Rotational Kinetic Energy - NEET Physics Questions
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Rotational Kinetic Energy

Question 1: easy

A circular ring of wire of mass M and radius R is making n revolutions/sec about an axis passing through a point on its rim and perpendicular to its plane. The kinetic energy of rotation of the ring is given by

1. 4π²MR²n²
2. 2π²MR²n²
3. π²MR²n²/2
4. 8π²MR²n²
View Answer

Rotational Kinetic Energy = ½ I ω²= ½(2MR²)(2πn)²= 4π²MR²n²

Question 2: easy

What is moment of inertia in terms of angular momentum (L) and kinetic energy (K)

1. \[ \frac{L²}{K} \]
2. \[ \frac{L²}{2K} \]
3. \[ \frac{L}{2K^{2}} \]
4. \[ \frac{L}{2K} \]
View Answer

\[ K = \frac{1}{2}I \omega^{2} \]

and

L = I ω ⇒  Substituting we get,

\[ I= \frac{L²}{2K} \]

Question 3: easy

If moment of inertia of a spinning object drops to \(\left(\frac{1}{4}\right)^{\text{th}}\) of its initial value, the ratio of new rotational kinetic energy to initial rotational kinetic energy will be (Assume net external torque about the axis of rotation is zero)

1. 1 : 4
2. 4 : 1
3. 2 : 1
4. 1 : 2
View Answer

Since external torque is zero, angular momentum \(L = I\omega\) is conserved. Rotational kinetic energy is \(K = \frac{L^2}{2I}\). If \(I' = I/4\), then \(K' = 4K\), so \(K' : K = 4 : 1\).

Question 4: easy

Assertion (A): Kinetic energy of a rigid body can be greater than \( \frac{1}{2}mv^2 \), where \( m \) is mass of rigid body & \( v \) is speed of centre of mass of body.


Reason (R): Kinetic energy of a particle (point mass) cannot be greater than \( \frac{1}{2}mv^2 \), where \( m \) is mass of particle & \( v \) is speed of particle.

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

The total kinetic energy of a rigid body is \( K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \), where the second term is rotational KE. For a particle, \( K = \frac{1}{2}mv^2 \) only. Therefore, a rigid body's KE can be greater than \( \frac{1}{2}mv^2 \).


Both A and R are true, and R explains A by highlighting the difference in KE components.

Question 5: easy

If moment of inertia of a spinning object drops to \( \left(\frac{1}{4}\right)^{\text{th}} \) of its initial value, the ratio of new rotational kinetic energy to initial rotational kinetic energy will be (Assume net external torque about the axis of rotation is zero)

1. 1 : 4
2. 4 : 1
3. 2 : 1
4. 1 : 2
View Answer

Under zero external torque, angular momentum is conserved: \( I_1 \omega_1 = I_2 \omega_2 \). If \( I_2 = I_1/4 \), then \( \omega_2 = 4\omega_1 \). The ratio of kinetic energy is \( \frac{K_2}{K_1} = \frac{\frac{1}{2}I_2\omega_2^2}{\frac{1}{2}I_1\omega_1^2} = \frac{1}{4} \times 16 = 4 \implies 4 : 1 \).