To start a lawn mower, you must pull on a rope wound around the perimeter of a f
ID: 1324007 • Letter: T
Question
To start a lawn mower, you must pull on a rope wound around the perimeter of a flywheel. After you pull the rope for 0.99 s, the flywheel is rotating at 3.6 revolutions per second, at which point the rope disengages. This attempt at starting the mower does not work, however, and the flywheel slows, coming to rest 0.24 s after the disengagement. Assume constant acceleration during both spin up and spin down.
(a) Determine the average angular acceleration during the 0.99-s spin-up and again during the 0.24-s spin-down.
(b) What is the maximum angular speed reached by the flywheel?
rad/s
(c) Determine the ratio of the number of revolutions made during spin-up to the number made during spin-down.
Explanation / Answer
(a)
Spin up
alpha = 3.6 * 2 * pi / 0.99
alpha = 22.836 rad/s^2
Spin down
alpha= -3.6 * 2 * pi / 0.24
alpha = 94.2 rad/s^2
(b)
wmax = 3.6 * 2 * pi
wmax = 22.61 rad/s
(c)
spin-up
0.5 * 3.6 * 2 * pi * 0.99 = 11.19
down
0.5 * 3.6 * 2 * pi * 0.24 = 2.713
the ratio is = 11.19 / 2.713 = 4.125
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