How Do You Calculate Stopping Time and Deceleration for a Grinding Wheel?

[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?
Express your answer in radians.

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

b. At what time does the wheel stop?
Express your answer in seconds.

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

c. What was the wheel's angular acceleration as it slowed down?
Express your answer in radians per second per second.

Would I use this equation [tex] \omega_{f} = \omega_{i} + \alpha t [/tex]
and just solve for [tex]\alpha[/tex]?
[tex] \omega_{f} = 0
\omega_{i} = 22.0
t = time solved in part b [\tex]

 

[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[/tex]
[tex]\theta = \theta _0 + \omega _0 t + \frac{1}{2} \alpha t^2[/tex]
Now with two equations and two unknowns, can you solve for [itex]t[/itex] ?

 

[quicknote]

Yes, you are on the right track. To find the angular acceleration, you can use the equation \omega_{f} = \omega_{i} + \alpha t and solve for \alpha. Since the final angular velocity is 0, you can plug in the values for \omega_{f} and \omega_{i} and solve for \alpha. This will give you the angular acceleration during the deceleration phase of the wheel.