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 seconds,the flywheel is rotation at 4.5 revolutions per second, at whichpoint the rope disengages. This attempt at starting the mower doesnot work, however, and the flywheel slows, coming to rest 0.24seconds after the disengagement. Assume constant accelerationduring both spin-up and spin-down.

(a) Determine the average angular acceleration during the 4.5second spin-up and again during the 0.24 second spin-down.

(b) What is the maximum angular speed reached by the flywheel?

(c) Determine the ratio of the number of revolutions made duringspin-up to the number made during spin-down.

Explanation / Answer

part a: during spin up, initial = 0 final = 4.5 rps time = .95 s average acceleration = (4.5-0)/.95 = 4.74 rev/s2 during spin down, initial = 4.5 rps final = 0 time = .24 s average acceleration = (0-4.5)/.24 = -18.75 rev/s2>> negative means deceleration part b: maximum angular velocity = 4.5 rps part c: revs during spin up = (1/2) (acceleration)(time)2 = .5(4.74)(.952) revs during spin down = .5(18.75)(.242) ratio = 3.96 hope this helps!

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