# Physics

posted by Happinessdelight

A wheel mounted on a stationary axle starts at rest and is given the following angular acceleration: 9-12t where t is the time after the wheel begins to rotate. Find the number of revolutions that the wheel turns before it stops ( and begins to turn in the opposite direction).

1. Damon

dv/dt = 9 - 12 t and v(0)= 0

v = 9 t - 6 t^2 + constant
when t = 0, v = 0 so constant = 0

v = 9 t - 6 t^2
at v = 0
0 = t(9-6t)
so v = 0 at t = 0 as we knew
and
v = 0 at t=9/6 = 3/2 = 1.5
so how far did it go?
x = (9/2) t^2 - (6/3)t^3 + 0
x = (9/2)(9/4) - (6/3)(9/4)(3/2)
= 81/8 - 54/12
= (243 - 108)/24
check my arithmetic
divide by 2 pi to get revolutions

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