A 160 g, 47.0-cm-diameter turntable rotates onfrictionless bearings at 51.0 rpm.

Question

A 160 g, 47.0-cm-diameter turntable rotates onfrictionless bearings at 51.0 rpm. A 25.0 g block sits atthe center of the turntable. A compressed spring shoots the blockradically outward along a frictionless groove in the surface of theturntable. What is the turntable's rotation angularvelocity when the block reaches the outer edge? A 160 g, 47.0-cm-diameter turntable rotates onfrictionless bearings at 51.0 rpm. A 25.0 g block sits atthe center of the turntable. A compressed spring shoots the blockradically outward along a frictionless groove in the surface of theturntable. What is the turntable's rotation angularvelocity when the block reaches the outer edge? What is the turntable's rotation angularvelocity when the block reaches the outer edge?

Explanation / Answer

It = (1/2)mr2    = (1/2)160(47.0/2)2    =44180 g*cm2 Ib = mr2     = 25.0(47.0/2)2     = 13806.25 g*cm2 I = It + Ib = 57986.25g*cm2 2 = 1(44180/57986.25)       = 51(44180/57986.25)       = 38.857 rpm
2 = 1(44180/57986.25)       = 51(44180/57986.25)       = 38.857 rpm

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