Question

The angular position of a point on the rim of a rotating wheel is given by...

The angular position of a point on the rim of a rotating wheel is given by θ = 7.88t - 3.72t2 + 2.28t3, where θ is in radians and t is in seconds. What are the angular velocities at (a) t = 1.42 s and (b) t = 6.30 s? (c) What is the average angular acceleration for the time interval that begins at t = 1.42 s and ends at t = 6.30 s? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

Homework Answers

Answer #1

θ = 7.88t - 3.72t^2 + 2.28t^3

angular speed, w = d/dt(7.88t - 3.72t^2 + 2.28t^3)

w = 7.88 - 7.44 * t + 6.84 * t^2

a) at t = 1.42 s

w = 7.88 - 7.44 * 1.42 + 6.84 * (1.42)^2

w = 11.11 rad/s

the angular speed at t = 1.42 s is 11.11 rad/s

b) at t = 6.3 s

w = 7.88 - 7.44 * 6.3 + 6.84 * (6.3)^2

w = 232 rad/s

angular speed at 6.3 s is 232 rad/s

c) for the average acceleration

average acceleration = change in angular speed/time

average acceleration = (232 - 11.11)/(6.3 - 1.42)

average acceleration = 45.3 rad/s^2

d) for the angular acceleration

a = d/dt(7.88 - 7.44 * t + 6.84 * t^2 )

a = 13.68 * t - 7.44

Now, at t = 1.42 s

a = 13.68 * 1.42 - 7.44

a = 12 rad/s^2

the angular acceleration at the begining of the interval is 12 rad/s^2

e) at t = 6.3 s

a = 13.68 * 6.3 - 7.44

a = 78.7 rad/s^2

the angular acceleration at the end is 78.7 rad/s^2

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