NEET Angular Momentum and Conservation of Angular Momentum MCQs & PYQs

If We take Ring and 4 small blocks as one system net Torque will be zero, Using Principal of conservation of angular momentum.

\[ I_{1}\omega= \left( I_{1}+ I_{2} \right)\omega_{1} \]

\[ MR^{2}\omega= \left( MR^{2}+ 4mR^{2} \right)\omega_{1} \]

\[ \omega_{1}= \left( \frac{M}{M+4m} \right)\omega \]

Angular Momentum = momentum × ( Perpendicular distance of momentum from axis of rotation )

Angular Momentum = mv cos (45º) × h = mvh/√2

Distance of line 

\[ ax+by+c=0 \]  from point (x1,y1) is given by 

\[ d = \left( \frac{ax_{1}+ by_{1}+c}{\sqrt{a^{2}+b^{2}}} \right) \]

So, distance of direction of velocity from origin is d= 2√2 

Angular momentum = Perpendicular distance of momentum × momentum = 2√2 × 5 ×3√2= 60 Unit

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.

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 \(\omega = \frac{6v}{7\ell}\).

By conservation of angular momentum, \(I_1 \omega_1 = I_2 \omega_2\). Since \(I \propto R^2\) and \(\omega \propto 1/T\), we get \(T_2 = T_1 \left(\frac{R_2}{R_1}\right)^2 = 24 \left(\frac{1}{\sqrt{3}}\right)^2 = 8\text{ hrs}\).

Rotational kinetic energy is related to angular momentum by the formula $K = \frac{L^2}{2I}$, which yields $L = \sqrt{2KI}$. Since $K$ is the same for both bodies and it is given that $I_B > I_A$, it directly follows that $L_B > L_A$.

For angular momentum to be conserved, torque $\vec{\tau} = \vec{r} \times \vec{F}$ must be zero, meaning $\vec{r}$ and $\vec{F}$ are collinear. Taking the ratio of their components: $\frac{2}{\alpha} = \frac{-6}{3} = \frac{-12}{9}$, which simplifies to $\frac{2}{\alpha} = -2$, giving $\alpha = -1$.