The angular position of a point on the rim of a rotating wheel is given by = 5.0

Question

The angular position of a point on the rim of a rotating wheel is given by = 5.0 t + (-7.0)t2 + +(1.2)t3, where is in radians if t is given in seconds. What is the angular velocity at t = 2.0 s? What is the angular velociy at t = 4.0 s? What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s? What is the instantaneous angular acceleration at the beginning of this time interval? What are the instantaneous angular accelerations at the end of this time interval?

Explanation / Answer

w = dtheta/dt = 5 - 14*t + 3.6*t^2
a) w(2) = 5 - 14*2 + 3.6*2^2 = -8.6 rad/s

b) w(4) = 5 - 14*4 + 3.6*4^2 = 6.6 rad/s

c) ave alpha = delta w/ delta t = (6.6 +8.6)/2 = 7.6 rad/s^2

d) a = dw/dt = - 14 + 7.2*t

a(2) = - 14 + 7.2*2 = 0.4 rad/s^2

e) a(4) = - 14 + 7.2*4 = 14.8 rad/s^2

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