As object is in circular motion angular momentum

\[ \vec L=I\vec\omega \]

Direction of omega is along the axis so, L will have direction along axis OC. So both magnitude

and direction of angular momentum L is constant.

Taking rectangle and the object as one system angular momentum is conserved.

so,  mva= ( m(√5a/2)² + m((2a)²+ a²)/12) ω

\[ \omega = \frac{3u}{5a} \]

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

Conserving angular momentum about pivot: \(L_i = mv\frac{\ell}{2}\). Total moment of inertia is \(I = \frac{1}{3}m\ell^2 + m(\ell/2)^2 = \frac{7}{12}m\ell^2\). Equating \(I\omega = L_i\) yields \(\omega = \frac{6v}{7\ell}\).

For translational and rotational equilibrium of the ladder, taking torque about the base gives \(N_{\text{wall}} \ell \sin\theta = mg \frac{\ell}{2} \cos\theta\). With \(N_{\text{wall}} = f \le \mu mg\), we get the minimum angle for equilibrium to be \(\tan\theta = \frac{1}{2\mu}\).

The distance of masses \(m\) and \(2m\) from the midpoint of their side is \(L/2\). The third mass \(3m\) lies at a distance of \(h = \frac{\sqrt{3}}{2}L\) (the height of the triangle). The total moment of inertia is \(I = m\left(\frac{L}{2}\right)^2 + 2m\left(\frac{L}{2}\right)^2 + 3m\left(\frac{\sqrt{3}}{2}L\right)^2 = \frac{mL^2}{4} + \frac{2mL^2}{4} + \frac{9mL^2}{4} = 3mL^2\).