The angular position of a point on the rim of a rotating wheel is given by ?=4.0

Question

The angular position of a point on the rim of a rotating wheel is given by ?=4.0t- 3.0t2 + t3, where ? is in radians if t is given in seconds.
What is the angular velocity at t = 4.5 s?
What is the angular velocity at t = 6.0 s?

What is the average angular acceleration for the time interval that begins at t = 4.5 s and ends at t = 6.0 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

Given ? = 4.0 t-3.0 t^2+t^3 Angular velocity ,? = d?/ dt    = d(4.0 t-3.0 t^2+t^3)/ dt ? = 4.0-6 t+3t^2 At t1 = 4.5 s, ?1 = 4 -6(4.5)+3(4.5)^2 = 37.75 rad /s At t2 = 6.0 s ?2 = 4 - 6(6.0) +3(6.0)^2 = 76 rad/s --------------------------------------------------------------------------- Average angular acceleration is aavg = ?2-?1/(t2 -t1)       = 38.25/1.5 =25.5 rad/s^2 ------------------------------------------------------------------------------ Instantaneous angular acceleration at the beginning of the time interval is a = d?/dt = d(4.0-6 t+3t^2)/dt               = -6 +6 t               = -6 + 6(4.5 ) = 21 rad/s^2 Instantaneous angular acceleration at the end of the time interval is a = d?/dt = d(4.0-6 t+3t^2)/dt               = -6 +6 t               = -6 + 6(6.0 ) = 30  rad/s^2 Instantaneous angular acceleration at the end of the time interval is a = d?/dt = d(4.0-6 t+3t^2)/dt               = -6 +6 t               = -6 + 6(6.0 ) = 30  rad/s^2 a = d?/dt = d(4.0-6 t+3t^2)/dt               = -6 +6 t               = -6 + 6(6.0 ) = 30  rad/s^2

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