(a) is not 14.8 or 17.9 Will rate the best answer. A bug of mass 0.028 kg is at
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Question
(a) is not 14.8 or 17.9 Will rate the best answer.
A bug of mass 0.028 kg is at rest on the edge of a solid cylindrical disk (M 0.10 kg, R 0.20 m) rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 14.0 rad/s. The bug crawls to the center of the disk (a) What is the new angular velocity of the disk (in rad/s)? (Enter the magnitude. Round your answer to at least one decimal place.) X rad/s (b) What is the change in the kinetic energy of the system (in 3)? (c) If the bug crawls back to the outer edge of the disk, what is the angular velocity of the disk (in rad/s) then? (Enter the magnitude.) 14 rad/s (d) What is the new kinetic energy of the system (in J)?Explanation / Answer
(A) Applying angular momentum conservation,
Li = Lf
I wi = If wf
((0.10 x 0.20^2 /2) + (0.028 x 0.20^2))(14) = (0.10 x 0.20^2 /2 + 0 ) w
((2 x 10^-3) + (1.12 x 10^-3)) (14) = (2 x 10^-3) w
w = 21.84 rad/s
(b) KE = I w^2 /2
Ki = (3.12 x 10^-3)(14^2)/2 = 0.30576 J
Kf = (2 x 10^-3)(21.84^2)/2 = 0.477 J
delta(K) = Kf - Ki = 0.171 J
(C) w = 14 rad/s
(D) new KE = ki = 0.306 J
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