Question

A circular platform is mounted on a vertical frictionless axle. Its radius is $r = 2\, m$ and its moment of inertia $I = 200\, kg\, m^2$. It is initially at rest. A $70\, kg$ man stands on the edge of the platform and begins to walk along the edge at speed $\upsilon_{0} = 1.0 m\, s^{-1}$ relative to the ground. The angular velocity of the platform is

• A. $1.2$ rad $s^{-1}$
• B. $0.4$ rad $s^{-1}$
• C. $0.7$ rad $s^{-1}$
• D. $2.0$ rad $s^{-1}$

Answer 1

Answer: (C)

As the system is initially at rest, therefore, initial angular momentum, $L_i = 0.$
According to the law of conservation of angular momentum, final angular momentum, $L_f = 0$
$\therefore $ Angular momentum of man = Angular momentum of platform in opposite direction
i.e., $m\upsilon_{0}r =I\omega$
or $\omega=\frac{m\upsilon_{0}r}{I}=\frac{70\left(1.0\right)\left(2\right)}{200}=0.7$ rad $s^{-1}$