14) A uniform solid brass cylinder of mass 0.50 kg and radius of 0.030 m starts
ID: 3308716 • Letter: 1
Question
14) A uniform solid brass cylinder of mass 0.50 kg and radius of 0.030 m starts to rotate about its cylinder axis with an angular velocity of 60 rad/s. Answer the questions below and show all work.
a. What is the angular momentum of the cylinder and how much work is required to reach this rate of rotation, starting from rest?
b. After the cylinder has reached 60 rad/s, it is heated up from room temperature (20oC) to 100oC. The mean coefficient of linear expansion of brass is 2.0 x 10-5 1/Co. What is the angular velocity after it is heated? What is the fractional change in angular velocity?
c. What is the angular momentum after it is heated? Explain why.
d. What is the kinetic energy after it is heated?
Explanation / Answer
(A) I = m r^2 / 2 = 0.50 x 0.03^2 / 2
I = 2.25 x 10^-4 kg m^2
and w = 60 rad /s
momentum, L = I w = 0.0135 kg m^2 / s ....Ans
Work done = change in KE
= I w^2 / 2 - 0
= (2.25 x 10^-4) (60^2)/ 2
= 0.405 J ........Ans
(B) r' = r [1 + alpha deltaT ]
r' = (0.030)[1 + (2 x 10^-5 x 80)]
r' = 0.030048 m
applying angular momentum conservation,
I w = I' w'
w' = I w / I'
and delta(w) = w' - w
= w (I - I') / I'
= w (-2 alpha deltaT )
= -60 x 2 x 2 x 10^-5 x 80
= -0.192 rad/s
(C) final angular mometum will be same as initial.
because expansion happen due to some internal forces.
such that net torque = 0
(d) delta(KE) = I w^2 / 2
= (0.50 x 0.030048^2 / 2) (60 - 0.192)^2 / 2
= 0.4037 J
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