(6.1.60) A gold cylinder with thickness 2.00 cm and radius 10.0 cm (the lower ma

Question

(6.1.60) A gold cylinder with thickness 2.00 cm and radius 10.0 cm (the lower mass in the pictures below) is rotating counterclockwise at 700 rev/s. A silver cylinder with 4.00 cm thickness, radius 5.00 cm (the upper cylinder in the pictures below) , and rotating clockwise at 500 rev/s is dropped onto the gold cylinder and sticks toil in such a way that the two cylinders are coaxial after the collision. (a) Friction between the gold and silver cylinders causes them to come to a common final angular velocity. What is this final angular velocity? (b) Both cylinders begin the process with a temperature of 30 degree C. What is the final temperature?

Explanation / Answer

a) conservation of angular momentum

Iw + I w = ( I + I ) w

I = 1/2 M R^2
M = density * volume

density gold = 19300 kg/m^3
density of silver = 10490 kg/m^3

(0.5*(2.0E-2*pi*10.0E-2^2*19300)*10.0E-2^2)*700 + (0.5*(4.0E-2*pi*5.0E-2^2*10490)*5.0E-2^2)*-500 = ((0.5*(2.0E-2*pi*10.0E-2^2*19300)*10.0E-2^2) + (0.5*(4.0E-2*pi*5.0E-2^2*10490)*5.0E-2^2))*w

w= 623 rad/s

b)

first find change in kinetic energy

dKE = 0.5*(0.5*(2.0E-2*pi*10.0E-2^2*19300)*10.0E-2^2)*700^2 + 0.5(0.5*(4.0E-2*pi*5.0E-2^2*10490)*5.0E-2^2)*500^2- 0.5*(0.5*(2.0E-2*pi*10.0E-2^2*19300)*10.0E-2^2)*623^2 - 0.5(0.5*(4.0E-2*pi*5.0E-2^2*10490)*5.0E-2^2)*623^2

dKE = 2804 J

this will go into Q

Q = m silver c silver dT + m gold c gold dT

2804 = (2.0E-2*pi*10.0E-2^2*19300)*129*dT + (4.0E-2*pi*5.0E-2^2*10490)*240*dT

dT=1.19 C

Tfinal = 31.19 C

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