A small ball (point mass) of mass m = 0.23 kg is moving in perfectly circular or

Question

A small ball (point mass) of mass m = 0.23 kg is moving in perfectly circular orbit of radius r = 1.2 m. At an instant t = 1s, it is found to be traveling with an angular velocity of 0.8 rad/s (+ki). From that point on wards another force acts on it, and it begin accelerating with an angular acceleration given by vector a (t) = C square root of t (+k) where C = 1.7 rad/s^5/2 (while still remaining in the same circular orbit). By how many radians will this ball move from its location t = 1 s, 3 seconds later (i.e. at t = 4 s)?

Explanation / Answer

alpha = 1.7 sqrt (t)

dw/dt = 1.7 sqrt(t)

dw = 1.7 sqrt (t) dt

at t = 1s w = 0.8 rad/s

w = (1.7*(2/3)* t^(3/2)) + (0.8 - 1.7*(2/3))

w = (1.7*(2/3)* t^(3/2)) - 0.33333333333

d(theta)/dt = (1.7*(2/3)* t^(3/2)) - 0.33333333333

theta(t) = ( (1.7*(2/3)*(2/5)* t^(5/2)) - 0.33333333333 t )

theta(4) - theta(1) = ( (1.7*(2/3)*(2/5)* 4^(5/2)) - 0.33333333333*(4) ) -(( (1.7*(2/3)*(2/5)* 1^(5/2)) - 0.33333333333 *1 ))

= 13.0533333333 radians

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