A child is standing with folded hands at the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The initial kinetic energy of the system is given by $K = \frac{1}{2} I \omega^2$,where $I$ is the initial moment of inertia and $\omega$ is the i

Vedclass · Accepted Answer

$\frac{K}{2}$

Vedclass · Answer

(A) The initial kinetic energy of the system is given by $K = \frac{1}{2} I \omega^2$,where $I$ is the initial moment of inertia and $\omega$ is the initial angular velocity.According to the law of conservation of angular momentum,since no external torque acts on the system,the angular momentum $L$ remains constant: $L = I \omega = I^{\prime} \omega^{\prime}$.Given that the new moment of inertia is $I^{\prime} = 2I$,we have $2I \omega^{\prime} = I \omega$,which implies $\omega^{\prime} = \frac{\omega}{2}$.The new kinetic energy $K^{\prime}$ is given by $K^{\prime} = \frac{1}{2} I^{\prime} \omega^{\prime 2}$.Substituting the values,$K^{\prime} = \frac{1}{2} (2I) \left( \frac{\omega}{2} \right)^2 = I \left( \frac{\omega^2}{4} \right) = \frac{1}{2} \left( \frac{1}{2} I \omega^2 \right) = \frac{K}{2}$.

$A$ child is standing with folded hands at the centre of a platform rotating about its central axis. The kinetic energy of the system is $K$. The child now stretches his arms so that the moment of inertia of the system becomes double. The kinetic energy of the system now is

Similar Questions

If the torque applied on a particle is zero,then its angular momentum will be:

$A$ person is rotating on a rotating table with their arms held down. Suddenly,they spread their arms out. What happens to their angular momentum?

$A$ force $\overrightarrow{F} = \alpha \hat{i} + 3\hat{j} + 6\hat{k}$ is acting at a point $\overrightarrow{R} = 2\hat{i} - 6\hat{j} - 12\hat{k}$. The value of $\alpha$ for which angular momentum about the origin is conserved is

Vedclass Test Series

Exam Paper Generator

Online Exam Module