Merry go round with a moment of inertia

AI Thread Summary
To solve the problem, start by recognizing that the total angular momentum of the system is conserved. Initially, the angular momentum of the merry-go-round is calculated using its moment of inertia (1000 kg m²) and its angular velocity (2.20 rad/s). When the 80-kg man steps onto the rim, the new moment of inertia becomes the sum of the merry-go-round's moment of inertia and the man's contribution, which is calculated as 80 kg multiplied by the square of the distance from the axis (2 m). Set the initial angular momentum equal to the final angular momentum to find the new angular velocity after the man steps on. This approach will lead to the solution for the angular velocity after the man steps on the merry-go-round.
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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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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