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; i. the disc+putty after the putty sticks? b) What is the angular velocity of the disk (in rpm) after the putty sticks? to c) Determine the angular momentum about the rotational axis of only the disc i. before the putty is dropped; ii, 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. Grading Codes CN (TA use only)Conceptual Error MA CL Calculation error UN Wrong units Wrong equation Math/algebra error

Explanation / Answer

Given

mass of disc is M = 4 kg , radius r = 1 m

initial angular speed is w1 = 36 rpm = 36*2pi/60 rad/s = 3.77 rad/s

mass of putty is m = 0.5 m

position of hte putty is x = 0.85 m from axis of rotation passing through the center of the disc

wee know that the moment of inertia of the disc is I = M*r^2/2

a)

the moment of inertia of the disc is I = 4*1^2/2 = 2 kg m^2

of the putty is ip = m8x^2 = 0.5*0.85^2 kg m^12 = 0.36125 kg m^2

b) angular velocity of the disc after the putty sticks  

by conservation of angular momentum  

L1 = L2

I1*W1 = I2*W2

(0.5*M*r^2)(W1) = (0.5*M*r^2+m*x^2)(W2)

W2 = (0.5*M*r^2)(W1) / (0.5*M*r^2+m*x^2)

W2 = (0.5*4*1^2)(3.77)/(0.5*4*1^2+0.5*0.85^2)

W2 = 3.193224 rad/s

W2 = 30.49 rpm

c) angular momentum L = I*W

before putty dropped  

L1 = I1*W1 = (0.5*M*r^2)(W1) = (0.5*4*1^2)(3.77) = 7.54 kg m2/s

after  

L2 = (0.5*M*r^2+m*x^2)(W2)

L2 = (0.5*4*1^2+0.5*0.85^2)(3.193224) = 7.54 kg m2/s

d)

time taken is t = 12 ms = 12*10^-3 s

we know that the impulse like chang ein angular momentum is = F*T

F*t = 7.54

F = 7.54/(12*10^-3) N = 628 N

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