a solid uniform horizontal disk with a diameter of 2.0m and a mass of 4.0kg free
ID: 1292341 • Letter: A
Question
a solid uniform horizontal disk with a diameter of 2.0m and a mass of 4.0kg freely rotates at 36 rpm about a vertical axis through its center. a small 0.50 kg stone is suddenly dropped onto the disk and sticks to the distance of 80cm from the axis of rotation. the figure shows before and after views. (a) before the stone fell what was the moment of inertia of the disk about its central axis? (b) after the stone stuck what was the moment of inertia of te system about the same axis? (c) what is the angular velocity of the disk (in rmp) after the stone fell onto the disk?
Explanation / Answer
a) I1 = (1/2)*m*R^2 = (1/2)*4*1*1 = 2 kg m^2
b) I2 = (I1 + (m*r^2)) = 2+ (0.5*0.8*0.8) = 2.32 kg m^2
c)
initial angualr momentum Li = I1*w1 = (1/2)*m*r^2 *w
Li = (1/2)*4*1*1*36 = 72
after dropping of stone
final angular momentum Lf = I2*w2 = (I + (m*r^2)) *W2
Lf = 2.32*w2
from conservation of angualar momentum Lf = Li
2.32*w2 = 72
w2 = 31.03 rpm
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