incline, there are no dissipative forces (assuming a smooth incline and no slipping), so energy is conserved. The total kinetic energy at the bottom will be converted entirely into gravitational potential energy at the maximum height $ h $:

$ K_{\text{total}} = M g h. $

Substitute $ K_{\text{total}} = \frac{3}{4} M v^2 $:

$ \frac{3}{4} M v^2 = M g h. $

Canceling $ M $ from both sides:

$ \frac{3}{4} v^2 = g h. $

Thus,

$ h = \frac{3}{4} \frac{v^2}{g}. $

Final Answer:

$ \boxed{\frac{3}{4} \frac{v^2}{g}} $

This corresponds to Option A.

$\begin{aligned} & \mathrm{L}=m u \cos \theta H \\ & =m u \cos \theta \times \frac{u^2 \sin ^2 \theta}{2 g} \\ & =\frac{m u^3}{2 g} \times \frac{\sqrt{3}}{2} \times\left(\frac{1}{2}\right)^2=\frac{\sqrt{3} m u^3}{16 g} \end{aligned}$

First, let us restate the problem in our own words:

We have a uniform iron bar of weight $12\,\mathrm{kgf}$ (meaning its mass is $12\,\mathrm{kg}$, so its weight is $12g\,\mathrm{N}$ in SI units).

One end of the bar is on the ground (assume a frictionless contact), and the other end is on a man’s shoulder (also assume frictionless contact).

The bar makes an angle of $60^\circ$ with the horizontal.

We want to find how much load (“effective weight”) the man feels on his shoulder.

Under these (common) assumptions of frictionless contacts at both ends, each contact force is perpendicular to the bar. We can solve the problem by taking torques about the ground contact point.

1. Labeling and geometry

Let the length of the bar be $L$.

The bar is inclined at $60^\circ$ from the horizontal.

Because the bar is uniform, its center of gravity (C.O.G.) lies at its midpoint, i.e., at a distance $L/2$ from either end.

Let the weight of the bar be $W = 12g$.

Let the reaction force from the man’s shoulder be $R$. (Magnitude unknown.)

Let the reaction force at the ground be $R_g$. (We do not actually need its exact value to answer the question.)

We will place our origin (for torque calculation) at the ground contact point.

2. Torques about the ground contact

Torque due to the bar’s weight

The weight $W = 12g$ acts downward at the midpoint of the bar.

The position vector from the pivot (ground) to the bar’s midpoint is $\tfrac{L}{2}$ long and makes $60^\circ$ with the horizontal.

The weight is vertical downward, i.e., $90^\circ$ from the horizontal.

The angle between the position vector ($60^\circ$ from horizontal) and the weight ($90^\circ$ from horizontal) is

$ 90^\circ - 60^\circ \;=\; 30^\circ. $

Hence, the torque magnitude from the weight is

$ \tau_W \;=\; \bigl(12g\bigr)\,\bigl(\tfrac{L}{2}\bigr)\,\sin(30^\circ) \;=\; 12g \cdot \tfrac{L}{2}\cdot \tfrac12 \;=\; 12g \cdot \tfrac{L}{4}. $

Torque due to the man’s reaction force $R$

The man’s shoulder is at the other end of the bar, i.e., $L$ from the pivot.

For frictionless contact, $R$ is perpendicular to the bar. Since the bar is at $60^\circ$ from horizontal, a line perpendicular to the bar is $90^\circ$ away from the bar’s direction, so the angle between the bar’s position vector ($\vec{r}$) and $R$ is $90^\circ$.

Thus, the torque from $R$ about the ground contact is

$ \tau_R \;=\; R \,\times\, L \,\times\, \sin(90^\circ) \;=\; R \cdot L. $

Because the bar is in static equilibrium, the net torque about the ground contact must be zero. Taking the clockwise direction (by weight) as positive, we have

$ \tau_W \;-\; \tau_R \;=\; 0 \quad\Longrightarrow\quad 12g \cdot \frac{L}{4} \;=\; R \cdot L. $

Solving for $R$,

$ R \;=\; \frac{12g}{4} \;=\; 3g. $

Hence, the force on the man’s shoulder has magnitude $3g$ in newtons.

3. Converting to “kg weight”

Often, such problems phrase the answer in “kg of weight” rather than newtons. Since $W = mg$ for a mass $m$ in Earth’s gravity, a force of $3g\,\mathrm{N}$ corresponds to a “weight” of $3\,\mathrm{kgf}$.

So the man effectively feels 3 kg of load on his shoulder.