Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Rotational Kinematics & Dynamics›Question 19

MHT CET 2025 (25 Apr, Shift 2) · Rotational Kinematics & Dynamics

Rotational Kinematics & Dynamics question from MHT CET 2025 (25 Apr, Shift 2)

Q19MHT CET 2025 · 25 APR · SHIFT 2Medium

Rotational Kinematics & Dynamics

A bob of mass ' mm ' is tied by a massless string whose other end is wound on a flywheel (disc) of radius ' RR ' and mass ' mm '. When released from the rest, the bob starts falling vertically downwards. If the bob has covered a vertical distance ' hh ', then angular speed of wheel will be (There is no slipping between string and wheel, g - acceleration due to gravity)

Show answer and solution

Answer: (a)

  1. Energy: mgh=12mv2+12(12mR2)ω2mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)\omega^2 with v=ωRv = \omega R.
  2. mgh=34mω2R2⇒ω2=4gh3R2mgh = \frac{3}{4}m\omega^2R^2 \Rightarrow \omega^2 = \frac{4gh}{3R^2}
  3. ω=1R4gh3=2Rgh3\omega = \frac{1}{R}\sqrt{\frac{4gh}{3}} = \frac{2}{R}\sqrt{\frac{gh}{3}}
  4. So the answer is (a).
Report a mistake

More questions on Rotational Kinematics & Dynamics