A 1.50kg grinding wheel is in the form of a solid cylinder of radius 0.150m . A)
ID: 1288769 • Letter: A
Question
A 1.50kg grinding wheel is in the form of a solid cylinder of radius 0.150m .
A) What constant torque will bring it from rest to an angular speed of 1200rev/min in 2.70s ?
B) Through what angle has it turned during that time?
C) Use equation W=?z(?2??1)=?z?? to calculate the work done by the torque.
D) What is the grinding wheel's kinetic energy when it is rotating at 1200rev/min ?
E) Compare your answer in part (D) to the result in part (C).
The results are the same.The results are not the same.
Explanation / Answer
A) I = 0.5*m*r^2
= 0.5*1.5*0.15^2
= 0.0169 kg.m^2
w1 = 1200*2*pi/60 = 125.6 rad/s
w2 = 0
alfa = (w2-w1)/t
= (0-125.6)/2.7
= 46.52 rad/s^2
Torque = I*alfa
= 0.0169*46.52
= 0.78 N.m
B) theta = wo*t - 0.5*alfa*t^2
= 125.6*2.7 - 0.5*46.52*2.7^2
= 169.6 rad
c) W = -Torque*theta
= -0.78*0.169.6
= -132.2 J
D) KE = 0.5*I*w1^2
= 0.5*0.0169*125.6^2
= 132.3 J
E)
the relults are same in magnitude.
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