A 107 kg flywheel is accelerated from rest to a rotational speed of 2400 rpm by

Question

A 107 kg flywheel is accelerated from rest to a rotational speed of 2400 rpm by a couple of constant magnitude 724 N m, over which time it executes 500 revolutions. Answer all questions to at least 3 sf. If necessary, you may enter answers in standard form (e.g. answer = 7.12e7, for an answer of 71.2 million). Enter your answers in SI units but do not enter the units. If the radius of gyration of the flywheel is 698 mm, calculate the average magnitude of the couple exerted on the flywheel by friction within its bearing. (N m)

Explanation / Answer

= 500 revolution = 2 x 500 rad = 1000 rad

= 2400 rpm = 2400 x 2 / 60 = 251.22 rad/s

k = 698mm = 0.698m = (I/m) = (I/107)

moment of inertia I = 52.13 kg-m2

2 - 0 = 2

251.222 = 2 x x 1000

= 10.05 rad/s2

= t

t = 251.22/10.05 = 25.01 sec

couple = I

724 - = 52.13 x 10.05

= 200.09 N-m

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