Posted by winterWX on Saturday, March 10, 2007 at 12:31am.
The angular position of a point on the rim of a rotating wheel is given by = 4.0t - 1.0t2 + t3, where is in radians and t is in seconds.
(a) What is the angular velocity at t = 2 s?
(b)What is the angular velocity at t = 4.0 s?
(c) What is the average angular acceleration for the time interval that begins at t = 2 s and ends at t = 4.0 s?
(d) What is the instantaneous angular acceleration at the beginning of this time interval?
(e)What is the instantaneous angular acceleration at the end of this time interval?
(a) differentiate the theta(t) equation you provided. This will yield the following:
d(theta)/dt = omega(t) = 4 - 2t + 3 t^2
Then plug in t = 2 fr your answer.
The Greek letter "omega" is usually used for angular velocity.
(b) Use the same omega(t) formula but use t=4.
(c) Subtract the omega(2) from omega (4) for the angular velocity change, and divinde by 2 s.
(d) The formula for instantaneous angular acceleration is
alpha (t) = d(omega)/dt = -2 + 6t. Insert t = 2 s
(e) Insert t=4 into the alpha(t) equation above
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