Amerry-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-round’s axis of rotation is 5000 kg m2 . what is the moment of inertia of the merry-go-round?

  I₁ =1250  kg.m²

Explanation:

Given that

Angular speed of Merry ,ω₁= 0.2 rad/s

Angular speed of technician ,ω₂= 0.04 rad/s

Moment of the inertia of the technician ,I₂= 5000 kg.m²

Lets assume that

Moment of the inertia of merry with respected to the ground=I₁

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

Now by conserving angular momentum

I₁ ω₁=(I₁+I₂)ω₂

I₁  x 0.2  = (I₁ +5000 ) x 0.04

I₁ (0.2-0.04) = 5000 x 0.04

I₁ =1250  kg.m²

an energy transformation flow diagram is shown

where is it ?