To start a lawn mower, you must pull on a rope wound around theperimeter of a fl

Question

To start a lawn mower, you must pull on a rope wound around theperimeter of a flywheel. After you pull the rope for 0.95 s, theflywheel is rotating at 4.5 revolutions per second, at which pointthe rope disengages. This attempt at starting the mower does notwork, however, and the flywheel slows, coming to rest 0.24 s afterthe disengagement. Assume constant acceleration during both spin-upand spin-down. (a) Determine the average angularacceleration during the 4.5-s spin-up and again during the 0.24-sspin-down. (b) What is the maximum angular speed reachedby the flywheel? (c) Determine the ratio of the number ofrevolutions made during spin-up to the number made duringspin-down.

Explanation / Answer

= (f - i) / t = 2 (ff - fi) / t = 2 * 4.5 / .95 =29.8 / s2 during spin-up = - 2 * 4.5 / .24 = -118/ s2 during spin down maximum angular speed   = 2 f = 2 4.5 = 28.3 / s f = ft - 1/2 tf2 = 2 * 4.5 * .24 - 1/2 * 118 *.242 = 3.39 (total angle indeceleration) i = 1/2 ti2 =1/2 * 29.8 * .952 = 13.4 The ratio of the total angles traveled is the same as the sameas the ratio of the total revolutions traveled. c) 13.4 / 3.39 = 3.97     (since = 2 n   where n is the number ofrevolutions)

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