Question

A grinding wheel is in the form of a uniform solid disk of radius 7.07 cm and mass 2.06 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.590 N · m that the motor exerts on the wheel.

(a) How long does the wheel take to reach its final operating speed of 1 220 rev/min?
s

(b) Through how many revolutions does it turn while accelerating?
rev

Answer 1

Part A:

Torque = I*alpha

for grinding wheel, I = 0.5*M*R^2

alpha = Torque/(0.5*M*R^2)

R = 7.07 cm = 0.0707 m

M = 2.06 kg

alpha = 0.590/(0.5*2.06*0.0707^2)

alpha = angular acceleration = 114.60 rad/sec^2

Now given that

wi = initial angular velocity = 0 rad/sec

wf = 1220 rev/min = 1220*2*pi/60 = 127.76 rad/sec

wf = wi + alpha*t

t = (wf- wi)/alpha

t = (127.76 - 0)/114.60

t = 1.11 sec

Part B

theta = wi*t + 0.5*alpha*t^2

theta = 0*1.11 + 0.5*114.60*1.11^2 = 70.60 rad

theta = 70.60 rad/(2*pi) = 11.24 rev

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