Question

The angular position of a point on the rim of a rotating wheel is given by θ = 9.33t - 3.93t2 + 1.09t3, where θ is in radians and t is in seconds. What are the angular velocities at (a) t = 2.13 s and (b) t = 8.41 s? (c) What is the average angular acceleration for the time interval that begins at t = 2.13 s and ends at t = 8.41 s? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

Answer 1

=9.33t-3.93t²+1.09t³ radians

a) angular velocity is given by d/dt

d/dt=9.33-7.86t+3.27t² rad/s

At t=2.13 s,

Angular velocity=9.33-7.86*2.13+3.27*2.13²= 7.42 rad/s

b) at t=8.41 seconds,

Angular velocity=9.33-7.86*8.41+3.27*8.41²=174.51 rad/s

c) time interval is 2.13 s to 8.41 s

Initial angular velocity=7.42 rad/s

Final angular velocity=174.51 rad/s

Change in angular velocity=final-initial=174.51-7.42=167.09 rad/s

Time taken=8.41-2.13=6.28 seconds

Average angular acceleration=change in angular velocity/time taken=167.09/6.28=26.6 rad/s²

d) instantaneous angular acceleration=d²/dt²=-7.86+6.54t rad/s²

Instantaneous angular acceleration at t=2.13 second(beginning of time interval)=-7.86+(6.54*2.13)=6.07 rad/s²

e) instantaneous angular acceleration at t=8.41 seconds(end of time interval)=-7.86+(6.54*8.41)=47.14 rad/s²