The angular position of a point on the rim of a rotating wheel is given by theta

Question

The angular position of a point on the rim of a rotating wheel is given by theta = 6.0t - 1.0t^2 + t^3, where theta is in radians and t is in seconds. What is the angular velocity at t = 3 s? What is the angular velocity at t = 5.0 s? What is the average angular acceleration for the time interval that begins at t = 3 s and ends at t = 5.0 s? What is the instantaneous angular acceleration at the beginning of this time interval? What is the instantaneous angular acceleration at the end of this time interval?

Explanation / Answer

Angular velocity is the first derivative of the angular position
w=d theta/ dt = 6 - 2t + 3t^2
a) so at t= 3
w = 6 -2*3 + 3*3^2
= 27 radians / s

b) and at t= 5
w = 6-10+75
= 71 radians/s

c) ave alpha = delta w/ delta t = (71-27)/2 = 22 rad/s^2

d) Angular acceleration is the second derivative of angular position.( or the 1st derivative of angular velocity)
so derive once.
d theta/ dt = 6 - 2t + 3t^2
and again
a=d2 theta /dt2 = -2+6t

substitute t=3
a = -2 +18 = 16 radians/s^2

e) and at t=5
a = -2 +30 = 28 radians / s^2

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