The angular position of a point on a rotating wheel is given by ? = 6.37 + 3.23t

Question

The angular position of a point on a rotating wheel is given by ? = 6.37 + 3.23t2 + 3.07t3, where ? 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 = 6.34 s? (d) Calculate its angular acceleration at t = 1.25 s. I know angular v. is the derivative of the given equation and acceleration is the derivative of angular v., but some reason I get the wrong answer. I got A right, but the rest i dont know. I'm probably making a small mistake.

Explanation / Answer

If you are assuming "uniformly accelerated rotation" then the following apply; (please excuse the crude variables) w = d? / dt a = dw / dt = d^2 ? / dt^2 where "w" stands for omega

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