Rotational Motion

Initially, the wheel is at rest.
For the first 2 s :${\theta }_1=\frac{1}{2}\alpha (2{)}^2=2\alpha$
For the first 4 s :${\theta }^{\text{'}}=\frac{1}{2}\alpha (4{)}^2=8\alpha$
So, in the next 2 s :
${\theta }_2={\theta }^{\text{'}}-{\theta }_1=8\alpha -2\alpha =6\alpha$
Thus,
$\frac{{\theta }_2}{{\theta }_1}=\frac{6\alpha }{2\alpha }=3$

Given position vector of particle in circular motion.
$\overrightarrow{r}=a\cos \omega t\hat{i}+a\sin \omega t\hat{j}$
Velocity is time derivative of position.
$\overrightarrow{v}=\frac{d\overrightarrow{r}}{\mathrm{dt}}=-a\omega \sin \omega t\hat{i}+a\omega \cos \omega t\hat{j}$
Acceleration is time derivative of velocity.
$\overrightarrow{a}=\frac{d\overrightarrow{v}}{\mathrm{dt}}=-a{\omega }^2\cos \omega t\hat{i}-a{\omega }^2\sin \omega t\hat{j}$
Force is proportional to acceleration.
$\overrightarrow{F}=m\overrightarrow{a}$
Acceleration vector is opposite in direction to position vector.
Therefore acceleration is antiparallel to position vector.

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{L}$
This relation is incorrect in general because torque is not the cross product of position vector and angular momentum.

$\overrightarrow{\tau }=\frac{d}{dt}\left(\overrightarrow{r}\times \overrightarrow{p}\right)$
This is correct and represents the fundamental definition of torque.
Using Newton’s second law, the time derivative of momentum equals force.
$\overrightarrow{\tau }=\overrightarrow{r}\times \frac{d\overrightarrow{p}}{dt}=\overrightarrow{r}\times \overrightarrow{F}$
This form is correct and commonly used in mechanics.

$\overrightarrow{\tau }=I\overrightarrow{\alpha }$
This relation is correct for rigid bodies rotating about a fixed axis.

Required ratio $=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}I{\omega }^2}$

Using $(V=\omega R)$, we get

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

Moment of inertia of a uniform rod of mass M and length L about its center:
$$\alpha =\frac{M{L}^2}{12}$$
Moment of inertia of each half of the rod about its own center:
$$I=\frac{\left(\frac{M}{2}\right)\left(\frac{L}{2}\right){\mathrm{}}^2}{12}\times 2$$
Simplifying:
$$I=\frac{\alpha }{4}$$

Moment of inertia of a ring about its diameter is written as
$I=\frac{M{R}^2}{2}$
$M=\lambda L$
Substitute mass in the expression of moment of inertia
$I=\frac{\lambda L{R}^2}{2}$
$R=\frac{L}{2\pi }$
Substitute radius in terms of length
$I=\frac{\lambda L{\left(\frac{L}{2\pi }\right)}^2}{2}$
Simplifying the expression
$I=\frac{\lambda L^3}{8{\pi }^2}$
Final answer for moment of inertia
$I=\frac{\lambda L^3}{8{\pi }^2}$

Given: $\overrightarrow{r}=\left(10t^2\hat{i}+5t^3\hat{j}\right)$

Velocity: $\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}=\left(20t\hat{i}+15t^2\hat{j}\right)$

At t = 1s:

$\overrightarrow{v}=20\hat{i}+15\hat{j}$

Momentum:

$\overrightarrow{p}=m\overrightarrow{v}=0.1(20\hat{i}+15\hat{j})=2\hat{i}+1.5\hat{j}$

A is correct

Acceleration: $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}=(20\hat{i}+30t\hat{j})$

At t = 1s:

$\overrightarrow{a}=20\hat{i}+30\hat{j}$

Force: $\overrightarrow{F}=m\overrightarrow{a}=0.1(20\hat{i}+30\hat{j})=2\hat{i}+3\hat{j}$

B is correct

Angular momentum: $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$

At t = 1s

$\overrightarrow{r}=10\hat{i}+5\hat{j},\overrightarrow{p}=2\hat{i}+1.5\hat{j}$

$\overrightarrow{L}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 10& 5& 0\\ 2& 1.5& 0\end{array}\right|=(15-10)\hat{k}=5\hat{k}$

C is incorrect

Torque: $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

$\overrightarrow{F}=2\hat{i}+3\hat{j}$

$\overrightarrow{\tau }=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 10& 5& 0\\ 2& 3& 0\end{array}\right|=(30-10)\hat{k}=20\hat{k}$

D is correct

The torque is calculated using the cross product:

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

Expanding using the determinant method:

$$\begin{gathered} \overrightarrow{\tau }=\left|\begin{array}{lll}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& 1& 2\end{array}\right| \\ \overrightarrow{\tau }=\hat{i} - \hat{k} \end{gathered}$$

For a wheel in pure rolling, the speed at the highest point is twice the translational speed of its center of mass ($v_{cm}$).

$v_{top}=2v_{cm}=8\text{ m/s}$

$v_{cm}=4\text{ m/s}$

For a particle on the rim at the same level as the center, its velocity has two perpendicular components:

Translational velocity: $v_{cm}$ (horizontal)

Tangential (rotational) velocity: $\omega R=v_{cm}$ (vertical)

The resultant speed is the vector sum of these two perpendicular components:

$v=\sqrt{v_{cm}^2+v_{cm}^2}=\sqrt{2}{v}_{cm}$

Substitute $v_{cm}=4\text{ m/s}$:

$v=4\sqrt{2}\text{ m/s}$

Here,

$$\begin{gathered} \overrightarrow{\mathrm{F}}=2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+2\widehat{\mathrm{k}} \mathrm{N}, \overrightarrow{\mathrm{r}} = \hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{\tau }} = \overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}} = \left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 1& 1\\ 2& 1& 2\end{array}\right| = \hat{\mathrm{i}} - \hat{\mathrm{k}} \end{gathered}$$

For a rolling object, the acceleration a is given by the formula:

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

Since the cylinder rolls without slipping, the moment of inertia $I=\frac{1}{2}m{r}^2$. Substitute the known values:

$a=\frac{g\sin 45°}{1+\frac{1}{2}}=\frac{g\sin 45°}{\frac{3}{2}}=\frac{2g\sin 45°}{3}$

Since $\sin 45°=\frac{1}{\sqrt{2}}$ :

$a=\frac{2g·\frac{1}{\sqrt{2}}}{3}=\frac{\sqrt{2}g}{3}$

Let's first derive a general formula for the linear acceleration (\(a\)) of a round body of mass \(M\), radius \(R\), and moment of inertia $I_{cm}$ when a tangential force \(F\) is applied at its highest point.

Assume a static friction force \(f\) acts backward at the point of contact with the rough surface.

Translational Motion (Newton's Second Law):

Rotational Motion (about the Center of Mass):

Both the applied force \(F\) and the friction \(f\) produce a clockwise torque.

Since the body is rolling without slipping on a stationary surface, we use the constraint $a=\alpha R$, which gives $\alpha =\frac{a}{R}$. Substituting this in:

Solving for Acceleration (\(a\)):

Add Equation 1 and Equation 2 to eliminate \(f\):

Applying to Solid Sphere (A)

Mass:$M_A=5m$

Moment of Inertia: For a solid sphere, $I=\frac{2}{5}M{R}^2$.

Acceleration ($a_A$):

Applying to Spherical Shell (B)

Mass: $M_B=m$

Moment of Inertia: For a spherical shell (hollow sphere), $I=\frac{2}{3}M{R}^2$.

Acceleration ($a_B$):

Finding the Ratio

Now, we find the ratio of $a_A$ to $a_B$:

The ratio of their accelerations is 5 : 21.

• Mass of the cylinder: $m$
• Radius of the cylinder: $r$
• Inclined plane angle: $\theta = 45^\circ$

1. Forces acting on the cylinder:

   • Component of gravitational force along the incline: $m g \sin \theta$
   • Frictional force: $f$
   • Normal force: $N = m g \cos \theta$

2. Equations of motion: For rolling without slipping:

   • Translational motion: $$m a = m g \sin \theta - f$$
   • Rotational motion about the center of mass: $$f r = I \alpha$$ For a solid cylinder, $I = \frac{1}{2} m r^2$, and rolling without slipping implies $\alpha = \frac{a}{r}$.
     Substituting $I$ and $\alpha$: $$f r = \frac{1}{2} m r^2 \cdot \frac{a}{r}$$,​ $$f = \frac{1}{2} m a$$

3. Substitute $f$ into the translational motion equation:

   $$m a = m g \sin \theta - \frac{1}{2} m a$$

   Simplify:

   $$m a + \frac{1}{2} m a = m g \sin \theta$$, $$\frac{3}{2} m a = m g \sin \theta$$, $$a = \frac{2}{3} g \sin \theta$$

4. Substitute $\theta = 45^\circ$:

   $$a = \frac{2}{3} g \sin 45^\circ$$

   Since $\sin 45^\circ = \frac{1}{\sqrt{2}}$

   $$a = \frac{2}{3} g \cdot \frac{1}{\sqrt{2}} = \frac{2g}{3\sqrt{2}}$$​

The acceleration of the axis of the cylinder is:

$$a = \frac{2g}{3\sqrt{2}}$$​

Time is given by $t=\sqrt{\frac{2l}{a}}$ $l\to$ length of incline SS $\to$ Solid sphere

$$\begin{gathered} a=\frac{g\sin \theta }{1+\frac{I_{cm}}{M{R}^2}} \\ a_{SS}=\frac{5g\sin \theta }{7} \end{gathered}$$

$$\begin{gathered} a_{HS}=\frac{3g\sin \theta }{5} \\ a_{SS}>a_{HS} \\ t_1<t_2 \end{gathered}$$

1. Expression for Torque and Power: The power derived to the disc is given by:

   $$P = \tau \cdot \omega$$

   where $\tau$ is the torque, and $\omega$ is the angular velocity.

2. Moment of Inertia: For a disc rotating about its axis:

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

3. Angular Velocity ($\omega$): Angular velocity is the derivative of angle $\theta$ with respect to time:

   $$\omega = \frac{d\theta}{dt} = \frac{d}{dt} \left( at^2 + bt + c \right)$$, $$\omega = 2at + b$$

4. Angular Acceleration ($\alpha$): Angular acceleration is the derivative of angular velocity:

   $$\alpha = \frac{d\omega}{dt} = \frac{d}{dt} \left( 2at + b \right)$$, $$\alpha = 2a$$

5. Torque ($\tau$τ): Torque is related to angular acceleration:

   $$\tau = I \cdot \alpha$$

   Substitute $I = \frac{1}{2} M R^2$ and $\alpha = 2a$

   $$\tau = \frac{1}{2} M R^2 \cdot 2a = a M R^2$$

6. Power ($P$): Substitute $\tau = a M R^2$ and $\omega = 2at + b$ into the power formula:

   $$P = \tau \cdot \omega = (a M R^2) \cdot (2at + b)$$, $$P = a M R^2 (2at + b)$$

The power derived to the disc as a function of time is:

$$\boxed{a M R^2 (2at + b)}$$​

1. Expression for Torque and Power: The power derived to the disc is given by:

   $$P = \tau \cdot \omega$$

   where $\tau$ is the torque, and $\omega$ is the angular velocity.

2. Moment of Inertia: For a disc rotating about its axis:

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

3. Angular Velocity ($\omega$): Angular velocity is the derivative of angle $\theta$ with respect to time:

   $$\omega = \frac{d\theta}{dt} = \frac{d}{dt} \left( at^2 + bt + c \right)$$, $$\omega = 2at + b$$

4. Angular Acceleration ($\alpha$): Angular acceleration is the derivative of angular velocity:

   $$\alpha = \frac{d\omega}{dt} = \frac{d}{dt} \left( 2at + b \right)$$, $$\alpha = 2a$$

5. Torque ($\tau$τ): Torque is related to angular acceleration:

   $$\tau = I \cdot \alpha$$

   Substitute $I = \frac{1}{2} M R^2$ and $\alpha = 2a$

   $$\tau = \frac{1}{2} M R^2 \cdot 2a = a M R^2$$

6. Power ($P$): Substitute $\tau = a M R^2$ and $\omega = 2at + b$ into the power formula:

   $$P = \tau \cdot \omega = (a M R^2) \cdot (2at + b)$$, $$P = a M R^2 (2at + b)$$​

Radius of first disc, R₁ = R

Radius of ssecond disc, R₂ = 2 R

Let the surface mass density is d.

Mass of first disc is

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{first} \mathrm{disc} \mathrm{is} \mathrm{given} \mathrm{by} \\ {\mathrm{M}}_1 = \mathrm{d} \times {\mathrm{\pi R}}^2 \\ \mathrm{Mass} \mathrm{of} \mathrm{second} \mathrm{disc} \mathrm{is} \mathrm{given} \mathrm{by} \\ {\mathrm{M}}_2 = \mathrm{d} \times 4{\mathrm{\pi R}}^2 \\ \mathrm{So}, \mathrm{Moment} \mathrm{of} \mathrm{iinertia} \\ {\mathrm{I}}_1 = \frac{1}{2}{\mathrm{M}}_1{\mathrm{R}}_1^2 = \frac{1}{2}\times \mathrm{d} \times {\mathrm{\pi R}}^2\times {\mathrm{R}}^2=\frac{\mathrm{d} {\mathrm{\pi R}}^4}{2} \\ {\mathrm{I}}_2 = \frac{1}{2}{\mathrm{M}}_2{\mathrm{R}}_2^2 = \frac{1}{2}\times \mathrm{d} \times 4{\mathrm{\pi R}}^2\times 4{\mathrm{R}}^2=\frac{16 \mathrm{d} {\mathrm{\pi R}}^4}{2} \\ \mathrm{Thus}, \frac{{\mathrm{I}}_1}{{\mathrm{I}}_2} = \frac{1}{16} \end{gathered}$$

using the Work-Energy Theorem

$W=k_f-k_i$

The work done by gravity is given as:

$MgL\sin \theta =k_f-k_i$

the total kinetic

$K.E.=\frac{1}{2}m{V}_{cm}^2+\frac{1}{2}{I}_{cm}{\omega }^2$

For a solid sphere, the moment of inertia $I=\frac{2}{5}m{R}^2$, and since $\omega =\frac{V_{cm}}{R}$, the equation becomes:

$K.E.=\frac{1}{2}m{V}^2+\frac{1}{2}\times \frac{2}{5}m{R}^2{\left(\frac{V}{R}\right)}^2$

This simplifies to:

$\text{ K.E. }=\frac{7}{10}m{V}^2$

Using the work-energy principle, we equate the potential energy to the change in kinetic energy:

$mgL\sin \theta =\frac{7}{10}m{V}^2$

This gives: $V^2\propto \sin \theta$
Now, for the two different angles of inclination ( $30°$ and $45°$ ), the ratio of the velocities at the bottom of the incline is:

$$\begin{gathered} {\left(\frac{V_1}{V_2}\right)}^2=\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin 30°}{\sin 45°}=\frac{1}{\sqrt{2}} \\ V_1^2:V_2^2=1:\sqrt{2} \end{gathered}$$

The acceleration on the inclined plane is given by

$\mathrm{a} = \frac{\mathrm{g} \mathrm{sin\theta }}{1+{\displaystyle \frac{{\mathrm{K}}^2}{{\mathrm{r}}^2}}}$

where, K is the radius of gyration

$$\begin{gathered} \mathrm{For} \mathrm{solid} \mathrm{sphere}, \mathrm{K} = \sqrt{\frac{2}{5}\mathrm{R}} = 0.63 \mathrm{R} \\ \mathrm{For} \mathrm{hollow} \mathrm{sphere}, \mathrm{K} = \sqrt{\frac{2}{3}\mathrm{R}} = 0.81 \mathrm{R} \end{gathered}$$

So, $$\begin{gathered} {\mathrm{a}}_{\mathrm{solid}} >{\mathrm{a}}_{\mathrm{hollow}} \\ \mathrm{So}, \\ {\mathrm{t}}_{\mathrm{solid}} < {\mathrm{t}}_{\mathrm{hollow}} \\ {\mathrm{t}}_1 < {\mathrm{t}}_2 \end{gathered}$$

The most important detail is that \(r\) is given as the diameter. Therefore, the radius of the ring is $R=\frac{r}{2}$.

Moment of inertia about a diameter:

For a circular ring, the moment of inertia about an axis passing through its diameter (in its plane) is:

Apply the Parallel Axis Theorem:

To find the moment of inertia about a tangential axis lying in the same plane, shift the axis from the diameter to the edge by a distance equal to the radius (\(R\)):

Substitute the radius back in terms of \(r\):

Replace \(R\) with $\frac{r}{2}$:

Correct Option: B

Here,

$$\begin{gathered} \overrightarrow{\mathrm{F}}=2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+2\widehat{\mathrm{k}} \mathrm{N}, \overrightarrow{\mathrm{r}} = \hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{\tau }} = \overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}} = \left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 1& 1\\ 2& 1& 2\end{array}\right| = \hat{\mathrm{i}} - \hat{\mathrm{k}} \end{gathered}$$

The acceleration on the inclined plane is given by

$\mathrm{a} = \frac{\mathrm{g} \mathrm{sin\theta }}{1+{\displaystyle \frac{{\mathrm{K}}^2}{{\mathrm{r}}^2}}}$

where, K is the radius of gyration

$$\begin{gathered} \mathrm{For} \mathrm{solid} \mathrm{sphere}, \mathrm{K} = \sqrt{\frac{2}{5}\mathrm{R}} = 0.63 \mathrm{R} \\ \mathrm{For} \mathrm{hollow} \mathrm{sphere}, \mathrm{K} = \sqrt{\frac{2}{3}\mathrm{R}} = 0.81 \mathrm{R} \end{gathered}$$

So, $$\begin{gathered} {\mathrm{a}}_{\mathrm{solid}} >{\mathrm{a}}_{\mathrm{hollow}} \\ \mathrm{So}, \\ {\mathrm{t}}_{\mathrm{solid}} < {\mathrm{t}}_{\mathrm{hollow}} \\ {\mathrm{t}}_1 < {\mathrm{t}}_2 \end{gathered}$$

$$\begin{gathered} \mathrm{\theta } = 5{\mathrm{t}}^2 - 8\mathrm{t} \\ \mathrm{\omega } =\frac{\mathrm{d\theta }}{\mathrm{dt}}=10 \mathrm{t} - 8 \\ \mathrm{\alpha }= \frac{\mathrm{d\omega }}{\mathrm{dt}}=10 \\ \mathrm{P} =\mathrm{\tau } \mathrm{\omega } = \mathrm{I} \mathrm{\alpha } \mathrm{\omega } \\ \mathrm{P} = \left(\frac{{\mathrm{MR}}^2}{2}\right)\mathrm{\alpha \omega } = \left(\frac{{\mathrm{MR}}^2}{2}\right)\times 10\times \left(10\mathrm{t} - 8\right) \\ \mathrm{Put} \mathrm{t} = 2 \mathrm{s} \\ \mathrm{P} = 60 {\mathrm{MR}}^2 \end{gathered}$$

The torque is calculated using the cross product:

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

Expanding using the determinant method:

$$\begin{gathered} \overrightarrow{\tau }=\left|\begin{array}{lll}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& 1& 2\end{array}\right| \\ \overrightarrow{\tau }=\hat{i} - \hat{k} \end{gathered}$$

For a rolling object, the acceleration a is given by the formula:

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

Since the cylinder rolls without slipping, the moment of inertia $I=\frac{1}{2}m{r}^2$. Substitute the known values:

$a=\frac{g\sin 45°}{1+\frac{1}{2}}=\frac{g\sin 45°}{\frac{3}{2}}=\frac{2g\sin 45°}{3}$

Since $\sin 45°=\frac{1}{\sqrt{2}}$ :

$a=\frac{2g·\frac{1}{\sqrt{2}}}{3}=\frac{\sqrt{2}g}{3}$

Radius of first disc, R₁ = R

Radius of ssecond disc, R₂ = 2 R

Let the surface mass density is d.

Mass of first disc is

$$\begin{gathered} \mathrm{Mass} \mathrm{of} \mathrm{first} \mathrm{disc} \mathrm{is} \mathrm{given} \mathrm{by} \\ {\mathrm{M}}_1 = \mathrm{d} \times {\mathrm{\pi R}}^2 \\ \mathrm{Mass} \mathrm{of} \mathrm{second} \mathrm{disc} \mathrm{is} \mathrm{given} \mathrm{by} \\ {\mathrm{M}}_2 = \mathrm{d} \times 4{\mathrm{\pi R}}^2 \\ \mathrm{So}, \mathrm{Moment} \mathrm{of} \mathrm{iinertia} \\ {\mathrm{I}}_1 = \frac{1}{2}{\mathrm{M}}_1{\mathrm{R}}_1^2 = \frac{1}{2}\times \mathrm{d} \times {\mathrm{\pi R}}^2\times {\mathrm{R}}^2=\frac{\mathrm{d} {\mathrm{\pi R}}^4}{2} \\ {\mathrm{I}}_2 = \frac{1}{2}{\mathrm{M}}_2{\mathrm{R}}_2^2 = \frac{1}{2}\times \mathrm{d} \times 4{\mathrm{\pi R}}^2\times 4{\mathrm{R}}^2=\frac{16 \mathrm{d} {\mathrm{\pi R}}^4}{2} \\ \mathrm{Thus}, \frac{{\mathrm{I}}_1}{{\mathrm{I}}_2} = \frac{1}{16} \end{gathered}$$

Time is given by $t=\sqrt{\frac{2l}{a}}$ $l\to$ length of incline

(SS $\to$ Solid sphere)

$$\begin{gathered} a=\frac{g\sin \theta }{1+\frac{I_{cm}}{M{R}^2}} \\ a_{SS}=\frac{5g\sin \theta }{7} \end{gathered}$$

$$\begin{gathered} a_{HS}=\frac{3g\sin \theta }{5} \\ a_{SS}>a_{HS} \\ t_1<t_2 \end{gathered}$$

(3)

\(
\begin{aligned}
& \rho=\vec{\tau} \cdot \vec{\omega} \\
& \tau=l \alpha \\
& I=\frac{M R^2}{2}=1 \\
& \alpha=5 t^2-8 t \\
& \omega=\frac{5}{3} t^3-4 t^2 \\
& \text { At } t=1, \omega=\frac{5}{3}-4=-\frac{7}{3} \\
& t(3)=5-8=-3 \\
& \rho=\vec{\tau} \cdot \vec{\omega}=7 \mathrm{~W}
\end{aligned}
\)

Required ratio $=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}I{\omega }^2}$Using $(V=\omega R)$, we get

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

$\overrightarrow{{\mathrm{V}}_{\mathrm{A}}}={\mathrm{\alpha }}_1{\mathrm{t}}^2 \widehat{\mathrm{i}}+{\mathrm{\alpha }}_2\mathrm{t} \widehat{\mathrm{j}}+{\mathrm{\alpha }}_{3 }\widehat{\mathrm{k}},$ $\overrightarrow{{\mathrm{V}}_{\mathrm{B}}}={\mathrm{\alpha }}_1\widehat{ \mathrm{i}}+{\mathrm{\alpha }}_2 \widehat{\mathrm{j}}+{\mathrm{\alpha }}_3{\mathrm{t}}^2 \widehat{\mathrm{k}},$ ${\alpha }_1=2,{\alpha }_2=3n,{\alpha }_3=4p$

$$\begin{gathered} \mathrm{For} \mathrm{the} \mathrm{velocities} \mathrm{to} \mathrm{be} \mathrm{orthogonal}: \\ \overrightarrow{{\mathrm{V}}_{\mathrm{A}}}. \overrightarrow{{\mathrm{V}}_{\mathrm{B}}}=0 \\ \left({\mathrm{\alpha }}_1{\mathrm{t}}^2\right) {\mathrm{\alpha }}_1+\left({\mathrm{\alpha }}_2\mathrm{t}\right) {\mathrm{\alpha }}_2+\left({\mathrm{\alpha }}_3\right)\left({\mathrm{\alpha }}_3{\mathrm{t}}^2\right)=0 \\ \mathrm{At}, \mathrm{t}=1 \\ {\mathrm{\alpha }}_1^2+{\mathrm{\alpha }}_2^2+{\mathrm{\alpha }}_3^2=0 \\ \mathrm{Substitute} {\mathrm{\alpha }}_1=2,{\mathrm{\alpha }}_2=3\mathrm{n}, {\mathrm{\alpha }}_3=4\mathrm{p} \\ {2}^2+(3\mathrm{n}{)}^2+(4\mathrm{p}{)}^2=0 \\ 4+9{\mathrm{n}}^2+16{\mathrm{p}}^2=0 .... \left(1\right) \end{gathered}$$

$$\begin{gathered} \overrightarrow{{\mathrm{r}}_{\mathrm{A}}}=\int \overrightarrow{{\mathrm{V}}_{\mathrm{A}}} \mathrm{dt}=\int ({\mathrm{\alpha }}_1 {\mathrm{t}}^2 \hat{\mathrm{i}}+{\mathrm{\alpha }}_2 \mathrm{t} \hat{\mathrm{j}}+{\mathrm{\alpha }}_3 \hat{\mathrm{k}})\mathrm{dt} = {\mathrm{\alpha }}_1 \frac{{\mathrm{t}}^3}{3} \hat{\mathrm{i}}+{\mathrm{\alpha }}_2 \frac{{\mathrm{t}}^2}{2} \hat{\mathrm{j}}+{\mathrm{\alpha }}_3 \mathrm{t} \hat{\mathrm{k}} \\ \overrightarrow{{\mathrm{r}}_{\mathrm{B}}}=\int \overrightarrow{{\mathrm{V}}_{\mathrm{B}}} \mathrm{dt}=\int ({\mathrm{\alpha }}_1 \hat{\mathrm{i}}+{\mathrm{\alpha }}_2 \hat{\mathrm{j}}+{\mathrm{\alpha }}_3{\mathrm{t}}^2 \hat{\mathrm{k}})\mathrm{dt} = {\mathrm{\alpha }}_1 \mathrm{t} \hat{\mathrm{i}}+{\mathrm{\alpha }}_2 \mathrm{t} \hat{\mathrm{j}}+{\mathrm{\alpha }}_3\frac{{\mathrm{t}}^3}{3} \hat{\mathrm{k}} \\ \mathrm{At} \mathrm{t}=1 \mathrm{s} \\ \overrightarrow{{\mathrm{r}}_{\mathrm{A}}}= \frac{{\mathrm{\alpha }}_1}{3} \hat{\mathrm{i}}+ \frac{{\mathrm{\alpha }}_2}{2} \hat{\mathrm{j}}+{\mathrm{\alpha }}_3 \hat{\mathrm{k}} \\ \overrightarrow{{\mathrm{r}}_{\mathrm{B}}}= {\mathrm{\alpha }}_1 \hat{\mathrm{i}}+ {\mathrm{\alpha }}_2 \hat{\mathrm{j}}+\frac{{\mathrm{\alpha }}_3}{3} \hat{\mathrm{k}} \end{gathered}$$

$$\begin{gathered} \mathrm{Relative} \mathrm{displacement}: \\ r{⃗}_{AB}=r{⃗}_A-r{⃗}_B=\left(\frac{{\alpha }_1}{3}-{\alpha }_1\right)\hat{\mathrm{i}}+\left(\frac{{\alpha }_2}{2}-{\alpha }_2\right)\hat{\mathrm{j}}+\left({\alpha }_3\mathit{-}\frac{{\alpha }_3}{3}\right)\hat{\mathrm{k}} \\ r{⃗}_{AB}=\left(-\frac{\mathit{2}{\alpha }_1}{3}\right)\hat{\mathrm{i}}+\left(-\frac{{\alpha }_2}{2}\right)\hat{\mathrm{j}}+\left(\frac{\mathit{2}{\alpha }_3}{3}\right)\hat{\mathrm{k}} \\ \mathrm{Substitute} {\mathrm{\alpha }}_1=2,{\mathrm{\alpha }}_2=3\mathrm{n},{\mathrm{\alpha }}_3=4\mathrm{p} \\ r{⃗}_{AB}=\left(-\frac{4}{3}\right)\hat{\mathrm{i}}+\left(-\frac{\mathit{3}\mathrm{n}}{2}\right)\hat{\mathrm{j}}+\left(\frac{8\mathrm{p}}{3}\right)\hat{\mathrm{k}} \end{gathered}$$

Relative velocity is given by

$$\begin{gathered} \overrightarrow{{\mathrm{V}}_{\mathrm{AB}}}=\overrightarrow{{\mathrm{V}}_{\mathrm{A}}} - \overrightarrow{{\mathrm{V}}_{\mathrm{B}}} = \\ \overrightarrow{{\mathrm{V}}_{\mathrm{AB}}}={\mathrm{\alpha }}_1{\mathrm{t}}^2 \hat{\mathrm{i}} + {\mathrm{\alpha }}_2 \mathrm{t} \hat{\mathrm{j}} + {\mathrm{\alpha }}_3 \hat{\mathrm{k}} -{\mathrm{\alpha }}_1 \hat{\mathrm{i}} - {\mathrm{\alpha }}_2 \hat{\mathrm{j}} - {\mathrm{\alpha }}_3 {\mathrm{t}}^2 \hat{\mathrm{k}} \\ \mathrm{At} \mathrm{t} = 1 \mathrm{s} \\ \overrightarrow{{\mathrm{V}}_{\mathrm{AB}}} = 0 \hat{\mathrm{i}} + 0 \hat{\mathrm{j}} + 0 \hat{\mathrm{k}} \end{gathered}$$

The angular momentum of A with respect to B s givne by

$\overrightarrow{\mathrm{L}} = \mathrm{m}\left(\overrightarrow{{\mathrm{r}}_{\mathrm{AB}}} \times \overrightarrow{{\mathrm{V}}_{\mathrm{AB}}} \right) = \overrightarrow{0} \mathrm{As} \overrightarrow{{\mathrm{V}}_{\mathrm{AB}}} = 0$

using the Work-Energy Theorem

$W_8=k_f-k_i$

The work done by gravity is given as:

$MgL\sin \theta =k_f-k_i$

the total kinetic

$K.E.=\frac{1}{2}m{V}_{cm}^2+\frac{1}{2}{I}_{cm}{\omega }^2$

For a solid sphere, the moment of inertia $I=\frac{2}{5}m{R}^2$, and since $\omega =\frac{V_{cm}}{R}$, the equation becomes:

$K.E.=\frac{1}{2}m{V}^2+\frac{1}{2}\times \frac{2}{5}m{R}^2{\left(\frac{V}{R}\right)}^2$

This simplifies to:

$\text{ K.E. }=\frac{7}{10}m{V}^2$

Using the work-energy principle, we equate the potential energy to the change in kinetic energy:

$mgL\sin \theta =\frac{7}{10}m{V}^2$

This gives: $V^2\propto \sin \theta$
Now, for the two different angles of inclination ( $30°$ and $45°$ ), the ratio of the velocities at the bottom of the incline is:

$$\begin{gathered} {\left(\frac{V_1}{V_2}\right)}^2=\frac{\sin {\theta }_1}{\sin {\theta }_2}=\frac{\sin 30°}{\sin 45°}=\frac{1}{\sqrt{2}} \\ V_1^2:V_2^2=1:\sqrt{2} \end{gathered}$$

Given: $\theta =\frac{5t^4}{40}-\frac{t^3}{3}$

Angular velocity: $\omega =\frac{d\theta }{dt}=\frac{t^3}{2}-t^2$

Angular acceleration: $\alpha =\frac{d\omega }{dt}=\frac{3t^2}{2}-2t$

At t = 10s

$\alpha =\frac{3\left(100\right)}{2}-20=150-20=130rad/s^2$

$$\begin{gathered} \mathrm{The} \mathrm{velocity} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{v} = \mathrm{R\omega } \\ \mathrm{So}, \mathrm{the} \mathrm{ratio} \mathrm{is} \\ \frac{{\mathrm{K}}_{\mathrm{Trans}}}{{\mathrm{K}}_{\mathrm{Rot}}}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} {\mathrm{mv}}^2}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. {\mathrm{I\omega }}^2}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. {\mathrm{mv}}^2}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times {\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.} {\mathrm{mv}}^2}=\frac{5}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{The} \mathrm{velocity} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{v} = \mathrm{R\omega } \\ \mathrm{So}, \mathrm{the} \mathrm{ratio} \mathrm{is} \\ \frac{{\mathrm{K}}_{\mathrm{Trans}}}{{\mathrm{K}}_{\mathrm{Rot}}}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} {\mathrm{mv}}^2}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. {\mathrm{I\omega }}^2}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. {\mathrm{mv}}^2}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times {\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.} {\mathrm{mv}}^2}=\frac{5}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{The} \mathrm{velocity} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{v} = \mathrm{R\omega } \\ \mathrm{So}, \mathrm{the} \mathrm{ratio} \mathrm{is} \\ \frac{{\mathrm{K}}_{\mathrm{Trans}}}{{\mathrm{K}}_{\mathrm{Rot}}}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} {\mathrm{mv}}^2}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. {\mathrm{I\omega }}^2}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. {\mathrm{mv}}^2}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times {\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.} {\mathrm{mv}}^2}=\frac{5}{2} \end{gathered}$$