Question

A 30-cm-diameter, 1.4 kg solid turntable rotates on a 1.8-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?

Answer 1

Answer:

f = 0.40N

Step-by-step explanation:

Diameter of turntable = 30cm

Radius = 15cm = 0.15m

Mass of turntable = 1.4kg

Diameter of shaft = 1.8cm

Radius = 0.9cm = 0.009m

Angular velocity (w) = 33 rpm

t = 15secs

wi = initial angular velocity

wf = final angular velocity

Convert wi from rev/min to rad/s

wi = 33*2π/60

= 3.5rad/s

wf = 0

We will use the rotational kinematic equation

wf = wi + αΔt

0 = 3.5 + (15)α

α = -3.5/15

α = -0.23rad/s^2

Recall that

I = 1/2MR^2

I = moment of inertia

M = mass

R= radius

for turntable

I= 1/2 *1.4*0.15^2

= 15.75*10^-3 kg. m^2

For shaft

I = 1/2* 0.45*0.009^2

= 1.822*10^-5 kg. m^2

Total moment of inertia = 15.75*10^-3 + 1.822*10^-5

= 15.77 * 10^-3 kg.m^2

We then calculate the torque exerted on the shaft when brake is applied

α = Tnet/I

-0.23 = T/ 15.77*10^-3

T = -0.23 * 15.77*10^-3

T = -3.63*40^-3 Nm

Since friction force is always opposite to the motion Direction, it would be in the clockwise direction and tangential to the shaft

β = -90°

Also recall that

T = rFsinβ

F= frictional force

F = T /rsinβ

F = -3.63*10^-3/0.009sin(-90)

F = 0.40N