1.60 kg grinding wheel is in the form of a solid cylinder of radius 0.180 m . A)

ID: 1462315 • Letter: 1

Question

1.60 kg grinding wheel is in the form of a solid cylinder of radius 0.180 m .

A)What constant torque will bring it from rest to an angular speed of 1100 rev/min in 2.80 s ?

B)Through what angle has it turned during that time?

C)Use equation W=z(21)=z to calculate the work done by the torque.?

D)What is the grinding wheel's kinetic energy when it is rotating at 1100 rev/min ?

E) choose one of these ---> Compare your answer in part (D) to the result in part (C)? (*)The results are the same. (*)The results are not the sa

torque = I*alpha

alpha = wf -wi /t

wf = 1100 rev/min = 115.19 rad/s

alpha = 115.19- 0 /2.80 = 41.14 rd/s^2

I = inetria of solid cylinder = 1/2 * mr^2 = 0.02592 kg.m^2

torque = 1.066 N-m

part b )

dtheta = 1/2 * alpha*t^2

dtheta = 161.2688 rad

part c )

w = torque * dtheta

w = 1.066 * 161.2688 = 171.9 J

part d )

KE = 1/2 * Iw^2

KE = 1/2 * 0.02592 * (115.19)^2

KE = 171.9 J

results are same

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.