Question

A man stands on a platform that is rotating (without friction) with an angular speed of 2.4 rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.2 kg · m2. By moving the bricks the man decreases the rotational inertia of the system to 2.0 kg · m2. (a) What is the resulting angular speed of the platform? rev/s

Answer 1

Answer:

The resulting angular speed of the platform is 7.44 rev/s.

Explanation:

Given that,

Speed = 2.4 rev/s

Moment of inertia consist of the man  = 6.2 kg-m²

Moment of inertia by the bricks= 2.0 kg-m²

We need to calculate the resulting angular speed of the platform

Using law of conservation of momentum

$L_{1}=L_{2}$

$I\omega_{1}=I\omega_{2}$

$\omega_{2}=\dfrac{I_{1}\omega_{1}}{I_{2}}$

Where,

$I_{1}$ = moment of inertia consist of  the man

$I_{2}$ = moment of inertia by the bricks

$\omega_{2}$ = angular speed of platform

Put the value into the formula

$\omega_{2}=\dfrac{6.2\times2.4\times2\pi}{2.0}$

$\omega_{2}=46.74\ rad/s$

$\omega_{2}=\dfrac{46.74}{2\pi}\ rev/s$

$\omega_{2}=7.44\ rev/s$

Hence, The resulting angular speed of the platform is 7.44 rev/s.