a 1000 kg flywheel with all its mass at a radius of .5m is spinning at a 1250 re

Question

a 1000 kg flywheel with all its mass at a radius of .5m is spinning at a 1250 revolution per minute. as an emergency brake a board is pressed radially against its rim with a force of 120 N. The coefficient of friction between the board and the rim is 0.4. You must give numerical answers. A formula will get you no credit.

a.) What is the moment of inertia of the flywheel. I=?

b.) What is the magnitude of the torque exerted on the flywheel by the board? T=?

c.) What is the magnitude of the angular acceleration of the flywheel?

d.) How long does it take for the wheel to stop?

Explanation / Answer

(a) supposing wheel to be disk,]

I = (1/2) M R^2 = .5*1000*.5^2

=125 kgm^2

(b) Torque produced by bored,

T = F*R = 120*.5 = 60 Nm

I alpha= T=60

Angular acceleration,

alpha=60/I =60/125 = 0.48 rad/s^2

Angular speed

wo = 1250*2*pi/60

=65.41 rad/s

Now as we know,

w-wo = (1/2)*(-alpha)*(t^2)

t= time need to for complete stop, w=0

65.41*2/alpha =t^2

t2=65.41*2/.48

t= 272.54 seconds

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