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a circular disc rotates on a thin air film with a period of 0.3s. its moment of inertia about its axis of rotation is 0.06kgm^2...Asked by Suroj
a circular disc rotates on a thin air film with a period of 0.3s. its moment of inertia about its axis of rotation is 0.06kgm^2. a small mass is dropped onto th disc and rotates with it . the moment of inertia of the mass about the axis of rotation is 0.04kgm^2.determine the final period of rotating the disc and mass?
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Answered by
drwls
The answer depends upon where the mass is dropped, but since they tell you the moment of inertia of the dropped mass about the axis, that is enough to answer the question.
Angular momentum about the axis is conserved. The original angular velocity of the disc is
w1 = 2 pi/(period1) = 20.9 rad/s
Period is inversely proportional to angyualr velocity
If w2 is the final angular velocity,
I1*w1 = (I1 + I2)*w2
w2 = (0.06/0/10)*w1 = (3/5)*w1
Period2 = (5/3)*w1 = 0.5 seconds
Angular momentum about the axis is conserved. The original angular velocity of the disc is
w1 = 2 pi/(period1) = 20.9 rad/s
Period is inversely proportional to angyualr velocity
If w2 is the final angular velocity,
I1*w1 = (I1 + I2)*w2
w2 = (0.06/0/10)*w1 = (3/5)*w1
Period2 = (5/3)*w1 = 0.5 seconds
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