A 1.50-kg grinding wheel is in the form of a solid cylinder of radius 0.500 m. (

Question

A 1.50-kg grinding wheel is in the form of a solid cylinder of radius 0.500 m. (a) What constant torque will bring it from rest to an angular speed of 1000 rev/min in 3.0 s? N·m (b) Through what angle has it turned during that time? rad c) Use Eq(10.21) to calculate the work done by the torque. w = 402-81)-Tz (work done by a constant torque) (10.21) (d) What is the grinding wheel's kinetic energy when it is rotating at 1000 rev/min? Compare your answer to the result in part (c) The answer to part (c) is greater. The answer to part (d) is greater. The two answers are the same.

Explanation / Answer

initial angular speed wi = 1000 rev/min = 1000*2pi/60 rad/s = 104.7 rad/s

final angular speed wf = 0

time interval t = 3 s

angular acceleration alpha = (wf -wi)/t = 104.7/3 = 34.9 rad/s^2

torque = I*alpha

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

torque = (1/2)*1.5*0.5^2*34.9

torque = 6.54 Nm <<<<------ANSWER

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(b)

angle theta = (wi + wf)*t/2

theta = (104.7+0)3/2 = 157.05 rad <<<<-----ANSWER

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(c)

work = 6.54*157.05 = 1027 J

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(d)

Ki = (1/2)*I*wi^2 = (1/2)*(1/2)*1.5*0.5^2*104.7^2 = 1027 J

the two are the same

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