Posted by **Scott** on Sunday, November 7, 2010 at 10:18am.

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?

- Physics -
**bobpursley**, Sunday, November 7, 2010 at 11:18am
w=dTheta/dtime. I assume you know calculus.

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

- Physics -
**Scott**, Sunday, November 7, 2010 at 11:39am
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 dont know how to figure out part d and e. Thanks for the help.

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