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soldi70 [24.7K]
3 years ago
14

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
Physics
1 answer:
andrey2020 [161]3 years ago
4 0

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.

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