Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Rotational Kinetic Energy, Work & Power›Question 17

MHT CET 2025 (22 Apr, Shift 1) · Rotational Kinetic Energy, Work & Power

Rotational Kinetic Energy, Work & Power question from MHT CET 2025 (22 Apr, Shift 1)

Q17MHT CET 2025 · 22 APR · SHIFT 1Medium

Rotational Kinetic Energy, Work & Power

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

Show answer and solution

Answer: (b)

  1. Sphere: Ks=12(25mR2)ω2=mR2ω25K_s = \frac{1}{2}\left(\frac{2}{5}mR^2\right)\omega^2 = \frac{mR^2\omega^2}{5}
  2. Cylinder at 2ω2\omega: Kc=12(12mR2)(4ω2)=mR2ω2K_c = \frac{1}{2}\left(\frac{1}{2}mR^2\right)(4\omega^2) = mR^2\omega^2
  3. KsKc=15\frac{K_s}{K_c} = \frac{1}{5}
  4. So the answer is (b) 1:51 : 5.
Report a mistake

More questions on Rotational Kinetic Energy, Work & Power