\[ L = I \omega \]

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.

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 \(\omega_f = \frac{1/2}{1/2+1/8}\omega = \frac{4}{5}\omega\).

As polar ice melts and water flows towards the equator, mass is distributed further from the rotational axis, increasing the moment of inertia \(I\). Due to conservation of angular momentum, the angular velocity \(\omega\) decreases, which increases the duration of the day.

Kinetic energy \(K = \frac{p^2}{2m}\) gives momentum \(p = \sqrt{2mK}\) . Angular momentum is given by \(L = pR = \sqrt{2mK} R\).

No external torque means angular momentum is conserved (Reason is true). Kinetic energy may not be conserved as internal forces can change it (Assertion is false).

The angular momentum is \(L = m v d\), where \(d\) is the constant perpendicular distance of the line of motion from the origin. Since \(m\), \(v\), and \(d\) are all constant, \(L\) remains constant.

Since rotational kinetic energy \(K = \frac{L^2}{2I}\), we have \(L \propto \sqrt{K}\). An increase of 300% means \(K_f = 4K_i\), so \(L_f = 2L_i\). The percentage increase in angular momentum is 100%.