Angular Velocity of a turntable-man system

AI Thread Summary
The discussion revolves around calculating the angular velocity of a turntable-man system where a turntable makes one revolution every 10 seconds, and a man runs outward from the center. The initial angular velocity is calculated by converting the revolution time into radians per second. The initial angular momentum is derived using the formula L = Iw, where I is the moment of inertia. As the man moves 2 meters from the center, the new moment of inertia is calculated by adding the man's contribution to the turntable's inertia. The principle of conservation of angular momentum is emphasized, stating that since no external torque acts on the system, angular momentum remains constant.
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Homework Statement


A turntable rotates about a fixed vertical axis, making one revolution in 10s. The moment of inertia of the turntable about the axis is 1600kgm^2. A man, James Bond, of mass 120kg, is at the center of the turntable and begins to run out along a radius of the device. Treat the man as a point object.


Homework Equations


L = Iw


The Attempt at a Solution


There are three parts. The first part asks for the initial angular velocity, which is simply converting 1rev/10s into rad/s. The second part asks for the initial angular momentum, which would be L = Iw?

The third part is asking for the angular velocity of the turntable-man system when the man is 2m from the center of the turntable. It also says at this point the man is at rest with respect to the turntable.

Now I run into my problems. I'm not too sure about how to calculate this, but one method I'm thinking of is recalculating the inertia to be 1600kgm^2 + mr^2 where m and r are referring to the man (so 120kg and 2m respectively). This would give an inertia of 1840kgm^2.

I thought about using L = r (cross) p, but solving for the linear momentum would mean finding Vtan = wr? How would I find w in this case?
 
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Since there is no external torque acting on the tern table, angular momentum remains constant.
 
I was considering that approach but wasn't sure. Thanks. However, my approach to the new inertia would be correct, I assume?
 
Yes.
 
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