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Angular Velocity Problem

  1. 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. jcsd
  3. Päällikkö

    Päällikkö 500
    Homework Helper

    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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