Angular Acceleration of Rotating Object: $\frac{\pi}{8}$

[UrbanXrisis]

If a rotating object starts at rest and completes one rotation in 4s, what is its angular acceleration assuming its angular acceleration is constant?

$$\omega=\frac{2\pi}{t}$$

$$\omega=\frac{2\pi}{4}$$

$$\omega=\frac{\pi}{2}$$

$$\alpha=\frac{\omega}{t}$$

$$\alpha=\frac{\frac{\pi}{2}}{4}$$

$$\alpha=\frac{\pi}{8}$$

the book says the answer is pi/4, where did I go wrong?

 

[kevinalm]

theta=1/2 a t*t
2*pi = 1/2 a 16
4/16 pi =a
a=pi/4

 

[SpaceTiger]

The above solution is correct. Your mistake was in assuming that the final angular speed was one rotation every four seconds. In reality, if it took four seconds to go from rest to one rotation (assuming constant acceleration), then its final rotation rate is faster than one every four seconds (to be precise, twice as fast). One every four seconds is its average rotation rate in that interval.