The angular position of a point on the rim of a rotating wheel is given by = 3.5

Question

The angular position of a point on the rim of a rotating wheel is given by = 3.5 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

for first two parts, take the derivative of theta and substitute the appropriate time

theta (I will call this x for ease of typing): x = 3.5t - 7t^2 + 1.2t^3

angular velocity = dx/dt = 3.5 - 14t + 3.6t^2......................(1)

when t = 2, angular velocity = 3.5 - 14(2) +3.6(2)^2 = -10.1 rad/s

when t = 4, angular velocity = 3.5 - 14(4) +3.6(4)^2 = 5.1 rad/s


for next two parts, recall that angular acceleration is the derivative of angular velocity velocity (and the second derivative of angular displacement), so

alpha = angular acceleration = d/dt(dx/dt) = -14 + 7.2t

at t = 2, angular acceleration = -14 + 14.4 = -0.4 rad/s2

at t = 4 angular acceleration =-14 + 28.8 = 14.8 rad/s2

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