Chapter 12: Q. 66 (page 333)

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

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

Explanation

Calculate moment of inertia of the disk by

$I = \frac{M R^{2}}{2} . . . . . . . . . . . . . . . . . . . . . . . . . (1)$

Substitute the values we get

$I = \frac{(2 k g) {(0.15 m)}^{2}}{2} = 0.0225 k g . m^{2}$

Find the angular velocity using

ω_f = ω_i + αt ...................................(2)

So the angular acceleration is

$α = \frac{ω_{f} - ω_{i}}{t} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)$

Substitute the values in equation (3), we get

$α = \frac{(0 r a d / s - 10 πrad / s)}{3 s} = - 10.47 r a d / s^{2}$

Frictional force can be calculated by

$f = \frac{I α}{r} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4) Substitute values, f = \frac{(0.0225 kg . m^{2}) (10.47 rad / s^{2})}{0.15 m} = 1.57 N$