Question

A uniform circular disc of radius R, lying on a frictionless horizontal plane is rotating with an angular velocity '$\omega$' about its own axis. Another identical circular disc is gently placed on the top of the first disc coaxially. The loss in rotational kinetic energy due to friction between the two discs, as they acquire common angular velocity is:
(I is Moment of Inertia of the disc)

• A. $\frac{1}{8}I{\omega }^2$
• B. $\frac{1}{4}I{\omega }^2$
• C. $\frac{1}{2}I{\omega }^2$
• D. $I{\omega }^2$

Answer 1

Answer: (B)

$\frac{1}{4}I{\omega }^2$

Since no external torque acts on the entire system, angular momentum will remain conserved. Hence,

$I\omega =2I{\omega }^{\prime }$

${\omega }^{\prime }=\frac{\omega }{2}$

Loss in kinetic energy is $\Delta KE=\frac{1}{2}I{\omega }^2-\frac{1}{2}\left(2I\right){\left(\frac{\omega }{2}\right)}^2=\frac{1}{4}I{\omega }^2$.