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A 55kg runner runs around the endge of a horizontal turn table mounted on a vertical frictionless axis through its center. The runners velocity relative to the earth is 2.8m/s the turntable is rotating in the opposite direction with magnitude angular velocity 0.20rad/sec relative to the earth. The radius is 3.0m and the moment of inertia about the axis of rotation is 80 kg m^2. Find the final angular velocuty of the system if the runner comes to rest relative to the turn table.

I was thinking that I would start by finding the combined moment of inertia.

I_t = I + I_r = 80kg m^2 + (55kg)(3.0m)^2 = 575kg m^2

So the initial angular momentum is equal to the final angular momentum.

L_1 = L_2

But I guess I have to figure out his velocity relative to the turn table to be able to apply conservation of angular momentum. All I know is that the tangential accleration of the turn table is equal to ra (a is alpha).

So we know he has acceleration, and his final velocity relative to the turn table is zero, but how do I figure out his velocity relative to the turn table? I was thinking Galieliean Transformations of Velocitys, but does that apply with angular and linear velocities?

I have to figure this out.

Brian

I was thinking that I would start by finding the combined moment of inertia.

I_t = I + I_r = 80kg m^2 + (55kg)(3.0m)^2 = 575kg m^2

So the initial angular momentum is equal to the final angular momentum.

L_1 = L_2

But I guess I have to figure out his velocity relative to the turn table to be able to apply conservation of angular momentum. All I know is that the tangential accleration of the turn table is equal to ra (a is alpha).

So we know he has acceleration, and his final velocity relative to the turn table is zero, but how do I figure out his velocity relative to the turn table? I was thinking Galieliean Transformations of Velocitys, but does that apply with angular and linear velocities?

I have to figure this out.

Brian

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You firstly need to figure out the total angular momentum before they come to rest. That is done by adding the momentum of each seperate particle. The turn table have angular momentum L = Iw = 80 * 0,2 = 16,

while the runner have angular momentum L = Iw = 55*3^2*-0,8 = -396

(the angular velocity of the runner is calculated using v = wr, with tangential velocity 2,4 and radius 3, and remember to add the minus-sign, since the runner is running in the opossite direction).

When they come to rest, they are treated like a single object with a common moment of inertia of 575 (as you allready calculated). Then you just apply the law of conservation of angular momentum

L_1 = L_2

L_table + L_runner = L_rest

16-396 = 575 * w

w = (16-396) / 575

w = -0,68695

The minus-sign just shows that the table is now turning in the runners direction.