Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Angular Momentum & its Conservation›Question 56

JEE Main 2020 (3 Sep, Shift 1) · Angular Momentum & its Conservation

Angular Momentum & its Conservation question from JEE Main 2020 (3 Sep, Shift 1)

Q56JEE Main 2020 · 3 SEP · SHIFT 1Medium

Angular Momentum & its Conservation

A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its centre _________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 9

  1. Treat the platform as a disc: Ip=12(200)R2=100R2I_p = \frac{1}{2}(200)R^2 = 100R^2.
  2. Person on rim: Ii=100R2+80R2=180R2I_i = 100R^2 + 80R^2 = 180R^2; at centre: If=100R2I_f = 100R^2.
  3. Conservation of LL: Iini=Ifnf⇒nf=180100×5=9 rpmI_in_i = I_fn_f \Rightarrow n_f = \frac{180}{100}\times 5 = 9\ \text{rpm}
  4. Answer: 9 rpm
Report a mistake

More questions on Angular Momentum & its Conservation