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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

Respuesta :

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

[tex]L_{1}=L_{2}[/tex]

[tex]I\omega_{1}=I\omega_{2}[/tex]

[tex]\omega_{2}=\dfrac{I_{1}\omega_{1}}{I_{2}}[/tex]

Where,

[tex]I_{1}[/tex] = moment of inertia consist of  the man

[tex]I_{2}[/tex] = moment of inertia by the bricks

[tex]\omega_{2}[/tex] = angular speed of platform

Put the value into the formula

[tex]\omega_{2}=\dfrac{6.2\times2.4\times2\pi}{2.0}[/tex]

[tex]\omega_{2}=46.74\ rad/s[/tex]

[tex]\omega_{2}=\dfrac{46.74}{2\pi}\ rev/s[/tex]

[tex]\omega_{2}=7.44\ rev/s[/tex]

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

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