Merry go round with a moment of inertia

In summary, the question asks for a step-by-step solution to finding the final angular velocity of a merry go round with a moment of inertia of 1000 kg m^2 when an 80-kg man steps onto the rim, 2m from the axis of rotation. The solution involves using the conservation of angular momentum and finding the moment of inertia for the "merry-go-round + man" system.
  • #1
phy21050
Would someone work through a step by step solution...Thank you. A merry go round with a moment of inertia of 1000kgm^2 is coasting at 2,20 rad/s. When a 80-kg man steps onto the rim, a distance of 2m from the axisof rotation the angular velocity decreases to ? rad/s
 
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  • #2
Originally posted by phy21050
Would someone work through a step by step solution...Thank you. A merry go round with a moment of inertia of 1000 kg m^2 is coasting at 2,20 rad/s. When a 80-kg man steps onto the rim, a distance of 2m from the axis of rotation the angular velocity decreases to ? rad/s

You know that the total angular momentum is constant correct? Well write it out.

Write an expression for the total angular momentum of the merry go round before the man stepped on it. After the man steps on it then once more write an expression for total angular momentum. Then solve for the final angular momentum.

Hint - you're going to need to fine the moment of inertia for the "merry-go-round + man" system.

Give it a try and then we'll go from where you get confused - post the equations that you go to up until you get stuck. Good luck!

Pete
 
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