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A merry-go-round is rotating at an angular speed of 0.2 radians/s. Its motor falls off and it rotates freely. A technician jumps on the edge along the direction of the radius. The angular velocity after he lands is 0.04 radians/s. The moment of inertia of the technician, in (kg m2 ) with respect to the axis of the merry-go-roundā€™s axis of rotation is 5000 kg m2 . What is the moment of inertia of the merry-go-round?

Respuesta :

Answer:

Ā  Iā‚ =1250 Ā kg.mĀ²

Explanation:

Given that

Angular speed of Merry ,Ļ‰ā‚= 0.2 rad/s

Angular speed of technician ,Ļ‰ā‚‚= 0.04 rad/s

Moment of the inertia of the technician ,Iā‚‚= 5000 kg.mĀ²

Lets assume that

Moment of the inertia of merry with respected to the ground=Iā‚

There is no any external torque ,that is why angular momentum of the system will be conserve.

Now by conserving angular momentum

Iā‚ Ļ‰ā‚=(Iā‚+Iā‚‚)Ļ‰ā‚‚

Iā‚ Ā x 0.2 Ā = (Iā‚ +5000 ) x 0.04

Iā‚ (0.2-0.04) = 5000 x 0.04

[tex]I_1=\dfrac{5000\times 0.04}{0.2-0.04}[/tex]

Iā‚ =1250 Ā kg.mĀ²

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