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Nonuniform Circular Motion-Find time given r, a-sub-t, V-sub-0

  1. Sep 19, 2009 #1
    Nonuniform Circular Motion--Find time given r, a-sub-t, V-sub-0

    1. The problem statement, all variables and given/known data

    A car turns through 60 degrees while traveling on the circumference of a 35m radius circle. The care is traveling at 8.0 m/s as it enters the turn and maintains a constant tangential acceleration of .4m/s2 throughout the turn. How long did it take the car to make the turn?


    (A) 2.3s (B) 4.2s (C) 4.9s (D) 8.4s

    DATA SUMMARY:
    at: .4m/s2
    Vo: 8.0m/s
    R:35m

    2. Relevant equations

    at=d|v|/dt=.4m/s2

    |V|= [tex]\omega[/tex] *r

    [tex]\theta[/tex]= [tex]\omega[/tex]*r

    3. The attempt at solution

    at=d|v|/dt=.4m/s2

    |V|=.4t +8.0 (from integrating at)

    [tex]\omega[/tex]= (.4t+8.0)/35 (from |V|=[tex]\omega[/tex]*r)

    [tex]\theta[/tex]= [tex]\omega[/tex]*r

    ([tex]\pi[/tex]/3= ((.4t+8.0)/35)t= (.4/35) + (8.0t/35)

    t=4.6

    I get the feeling that there's one crucial equation I'm missing...
     
  2. jcsd
  3. Sep 19, 2009 #2
    Re: Nonuniform Circular Motion--Find time given r, a-sub-t, V-sub-0

    [tex]\theta = \omega_{0} t + \frac{1}{2} \alpha t^{2}[/tex]
    should be the equation to be used instead.

    [tex]\theta = \omega t[/tex] is true only under a constant angular velocity - the general form is [tex]\theta = \int \omega dt[/tex]
    which yields the equation above for constant angular acceleration.

    You could also use a translational kinematics approach instead, which I think should be less confusing than rotational dynamics.
     
    Last edited: Sep 19, 2009
  4. Sep 19, 2009 #3
    Re: Nonuniform Circular Motion--Find time given r, a-sub-t, V-sub-0

    Thank you!

    I got it using translational kinematics.
     
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