A 1.60-kg grinding wheel is in the form of a solid cylinder of radius 0.500 m. W

Question

A 1.60-kg grinding wheel is in the form of a solid cylinder of radius 0.500 m. What constant torque will bring it from rest to an angular speed of 1450 rev/min in 1.5 s? Through what angle has it turned during that time? Use Eq. (10.21) to calculate the work done by the torque. What is the grinding wheel's kinetic energy when it is rotating at 1450 rev/min? Compare your answer to the result in part (c). The two answers are the same. The answer to part (d) is greater O The answer to part (c) is greater.

Explanation / Answer

moment of inerta, I = 0.5*m*r^2
I = 0.5*1.6*(0.5)^2
= 0.2 Kg.m^2

a)
wi = 1450 rev / min
= 1450*2*pi/60 rad/s
= 151.84 rad/s
use:
wf = wi + a*t
0 = 151.84 + a*1.5
a = -101.23 rad/s^2
Torque = I*a
= -0.2*101.23
= - 20.25 Nm
Negative sign shows that torque causing retardation
Answer: 20.25 Nm

b)
use:
angle = wi*t + 0.5*a*t^2
= 151.84*1.5 + 0.5*(-101.23)*(1.5)^2
= 113.88 rad
Answer: 113.88 rad

c)
W = T* (delta angle)
= 20.25 Nm * 113.88 rad
= 2306 J
Answer: 2306 J

d)
w = 1450 rev/min
= 1450*2*pi/60 rad/s
= 151.84 rad/s
Kinetic energy = 0.5*I*w^2
= 0.5*0.2*(151.84)^2
= 2306 J

The two answers are same

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