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A 200 g, 40-cm-diameter turntable rotates on frictionless bearings at 60 rpm. A 20 g block sits at the center of the turntable. A compressed spring shoots the block radially outward along a frictionless groove in the surface of the turntable. What is the turntable’s rotation angular velocity when the block reaches the outer edge?

Short Answer

Expert verified

The turntable’s angular velocity is 5.76 rad / sec

Step by step solution

01

Given information

mass of turn table = 200 gm = 0.2 kg

Diameter of turntable = 40 cm = 0.4 m

so radius of turn table =0.2 m

mass of block = 20 gm =0.02 kg

rpm = 60.

covert rpm into rad/sec = 2π rad/sec

02

Explanation

Use the law of conservation for momentum

initial momentum = final momentum

Li= Lf..............................................(1)

Get initial momentum

Li= Iiωi.............................................(2)

Find moment of inertia

Ii=12(M+m)r2

Substitute this in equation(2)

Li=12(M+m)r2ωi........................(3)

Substitute the given value

Li=12×(0.2kg+0.02kg)(0.2m)2×2πrads-1=0.0044kgm2×2πrads-1.(4)

Now find the final momentum

Lf=IfωfIt+Iblockωf=12Mr2+mr2ωf(5)

Substitute the given values

Lf=12×0.2kg×(0.20m)2+0.02kg×(0.2m)2ωf=0.0048kgm2ωf...........................(6)

Equate (4) and (6) we get

ωf=0.0044kgm2×2πrads-10.0048kgm2=5.76rads-1

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