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 = 4.0t3t^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)^26(2)+4 = 4 rad/s
W4 = 3(4)^26(4)+4 = 28 rad/s
For average angular acceleration i got
W2W1 / T2T1 = 284/42 = 12 rad/s^2
But I don't know how to figure out part d and e. Thanks for the help.
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