Question

A particle performs uniform circular motion with an angular momentum L. If the frequency of particles motion is doubled and its KE is halved. the angular momentum becomes

• A. $\frac{L}{2}$
• B. $2L$
• C. $4L$
• D. $\frac{L}{4}$

Answer 1

Answer: (D)

The angular momentum is measure for the amount of torque that has been applied over time to the object. For a particle with a fixed mass that is rotating about a fixed symmetry axis, the angular momentum is expressed as $L=I\omega$ ....(i) where I is moment of inertia of particle and $\omega$ the angular velocity. Also, $K=\frac{1}{2}I{\omega }^2$ ...(ii) where K is kinetic energy of rotation. From Eqs. (i) and (ii), we get $L=\frac{2K}{{\omega }^2}\omega =\frac{2K}{\omega }$ $L=\frac{2(K/2)}{2\omega }=\frac{1}{4}\left(\frac{2K}{\omega }\right)=\frac{L}{4}$ Note: In a closed system angular momentum is constant.