PrepRung
Sign inSign up free

JEE Main 2016 · Physics · Rotational Motion · easy

A uniform disc of mass M and radius R rotates freely about its fixed symmetry axis (perpendicular to its plane) with angular speed ω. A lump of clay…

Single correct option

A uniform disc of mass MM and radius RR rotates freely about its fixed symmetry axis (perpendicular to its plane) with angular speed ω\omega. A lump of clay of mass M/3M/3 is dropped on the disc and sticks to its rim. The percentage of the initial kinetic energy of rotation that is lost is
  1. A. 25%25\%
  2. B. 33%33\%
  3. C. 40%40\%
  4. D. 60%60\%
Show answer and solution

Answer: (C) 40%40\%

No external torque acts about the axis, so angular momentum is conserved. Initially I1=12MR2I_1=\tfrac12MR^2. After the clay sticks at the rim, I2=12MR2+M3R2=56MR2I_2=\tfrac12MR^2+\dfrac M3R^2=\dfrac56MR^2. Then I1ω=I2ω′I_1\omega=I_2\omega' gives ω′=1/25/6ω=35ω\omega'=\dfrac{1/2}{5/6}\omega=\dfrac35\omega. The ratio of kinetic energies is 12I2ω′212I1ω2=5/61/2×925=35\dfrac{\tfrac12I_2\omega'^2}{\tfrac12I_1\omega^2}=\dfrac{5/6}{1/2}\times\dfrac{9}{25}=\dfrac35. So 60%60\% is retained and 40%40\% is lost, as heat in the sticking process. Note that the kinetic energy is not conserved, only L=IωL=I\omega.
Report a mistake

Original question written in the JEE Main 2016 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake