A free not spinning disk (m1 = 3.0 kg, r1 = 20 cm) is dropped from a height of 2

Question

A free not spinning disk (m1 = 3.0 kg, r1 = 20 cm) is dropped from a height of 25 cm symmetrically onto a freely spinning disk (m2 = 8 kg, r2 = 30 cm) the rim of which is spinning at a tangential velocity of Vt = 12m/s. The kinetic coefficient of friction between them is uk = 0.35.

a) What is their joint angular velocity after the top disk comes up to speed with the lower disk?

b) What is the final tangential speed of each disk at their rim?

c) Is the collision elastic? If not, how much energy is lost?

d) How much mechanical work (J) has the dropped disk done on the spinning disk after it comes up to speed?

Explanation / Answer

a) initial angular speed of spinning disk , w1

w1 = Vt/r2

w1 = 12/0.30 m/s = 40 rad/s

Now , for the joint angular speed wf

Using conservation of angular momentum

(0.5 * m1 * r1^2 + 0.5 * m2 * r2^2) * wf = 0.5 * m1 * r1^2 * w1

(3 * 0.20^2 + 8 * 0.30^2) * wf = 8 * 0.30^2 * 40

solving for wf

wf = 34.3 rad/s

their joint angular velocity after the top disk comes up to speed with the lower disk is 34.3 rad/s

part b)

final tangential speed of disk 1 = r1 * wf

final tangential speed of disk 1 = 34.3 * 0.20 m/s

the final tangential speed of disk 1 is 6.86 m/s

final tangential speed of disk 2 = r2 * wf

final tangential speed of disk 2 = 34.3 * 0.30 m/s

the final tangential speed of disk 2 is 10.3 m/s

part c)

NO , the collision is not elastic

part d)

mechanical work done on the rotating disk = kinetic energy of lost by rotating disk

mechanical work done on the rotating disk = 0.5 * 8 * 0.30^2 * (34.3^2 - 40^2)

mechanical work done on the rotating disk = -152.5 J

the mechanical work done on the rotating disk is -152.5 J

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