The angular position of a point on the rim of a rotating wheel is given by = 6.0

Question

The angular position of a point on the rim of a rotating wheel is given by = 6.0 t + (-7.5)t2 + +(1.3)t3, where is in radians if t is given in seconds. What is the angular velocity at t = 2.0 s?

a.What is the angular velociy at t = 4.0 s?

b.What is the average angular acceleration for the time interval that begins at t = 2.0 s and ends at t = 4.0 s?

c.What is the instantaneous angular acceleration at the beginning of this time interval?

d.What are the instantaneous angular accelerations at the end of this time interval?

Explanation / Answer

for a) and b), take the derivative of theta and substitute the appropriate time

theta (I will call this x for ease of typing): x=6t-7.5t^2+ 1.3t^3

ang vel = dx/dt = 6 - 15t + 3.9 t^2 (eq. 1)

when t=2, ang vel = 6 - 15 * 2 + 3.9 *(2)^2 = -8.4 rad/s

a) substitute t=4 in eq. 1 ang vel = 6 - 15 * 4 + 3.9 *(4)^2 = +8.4 rad/s

for part b) for parts c) and d), recall that ang accel is the derivative of ang velocity (and the second derivative of angular displacement), so

alpha = ang accel = d/dt(dx/dt)=-15+7.8t

B) Avg of ang acceleration = (0.6 + 16.2)/2 = 8.4 rad/s2
C) at t=2, ang accel = -15+15.6 = 0.6 rad/s2

D) at t=4 ang accel =-15+31.2 = 16.2 rad/s/s

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