A 48.0 kg runner is standing on the outer edge of a horizontal turntable mounted

Question

A 48.0 kg runner is standing on the outer edge of a horizontal turntable mounted on a frictionless axle through its center. The turntable has a radius of 2.90 m and moment of Inertia of 500 kg*m2 and is currently at rest. The runner starts his run along the outer edge and is observed to run with a speed, relative to the earth of 4.60 m/s. If he runs for 60.0 seconds, how many revolutions does the turntable make?
Please show HOW to solve in detail, not just an answer. Thanks!
I know the answer involves conservation of angular momentum, but I'm clueless on how to solve this.

Explanation / Answer

conservation of angular momentum. (let w be angular velocity). w*I=48*v*2.9 so that w=1.28(rad/s). in 60 sec. phi=1.28*60=76.8(rad)=12.2(rev)

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