# Angular Velocity Problem

1. Oct 27, 2005

### quicknote

I'm having problems with parts b and c...

At time t= 0 a grinding wheel has an angular velocity of 22.0 rad/s . It has a constant angular acceleration of 26.0 rad/s^2 until a circuit breaker trips at time t= 2.30 s . From then on, the wheel turns through an angle of 436 rad as it coasts to a stop at constant angular deceleration.

a. Through what total angle did the wheel turn between and the time it stopped?

$$\Delta \Theta = 13t^2 + 22t$$ at t=2.3s is 119.4rad
Therefore total angle is 119 + 436 = 555rad

b. At what time does the wheel stop?

So I know that $$\omega_{f} = 0$$ for the wheel to stop
$$\omega_{i} = 22.0 rad/s$$
That's as much as I understand...

c. What was the wheel's angular acceleration as it slowed down?
Would I use this equation $$\omega_{f} = \omega_{i} + \alpha t$$
and just solve for $$\alpha$$?
$$\omega_{f} = 0 \omega_{i} = 22.0 t = time solved in part b [\tex] 2. Oct 28, 2005 ### Päällikkö b) Supposing I understood the problem right (I'm unfamiliar with the term grinding wheel): The equations for constant acceleration are quite similar to the ones in kinematics. [tex]\omega = \omega _0 + \alpha t$$
$$\theta = \theta _0 + \omega _0 t + \frac{1}{2} \alpha t^2$$
Now with two equations and two unknowns, can you solve for $t$ ?