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 = 7.0 t + (-4.0)t^2 + +(1.0)t^3, where theta is in radians if t is given in seconds. What is the angular velocity at t = 2.0 s? What is the angular velocity 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 times interval? What are the instantaneous angular accelerations at the end of this time interval?

Explanation / Answer

w = d(theta) / dt

w = 7 - 8t + 3 t^2

at t =2 s

w = 7 - 16 + 12 = 3 rad/s ........Ans

at t = 4s

w = 7 - (8 x 4) + (3x 4^2)

w = 23 rad/s ......Ans

<alpha> = w at 4 - w at 2 / (4 - 2)

= (23 - 3) / 2

= 10 rad/s^2 .........Ans

alpha = dw /dt

= -8 + 6t

at t = 2

alpha = -8 + 12 = 4 rad/s^2 .....Ans

at t =4 s

alpha = -8 + (4 x 6) = 16 rad/s^2 ...........Ans

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