Question

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

The friction force applied by the break pad is 0.52 N

Step-by-step solution

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

Explanation

Moment of inertia of turn table ( consider disk) is

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

Substitute the values we get

$I=\frac{(1.2kg){(0.15m)}^2}{2}=0.0135kg.m^2$

Find the retardation using

$$\begin{gathered} \alpha =\frac{{\omega }_f-{\omega }_i}{t} \\ \mathrm{substitute}\mathrm{values}, \\ \alpha =\frac{0rad/s-3.45rad/s}{15s}=-0.23rad/s^2 \end{gathered}$$

Now find the frictional force by equating torque

$$\begin{gathered} f=\frac{I\alpha }{r} \\ \mathrm{substitute}\mathrm{values}, \\ f=\frac{(0.0135kg.m^2)(0.23rad/s^2)}{0.006m}=0.52N \end{gathered}$$