Question

The angular position of a point on the rim of a rotating wheel is given by Theta = 4.0t-3t^2+t^3, where theta is in radians and t is in seconds. At t=0, what are (a.) the point's angular position and (b.) its angular velocity? (c.) What is its angular velocity at t=4.0s? What are the instantaneous angular accelerations at (d.) the beginning and (e.) the end of this time interval?

Answer 1

w=dTheta/dtime. I assume you know calculus.

d) alpha=dw/dtime, at t=0 and at t=tfinal

Answer 2

So far i got angular velocities at t=2s and t=4s

W2 = 3(2)^2-6(2)+4 = 4 rad/s
W4 = 3(4)^2-6(4)+4 = 28 rad/s

For average angular acceleration i got

W2-W1 / T2-T1 = 28-4/4-2 = 12 rad/s^2


But I don't know how to figure out part d and e. Thanks for the help.