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?
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
In paragraph 2, I believe t is still in seconds, not minutes
Thanks for the catch re the units, sorry about my much too often carelessness in reading the question carefully.
To find the time elapsed after the flywheel reaches 100 r.p.m. until it stops, we need to use the concept of uniform retardation.
First, let's identify the given information:
Initial speed (u) = 234 r.p.m.
Final speed (v) = 100 r.p.m.
Time taken to decelerate (t) = 46 seconds
Since the retardation is uniform, we can use the equation of motion:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the retardation, and t is the time taken.
Given that v = 100 r.p.m., u = 234 r.p.m., and t = 46 seconds, we can rearrange the equation to solve for retardation (a):
a = (v - u) / t.
Plugging in the values, we get:
a = (100 r.p.m. - 234 r.p.m.) / 46 seconds.
a = -134 r.p.m. / 46 seconds.
To find the time it takes to reach zero r.p.m. from 100 r.p.m., we can use the equation:
v = u + at,
where v is the final velocity (zero in this case), u is the initial velocity (100 r.p.m.), a is the retardation (-134 r.p.m. per second), and t is the time taken.
0 = 100 r.p.m. + (-134 r.p.m./s) * t.
0 = 100 r.p.m. - 134 r.p.m./s * t.
Solving for t, we get:
134 r.p.m./s * t = 100 r.p.m.
t = 100 r.p.m. / (134 r.p.m./s).
t = 100 s / 134 s.
t ≈ 0.7463 seconds.
Therefore, approximately 0.7463 seconds will elapse after the flywheel reaches 100 r.p.m. before it stops.