Question: A solid, uniform, horizontal disc with a diameter of 2.0 m and a mass

Question

Question: A solid, uniform, horizontal disc with a diameter of 2.0 m and a mass of 4.0 kg rotates at 36 rpm about a vertical axis through its center. A small 0.50 kg piece of putty is dropped onto the disc and sticks at a distance of 85 cm from the axis of rotation. The figure shows before and after views. a) What is the moment of inertia about the rotational axis of: i. the disc before the putty sticks; ii. the disc+putty after the putty sticks? b) What is the angular velocity of the disk (in rpm) after the putty sticks? c) Determine the angular momentum about the rotational axis of only the disc i. before the putty is dropped i. after the putty has stuck d) What was the average frictional force exerted by the putty on the disc if it took 12 milliseconds for the disc to slow once the putty stuck to it? Hint: use your answer to (c) to help.

Explanation / Answer

M=4.0kg,R=1.0m, wi=36rpm = (36*2rad)/(60s) = 3.77rad/s , m=0.50kg,r=0.85m

a)

i)

Ii= 1/2MR^2=1/2*4.0*1.0^2 = 2.0 kg*m^2

ii) If = 1/2MR^2 +mr^2=1/2*4.0*1.0^2+0.50*0.85^2 = 2.36 kg*m^2

b)

By law of conservation of angular momentum

Lf=Li

If*wf=Ii*wi

Plug values,

2.36*wf=2.00*36

wf= 30.5rpm

c)

i)

Li(disk)= Ii*wi = 2.0kg*m^2*3.77rad/s = 7.54 kg*m^2/s

ii)

Lf(disk)= Ii*wi = 2.0kg*m^2*(30.5*2rad)/(60s) = 6.4 kg*m^2/s

d)

Let us first calculate angular acceleration

wf =wi + t

(30.5*2)/(60) = 3.77 + *12*10^-3

= -48.00rad/s^2

Now let us calculate angular displacement during this time.

wf^2=wi^2+2

Plug values,

[(30.5*2)/(60)]^2 = 3.77^2+2*-48.00*

= 0.042rad

Use eqn,

Wf = T

KE= Fr

KEf-KEi = Fr

½*If*wf^2 – ½*Ii*wi^2 = Fr

Plug values

½*2.36*[(30.5*2)/(60)]^2 – ½*2.00*3.77^2 = F*0.85*0.042

F= -60.93N

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