A circular platform is mounted on a vertical frictionless axle. Its radius is r = 2 m and its moment of inertia I = 200 kg m2. 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 upsilon0 = 1.0 m s-1 relative to the ground. The angular velocity of the platform is

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 m2. 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 upsilon0 = 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

Tardigrade Admin · Accepted Answer

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

∴

Angular momentum of man = Angular momentum of platform in opposite direction
i.e.,

m υ_{0} r = I ω

or

ω = \frac{m υ _{0} r}{I} = \frac{70 ( 1.0 ) ( 2 )}{200} = 0.7

rad

s^{- 1}

$1.2$ rad $s^{- 1}$

$0.4$ rad $s^{- 1}$

$0.7$ rad $s^{- 1}$

$2.0$ rad $s^{- 1}$

1.2 rad s−1

0.4 rad s−1

0.7 rad s−1

2.0 rad s−1

$1.2$ rad $s^{- 1}$

$0.4$ rad $s^{- 1}$

$0.7$ rad $s^{- 1}$

$2.0$ rad $s^{- 1}$