The 2.0 kg, 30-cm-diameter disk in Figure P12.66 is spinning at 300 rpm. How much friction force must the brake apply to the rim to bring the disk to a halt in 3.0 s?

Short Answer

Expert verified

Friction force must be applied by break to rim is : 1.57 N

Step by step solution

01

Given information

The mass of the disk i= 2 kg
The diameter of the disk i= 30 cm
This means radius = 15 cm = 0.15 m
Time taken to stop the disk is = 3 sec
Initial angular speed is = 300 rpm = 10 π rad/sec

Final speed =0

02

Explanation

Calculate moment of inertia of the disk by

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

Substitute the values we get

I=(2kg)(0.15m)22=0.0225kg.m2

Find the angular velocity using

ωf = ωi + αt ...................................(2)

So the angular acceleration is

α=ωf-ωit...............................(3)

Substitute the values in equation (3), we get

α=(0rad/s-10πrad/s)3s=-10.47rad/s2

Frictional force can be calculated by

f=Iαr..................................(4)Substitutevalues,f=(0.0225kg.m2)(10.47rad/s2)0.15m=1.57N

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