Rotational Kinetic Energy = ½ I ω²= ½(2MR²)(2πn)²= 4π²MR²n²

\[ K = \frac{1}{2}I \omega^{2} \]

and

L = I ω ⇒  Substituting we get,

\[ I= \frac{L²}{2K} \]

During pure rolling the point on ground acts as instantaneous axis of rotation. Distance from the point of contact with ground determines speed of point. so,  

\[ V_{Q}> V_{C}> V_{P} \]

\[ L = I \omega \]

and

\[ K = \frac{1}{2} I \omega^{2} \]

Squaring L and Dividing it with K we get,

\[ I=  \frac{L^{2}}{2K} \]

For Rolling L = MvR + Iω = M(ωR)R + (MR²/2)ω = (3/2)MR²ω

The angular momentum of a projectile about the point of projection when it reaches its maximum height is given by:

$$L = m v_x r$$

where:

• 
  $m$ 

  = mass of the projectile

• 
  $v_x$ 

  = horizontal component of velocity

• 
  $r$ 

  = perpendicular distance from the point of projection (which is the maximum height

  $h$ 

  )

Step 1: Horizontal Component of Velocity

The initial velocity components are:

$$v_x = v \cos 45^\circ = \frac{v}{\sqrt{2}}$$

$$v_y = v \sin 45^\circ = \frac{v}{\sqrt{2}}$$

Since there is no acceleration in the horizontal direction (ignoring air resistance),

$v_x$

remains constant throughout the motion.

Step 2: Maximum Height

Using the kinematic equation:

$$v_y^2 = u_y^2 - 2 g h$$

At maximum height, the vertical velocity becomes zero, so:

$$0 = \left(\frac{v}{\sqrt{2}}\right)^2 - 2 g h$$

Solving for

$h$

:

$$h = \frac{v^2}{2 g} \cdot \frac{1}{2} = \frac{v^2}{4g}$$

Step 3: Angular Momentum Calculation

The angular momentum at maximum height is:

$$L = m v_x h$$

Substituting values:

$$L = m \left(\frac{v}{\sqrt{2}}\right) \left(\frac{v^2}{4g}\right)$$

$$L = m \cdot \frac{v}{\sqrt{2}} \cdot \frac{v^2}{4g}$$

$$L = \frac{m v^3}{4g \sqrt{2}}$$

Thus, the magnitude of the angular momentum about the point of projection at maximum height is:

$$\frac{m v^3}{4g \sqrt{2}}$$

Explanation:

• Torque (
  $\mathrm{<strong\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;\; display:\; inline\; !important;"><span\; class="katex"><span\; class="katex-html"\; aria-hidden="true"><span\; class="base"><span\; class="mord\; mathnormal">\tau </span></span></span></span>)\; is\; given\; by</strong><span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;">:</span>}$ 

   

  $$\tau =\frac{dL}{d\mathrm{t<span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;"></span>}}$$where

  $\mathrm{L <span\; style="font-family:\; Georgia,\; \text{'}Times\; New\; Roman\text{'},\; \text{'}Bitstream\; Charter\text{'},\; Times,\; serif;">is\; the\; angular\; momentum.</span>}$ 

• If
  $\tau = 0$ 

  , then: 

  $$\frac{dL}{dt} = 0 \Rightarrow L = \text{constant}$$L=constantThis means angular momentum is conserved.

We are given \(I_{\text{solid}} = I_{\text{hollow}} ⇒ \frac{2}{5}M R_s^2 = \frac{2}{3}M R_h^2\). Thus, \(\frac{R_s}{R_h} = \sqrt{\frac{5}{3}}\).

Using conservation of angular momentum: \(I_1\omega = (I_1 + I_2)\omega_f\). Since \(I_1 = \frac{1}{2}MR^2\) and \(I_2 = \frac{1}{2}M(R/2)^2 = \frac{1}{8}MR^2\), we get \(\omega_f = \frac{1/2}{1/2+1/8}\omega = \frac{4}{5}\omega\).

Torque \(\tau = I\alpha ⇒ 3 = 1 \times \alpha ⇒ \alpha = 3\text{ rad/s}^2\). After 1 second, angular velocity is \(\omega = \alpha t = 3\text{ rad/s}\). The linear velocity is \(v = \omega R = 3 \times 1 = 3\text{ ms}^{-1}\).