We can solve this problem using the principle of conservation of angular momentum since no external torque acts on the system.

Step 1: Initial Angular Momentum

The moment of inertia of a thin circular ring about its axis is:

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

The initial angular momentum is given by:

$$L_{\text{initial}} = I_{\text{initial}} \cdot \omega = (M R^2) \cdot \omega$$

Step 2: Final Moment of Inertia

When two objects of mass m are attached to the ring, assuming they are symmetrically placed on the ring, their contribution to the moment of inertia is:

$$I_{\text{added}} = 2m R^2$$

Thus, the new total moment of inertia becomes:

$$I_{\text{final}} = M R^2 + 2m R^2 = (M + 2m) R^2$$

Step 3: Applying Conservation of Angular Momentum

Since no external torque acts on the system:

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

$$(M R^2) \cdot \omega = (M + 2m) R^2 \cdot \omega'$$

Canceling

$R^2$

from both sides:

$$M \omega = (M + 2m) \omega'$$

Solving for the new angular velocity

$\omega'$

:

$$\omega' = \frac{M \omega}{M + 2m}$$

Final Answer:

$$\omega' = \frac{M \omega}{M + 2m}$$

This shows that the angular velocity decreases after attaching the masses, as expected due to an increase in the moment of inertia.

The angular momentum

$L$

of a particle about the origin is given by:

$$L = m v r \sin\theta$$

where:

• 
  $m = 5$ 

  g (mass of the particle),

• 
  $v = 3\sqrt{2}$ 

  cm/s (speed of the particle),

• 
  $r = 2\sqrt{5}$ 

  cm (perpendicular distance from the origin),

• 
  $\theta = 90^\circ$ 

  (since the velocity is along a straight line parallel to the x-axis, the perpendicular distance is directly used).

Since

$\sin 90^\circ = 1$

, the equation simplifies to:

$$L = m v r$$

Substituting the given values:

$$L = (5) \times (3\sqrt{2}) \times (2\sqrt{5})$$

$$L = 5 \times 3 \times 2 \times \sqrt{10}$$

$$L = 30 \sqrt{10} \text{ g-cm²/s}$$

Thus, the magnitude of the angular momentum is:

$$\mathbf{30\sqrt{10} \text{ g-cm²/s}}$$

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.

The angular acceleration is \(\alpha = \frac{\tau}{I} = \frac{100}{300} = \frac{1}{3}\text{ rad/s}^2\). The final angular velocity starting from rest is \(\omega = \alpha t = \frac{1}{3} \times 3 = 1\text{ rad/s}\).

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 \(v = \omega R = 3 \times 1 = 3\text{ ms}^{-1}\).

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 \(L = pR = \sqrt{2mK} R\).