A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a r
ID: 1565149 • Letter: A
Question
A 1.15-kg grinding wheel 22.0 cm in diameter is spinning counterclockwise at a rate of 20.0 revolutions per second. When the power to the grinder is turned off, the grinding wheel slows with constant angular acceleration and takes 80.0 s to come to a rest. The moment of inertia of the wheel is 70. times 10 kg m^2. a. Assume the wheel is in same plane as the paper. What was the angular velocity of the wheel before the power is turned off (magnitude and direction)? b. What was the rotational kinetic energy of the wheel before the power was turned off? c. What was the angular acceleration after the power was turned off but before the wheel stopped (magnitude and direction)? d. What was the torque on the wheel while it was slowing down? e. How many revolutions did the wheel make after the power was turned off?Explanation / Answer
(a)
w = 20 rev/s or 20 rev/s ( 2pi rad/ rev) = 125.6 rad/s
(b)
KEr = 1/2 * I w^2 = 1/2 * 1/2 mR^2 w^2 = mR^2 w^2/4 = 1.15 ( 0.11)^2 ( 125.6)^2/4 = 54.87 J
(c)
alpha = wf- wi/del t = 0 - 125.6/80 = -1.57 rad/s^2 ( clockwise)
(d)
T = I alpha = 7 * 10^-3 ( 1.57) = 0.01099 N m
(e) theta = t ( wf + wi)/2
= 80 ( 20)/2
=800 rev
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