Question

A grinding wheel is in the form of a uniform solid disk of radius 7.07 cm and mass 1.97 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.594 N · m that the motor exerts on the wheel.

(a) How long does the wheel take to reach its final operating speed of 1 150 rev/min? ___s
(b) Through how many revolutions does it turn while accelerating? ___ rev

Answer 1

We know that torque is given by:

T = I*alpha

I = moment of inertia of wheel = m*r^2/2

I = 1.97*0.0707^2/2 = 4.92*10^-3 kg-m^2

Now

alpha = angular acceleration

alpha = Torque/I = 0.594/(4.92*10^-3) = 120.73 rad/sec^2

Now using equation:

wf = wi + alpha*t

wi = 0 rad/sec

wf = 1150 rev/min = 1150*2*pi/60 = 120.43 rad/sec

t = (120.43 - 0)/120.73 = 0.997 sec

Part B

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

theta = 0*0.997 + 0.5*120.73*0.997^2

theta = 60.00 rad = 60/(2*pi) = 9.55 rev

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