Angular Momentum and Conservation of Angular Momentum

Practice Questions:

Question 1: moderate

A thin circular ring of mass M and radius R rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity . Four small spheres each of mass m (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

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

2. \[ \left( \frac{M}{4m} \right)\omega \]

3. \[ \left( \frac{M+4m}{M} \right)\omega \]

4. \[ \left( \frac{M}{M-4m} \right)\omega \]

View Answer

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

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

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

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

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Question 2: moderate

A particle is projected with a speed v at 45° with the horizontal. The magnitude of angular
momentum of the projectile about the point of projection when the particle is at its maximum height h is

1. Zero

2. mvh²/√2

3. mv²h/2

4. mvh/√2

View Answer

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

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

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Question 3: moderate

A particle of mass m = 5 units is moving with a uniform speed v = 3√2 m in the XOY plane along the line Y = X + 4. The magnitude of the angular momentum about origin is

1. Zero

2. 60 unit

3. 7.5 unit

4. 40√2 unit

View Answer

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

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Question 4: moderate

A conical pendulum consists of a simple pendulum moving in a horizontal circle as shown. C is the pivot, O the centre of the circle in which the pendulum bob moves and ω the constant angular velocity of the bob. If L is the angular momentum about point C, then

1. L is constant ( magnitude as well as direction is constant )

2. Only direction of L is constant

3. Only magnitude of L is constant

4. None of the above.

View Answer

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.

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Question 5: moderate

A uniform rectangular plate of mass m which is free to rotate about the smooth vertical hinge passing through the centre and perpendicular to the plate, is lying on a smooth horizontal surface. A particle of mass m moving with speed ‘u’ collides with the plate and sticks to it as shown in figure. The angular velocity of the plate after collision will be

1. \[ \frac{12u}{5a} \]

2. \[ \frac{12u}{19a} \]

3. \[ \frac{3u}{2a} \]

4. \[ \frac{3u}{5a} \]

View Answer

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

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

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

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Question 6: moderate

A thin circular ring of mass M and radius R is rotating about its axis with a constant angular velocity ω. Two objects of mass ‘m’ are attached gently to the ring. The wheel now rotates with an angular velocity.

1. \[ \frac{\omega M }{(M+m)} \]

2. \[ \frac{\omega(M-2m)}{(M+ 2m)}\]

3. \[ \frac{\omega M}{M+2m}\]

4. \[ \frac{\omega (M+2m)}{M}\]

View Answer

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.

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Question 7: moderate

A uniform rod of mass $m$ and length $\ell$ is pivoted about one end and hung vertically. Another mass $m$ hits it perpendicular to its length with a velocity $v$ at its midpoint and sticks to it. The initial angular velocity of the rod is:

1. $\frac{v}{\ell}$

2. $\frac{v}{2\ell}$

3. $\frac{6v}{7\ell}$

4. $\frac{v}{3\ell}$

View Answer

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 $\omega = \frac{6v}{7\ell}$.

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