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A runner of mass 60.0kg runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner's velocity relative to the earth has magnitude 2.50m/s . The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.190rad/s relative to the earth. The radius of the turntable is 3.60m , and its moment of inertia about the axis of rotation is 81.0kg*m2 .

A) Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can treat the runner as a particle.)
answer in rad/s please

Answer :

Answer:

See explanation

Explanation:

In this exercise, we need to use the law of conservation of angular momentum which is:

I1*W1 + I2*W2 = (I1 + I2)*W2'

Where:

I: moment of innertia.

W: angular velocity

Now let's call 1 the runner and 2, the turntable. the system would be W2'.

The angular speed of the runner, we can calculate that with the following expression:

W = V/r

so:

W1 = 2.5 / 3.6 = 0.694 rad/s

The innertia is calculated with the expression:

I = m*r²

I = 60 * 3.6 = 216 kg.m²

I2 and W2 are provided in the exercise, so, replacing all the data in the conservation of angular momentum, let's solve for W2'

(216*0.694) + (-0.190*81) = (81 + 216)W2'

134.514 = 297W2'

W2' = 134.514 / 297

W2' = 0.453 rad/s

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