The angular position of a point on the rim of a ro...

Question

The angular position of a point on the rim of a rotating wheel is given by θ = 4.0 t — 3.0t² + t³, where θ is in radians and t is in seconds. What are the angular velocities at:

(a) t = 2.0 s and

(b) t = 4.0 s?

c) 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 are the instantaneous angular accelerations at:

(d) the beginning and

(e) the end of this time interval?

Patrick Ford · Accepted Answer

Angular velocity is given by:ω=dθdtAverage angular acceleration is expresed as:αavg=∆ω∆tInstantaneous angular acceleration:α=dωdt(a)Angular velocity is:ω=d(4.0t-3.0t2+t3)dt=4.0-6.0t+3t2[readmore]At t = 2.0s, ω = 4.0 - (6.0)(2.0)+(3)(2.02) = 4 rad/sAngular velocity at t = 2.0s is 4 rad/s.(b)At t = 4.0ω = 4.0 - (6.0)(4.0)+(3)(4.02) = 28 rad/sAngular velocity at t = 4.0s is 28 rad/s.(c)From the average angular acceleration equation:αavg=∆ω∆tα=ω4-ω2t4-t2=28-44-2α = 12 rad/s2The average angular acceleration for the time interval that begins at  t = 2.0 s and ends at t = 4.0 s is 12 rad/s2.(d)Using the instantaneous angular acceleration equation:α=dωdtWe had found angular velocity in part (a)We then differentiate it to get α:α=d(4.0-6.0t+3t2)dt=-6.0+6tAt the beginning, t = 2.0 sα = - 6.0 + (6)(2.0) = 6.0 rad/s2 The instantaneous angular accelerations at the beginning is 6.0 rad/s2.(e)At the end of the time interval, t = 4.0sα = - 6.0 + (6)(4.0) = 18 rad/s2 The instantaneous angular accelerations at the end of this time interval is 18 rad/s2.

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