Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Angular Momentum & its Conservation›Question 40

MHT CET 2023 (12 May, Shift 1) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from MHT CET 2023 (12 May, Shift 1)

Q40MHT CET 2023 · 12 MAY · SHIFT 1Medium

Angular Momentum & its Conservation

A thin uniform circular disc of mass 'MM' and radius 'RR' is rotating with angular velocity 'ω\omega', in a horizontal plane about an axis passing through its centre and perpendicular to its plane. Another disc of same radius but of mass (M2)\left(\frac{M}{2}\right) is placed gently on the first disc co-axially. The new angular velocity will be

Show answer and solution

Answer: (a)

  1. Angular momentum is conserved: I1ω=(I1+I2)ω′I_1\omega = (I_1+I_2)\omega'.
  2. Both discs have I=12(mass)R2I = \frac{1}{2}(\text{mass})R^2, so ω′=MM+M/2 ω\omega' = \frac{M}{M + M/2}\,\omega.
  3. ω′=23ω\omega' = \frac{2}{3}\omega
  4. So the answer is (a) 23ω\frac{2}{3}\omega.
Report a mistake

More questions on Angular Momentum & its Conservation