Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Rotational Kinetic Energy, Work & Power›Question 16

MHT CET 2025 (22 Apr, Shift 2) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from MHT CET 2025 (22 Apr, Shift 2)

Q16MHT CET 2025 · 22 APR · SHIFT 2Medium

Rotational Kinetic Energy, Work & Power

A solid cylinder of mass ' MM ' and radius ' RR ' is rotating about its geometrical axis. A solid sphere of same mass and same radius is also rotating about its diameter with an angular speed half that of the cylinder. The ratio of the kinetic energy of rotation of the sphere to that of the cylinder will be

Show answer and solution

Answer: (c)

  1. Cylinder: Kc=12(12MR2)ω2=MR2ω24K_c = \frac{1}{2}\left(\frac{1}{2}MR^2\right)\omega^2 = \frac{MR^2\omega^2}{4}
  2. Sphere at ω2\frac{\omega}{2}: Ks=12(25MR2)ω24=MR2ω220K_s = \frac{1}{2}\left(\frac{2}{5}MR^2\right)\frac{\omega^2}{4} = \frac{MR^2\omega^2}{20}
  3. KsKc=15\frac{K_s}{K_c} = \frac{1}{5}
  4. So the answer is (c) 1:51 : 5.
Report a mistake

More questions on Rotational Kinetic Energy, Work & Power