The angular position of a point on the rim of a rotating wheel is given by ? = 4

Question

The angular position of a point on the rim of a rotating wheel is given by ? = 4.0t - 2.0t2 + t3, where ? is in radians and t is in seconds.
(a) What is the angular velocity at t = 3 s?
rad/s

(b)What is the angular velocity at t = 5.0 s?
rad/s

(c) What is the average angular acceleration for the time interval that begins at t = 3 s and ends at t = 5.0 s?
rad/s2

(d) What is the instantaneous angular acceleration at the beginning of this time interval?
rad/s2

(e)What is the instantaneous angular acceleration at the end of this time interval?
rad/s2

Explanation / Answer

Derivative of angular position is angular velocity: 4.0t - 2.0t2 + t3 goes to: 4-4t+3t^2 at 3: 4-4(3)+3(9)=19 at 5: 4-4(5)+3(25)=59 average angular acceleration is change in v/ change in t. 40/2=20 instantaneous is alpha, the derivative of omega: omega=4-4t+3t^2 alpha=-4+6t at the start: -4+18=14 at the end: -4+30=26

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