A brake pad is used to give a normal force of 150 N to the outer surface of a 0.25 kg spinning disk, in order to slow it down from 250 rpm to rest. The disk's radius is 0.25 m. If the coefficient of friction between the pad and the disk is 0.45, how much time is required for the disk to slow from 250 rpm to rest?
A) 0.19 ms
B) 0.55 ms
C) 2.4 ms
D) 12 ms

Question

Quizplus · Accepted Answer

First, we need to find the initial angular velocity of the disk in radians per second:

w = (250 rpm) * (2*pi radians/rev) / (60 s/min) = 26.18 rad/s

The torque on the disk due to the brake pad can be found as:

τ = r*F*f = (0.25 m)*(150 N)*(0.45) = 16.88 N*m

We can use the torque and the moment of inertia of the disk to find the angular acceleration of the disk:

τ = I*α  =>  α = τ/I

The moment of inertia of a disk is given by:

I = (1/2)*m*r^2

Substituting the given values:

I = (1/2)*(0.25 kg)*(0.25 m)^2 = 0.0078 kg*m^2

α = (16.88 N*m) / (0.0078 kg*m^2) = 2164 rad/s^2

Now we can use kinematic equations to find the time required for the disk to stop rotating:

ωf = ωi + α*t

0 = 26.18 rad/s + (-2164 rad/s^2)*t

t = 0.012 s

So the time required for the disk to slow from 250 rpm to rest is 12 ms.

A Brake Pad Is Used to Give a Normal Force