
#1
Oct2705, 09:13 PM

P: 33

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] 



#2
Oct2805, 06:10 AM

HW Helper
P: 482

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] ? 


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