A rough-surfaced turntable mounted on frictionless bearings initially rotates at

Question

A rough-surfaced turntable mounted on frictionless bearings initially rotates at 1.8 rev/s about its vertical axis. The rotational inertia of the turntable is 0.020 kgm2. A 200-g lump of putty is dropped onto the turntable from 0.0050 m above the turntable and at a distance of 0.15 m from its axis of rotation. The putty adheres to the surface of the turntable. The putty is initially at rest.

Part A: Find the initial kinetic energy of the turntable.

Part B: What is the final rotational speed of the system (the lump of putty and turntable)?

Part C: What is the final linear speed of the lump of putty?

Part D: Find the change in kinetic energy of the turntable.

Part E: Find the change in kinetic energy of the putty. The putty is initially at rest 0.0050 m above the turntable.

Part F: Find the change in kinetic energy of the putty-turntable combination. The putty is initially at rest 0.0050 m above the turntable.

Explanation / Answer

A)

w = 1.8*2*pi = 11.31 rad/s

KEi = 0.5*I*w^2 = 0.5*0.02*11.31^2 = 1.28 J

B)

initial velocity of putty u1 = sqrt(2*g*h) = sqrt(2*9.8*0.005) =0.313 m/s

initial angualar momentum Li = I*w + m*v*r

final angulr momentum lf = (I+mr^2)*wf

frommoemtum conservation

Lf = Li

(0.02+(0.2*0.15^2))*wf = (0.02*11.31)+(0.2*0.313*0.15)

wf = 9.61 rad/s

final rotational speed = 9.61 rad/s

+++++++++

C)

v = r*wf = 1.4415 m/s

+++++++

D)

dKE = 0.5*I*(wf^2-w^2)

dKE = 0.5*0.02*(9.61^2-11.31^2)

dKE = -0.35564 J

++++++++

E)

dKE = 0.5*m*r^2*wf^2 - 0.5*m*u^2

dKE = (0.5*0.2*0.15^2)-(0.5*0.2*0.313^2)

dKE =-0.0075469 J

++++++++++++++

F)

dKE = -0.3631869 J

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