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Chapter 12: Q. 67 (page 333)

A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop switch, a brake pad presses against the shaft and brings the turntable to a halt in 15 seconds. How much friction force does the brake pad apply to the shaft?

Short Answer

Expert verified

The friction force applied by the break pad is 0.52 N

Step by step solution

01

Given information

Mass of the turntable = 12.kg
Mass of the shaft =450 gm = 0.45 kg
The diameter of the turntable = 30 cm =0.3 m
radius = 0.15 m
The diameter of the shaft = 1.2 cm = 0.012 m
So radius of shaft =0.006m
Time taken to stop = 15 sec

The initial angular speed of the turntable =33 rpm = 3.45 rad/s

Final speed =0

02

Explanation

Moment of inertia of turn table ( consider disk) is

I=MR22..............................(1)

Substitute the values we get

I=(1.2kg)(0.15m)22=0.0135kg.m2

Find the retardation using

α=ωf-ωitsubstitutevalues,α=0rad/s-3.45rad/s15s=-0.23rad/s2

Now find the frictional force by equating torque

f=Iαrsubstitutevalues,f=(0.0135kg.m2)(0.23rad/s2)0.006m=0.52N

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