Question
A flywheel is making 234 r.p.m. and after 46 seconds it is running at 100 r.p.m. What time will elapse after 100 r.p.m. before it stops, if the retardation is uniform?
Answers
mathhelper
let the retardation be a rpm/m
v = at + k
when t = 0 , v = 234
234 = 0 + k , so k= 234
when t = 46, v = 100
100 = 46a + 234
a = -134/46 = -67/23
v = (-67/23)t + 234
so we need the t when v = 0
0 = (-67/23)t + 234
67t/23 = 234
67t = 5382
t = appr 80.3 minutes
so the time it took from 100 rpm to a stop = 80.3 - 46 or 34.3 minutes
v = at + k
when t = 0 , v = 234
234 = 0 + k , so k= 234
when t = 46, v = 100
100 = 46a + 234
a = -134/46 = -67/23
v = (-67/23)t + 234
so we need the t when v = 0
0 = (-67/23)t + 234
67t/23 = 234
67t = 5382
t = appr 80.3 minutes
so the time it took from 100 rpm to a stop = 80.3 - 46 or 34.3 minutes
oobleck
In paragraph 2, I believe t is still in seconds, not minutes
mathhelper
Thanks for the catch re the units, sorry about my much too often carelessness in reading the question carefully.