A circular platform is mounted on a frictionless vertical axle. Its radius R=2 ~m and its moment of inertia about the axle is 200 ~kg ~m^2 . It is initially at rest. A 50 ~kg man stands on the edge of the platform and

Correct answer: C. 2 π sec From conservation of angular momentum aligned I ω& =m v r 200 × ω& =50 ×2 ×1 ω& =(1)/(2) rad/ s aligned aligned & =r ω=1 ~m/ s T & =(2 π r)/(1-(-1))=(2 π r)/(2)= =2 π s aligned

NEET 2012

Exam: NEET
Year: 2012
Subject: Physics
Chapter: Rotational Motion

NEET 2012 · Physics · Rotational Motion

A circular platform is mounted on a frictionless vertical axle. Its...

A circular platform is mounted on a frictionless vertical axle. Its radius

R=2 \mathrm{~m}

and its moment of inertia about the axle is

200 \mathrm{~kg} \mathrm{~m}^2

. It is initially at rest. A

50 \mathrm{~kg}

man stands on the edge of the platform and begins to walk along the edge at the speed of

1 \mathrm{~ms}^{-1}

relative to the ground. Time taken by the man to complete one revolution is

Options

A $\pi \sec$
B $\frac{3 \pi}{2} \sec$
C $2 \pi \mathrm{sec}$
D $\frac{\pi}{2} \sec$

Correct answer

C. $2 \pi \mathrm{sec}$

Step-by-step solution

From conservation of angular momentum

\begin{aligned}I \omega& =m v r \\ 200 \times \omega& =50 \times2 \times1 \\ \omega& =\frac{1}{2} \mathrm{rad}/ \mathrm{s} \end{aligned}

\begin{aligned} \quadv & =r \omega=1 \mathrm{~m}/ \mathrm{s}\\ \therefore \quad T & =\frac{2 \pi r}{1-(-1)}=\frac{2 \pi r}{2}=\pir=2 \pi \mathrm{s} \end{aligned}

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