Question

The angular position of a point on the rim of a rotating wheel is given by θ = 3.38t - 2.46t2 + 2.61t3, where θ is in radians and t is in seconds. What are the angular velocities at (a) t = 1.17 s and (b) t = 9.53 s? (c) What is the average angular acceleration for the time interval that begins at t = 1.17 s and ends at t = 9.53 s? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

Answer 1

Angular position of a point on the rim of a rotating wheel is given by

= 3.38 t - 2.46 t² + 2.61 t³

taking derivative both side relative to "t"

ddt = 3.38 - 2.46 (2t) + 2.61 (3 t²)

w = 3.38 - 4.92 t + 7.83 t²

a)

at t = 1.17 s

w = 3.38 - 4.92 (1.17) + 7.83 (1.17)²

w = 8.34 rad/s

b)

at t = 9.53 sec

w = 3.38 - 4.92 (9.53) + 7.83 (9.53)²

w = 667.62 rad/s

c)

w_i = initial angular velocity at t = 1.17 s , = 8.34 rad/s

w_f = final angular velocity at t = 9.53 s , = 667.62 rad/s

t = time interval = 9.53 - 1.17 = 8.36 sec

Angular acceleration is given as

= (w_f - w_i )/t = (667.62 - 8.34)/8.36

= 78.86 rad/s²

d)

w = 3.38 - 4.92 t + 7.83 t²

Taking derivative both side relative to "t"

dw/dt = 0 - 4.92 + 2 (7.83) t

= - 4.92 + 15.66 t

at t = 1.17 s

= - 4.92 + 15.66 (1.17)

= 13.4 rad/s²

e)

at t = 9.53 s

= - 4.92 + 15.66 (9.53)

= 144.32 rad/s²