Attached to each end of a thin steel rod of length 1.20 m and mass 6.40 kg is a

Question

Attached to each end of a thin steel rod of length 1.20 m and mass 6.40 kg is a small ball of mass 1.06 kg. The rod is constrained to rotate in a horizontal plane about a vertical axis through its midpoint. At a certain instant, it is rotating at 39.0 rev/s. Because of friction, it slows to a stop in 32.0 s. Assuming a constant retarding torque due to friction, compute (a) the angular acceleration, (b) the retarding torque, (c) the total energy transferred from mechanical energy to thermal energy by friction, (d) work done by the retarding torque to transfer the total energy to thermal, and (e) the number of revolutions rotated during the 32.0 s. (f) Now suppose that the retarding torque is known not to be constant. If any of the quantities (a), (b), (c), and (d) can still be computed without additional information, give its value.

Explanation / Answer

initial angular velocity w1 = 39 x 2 x pi rad/s
Final angular velocity w2 = 0 rad /sec
time = 32 sec

w2 = w1 + at
0 = 245.0442 + 32 a
a = - 7.657632 rad / sec^2 and is retardation ans (a)

Moment of Inertia = about axis of rotation = (1/12 x 6.4 x 1.2^2) + 1.06 x 0.6^2 = 0.768 + 0.3816 = 1.1496
torque = retardation x moi
= 7.658 x 1.1496 = 8.804 kg-m ans (b)

K.E. = 0.5 x MOI x w1^2 = 0.5 x 1.1496 x 245.0442^2 = 34514.82 J ans. (c)

w2^2 = w1^2 + 2 x a x s
0 = 245.0442^2 - 2 x 7.658 x s
s = 3920.52 radians = 623.97 revolutions ans (d)

(e) the question is not clear to me.

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