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Rotational Kinematics - angle between force and velocity

  1. Nov 7, 2015 #1
    Here is the problem I am working on. I have found answers for all of them except part (f), which is the one I need help with. I will report the answers I have so far:

    A classic 1957 Chevrolet Corvette of mass 1240 kg starts from rest and speeds up with a constant tangential acceleration of 2.00 m/s^2 on a circular test track of radius 60.0 m. Treat the car as a particle.

    (a) What is its angular acceleration? --> 0.0333 rad/s2
    (b) What is its angular speed 6.00 s after it starts? --> 0.2 rad/s
    (c) What is its radial acceleration at this time? --> 2.4 rad/s2
    (d) (I'll skip this one, since it's just a sketching problem)
    (e) What are the magnitudes of the total acceleration and net force for the car at this time? --> atot = 3.12 m/s2 and ΣF = 3874 N

    (f) What angle do the total acceleration and net force make with the car's velocity at this time?

    I'm wondering if I should use one of the rotational kinematics equations, but I'm still not sure how to go about it. Any ideas would be helpful. Thank you!
  2. jcsd
  3. Nov 7, 2015 #2
    Use your kinematics equations and substitute the variables. v=omega, alpha=a or acceleration, time=time, delta(x)=delta(theta).

    Then solve as though kinematics. Use correct units: rad/s rad/s^2 instead of m/s etc. If you need more help, let me know.
  4. Nov 7, 2015 #3
    This is what I did using a kinematics equation:

    Θ = Θo + ωozt + 1/2(αzt2

    Since it starts from rest,

    Θ = 1/2(0.0333 rad/s2)(6 s)2
    Θ = 0.594 rad = 107°

    This is not the right answer. Does Θ really represent the angle between the acceleration and the velocity? I thought it was just the angular displacement, in which I can't see a correlation between that and the acceleration and velocity angle.
  5. Nov 7, 2015 #4
    Delta Theta is Angular displacement. Starts from rest wi=0. Time=6.00s. Alpha=0.0333 rad/s^2. You now have 3 variables. Solve for wf.

    Use wf=wi+alpha(time) -----------> wf=0 + (0.0333 rad/s^2)(6.00sec) That should result in your angular speed of 0.1998 or 0.2rad/s

    Foe reg kinematics Vf=Vi+at
  6. Nov 7, 2015 #5
    Oh sorry, for part (f) try

    Delta Theta = ((wi+wf)/2 ))(time) wi+wf, divide by 2, multiply by time
  7. Nov 7, 2015 #6
    Alright, so when I use that equation I get:

    ΔΘ = ½(ωo + ωf)(t)
    ΔΘ = ½(0 + 0.2 rad/s)(6 s) = 0.6 rad = 34.4°

    The book says it should be 50.2°, which doesn't make sense to me. Is there a certain concept that I might be missing which could be keeping me from getting the right answer?
  8. Nov 7, 2015 #7
    You need to use Arctan(ar/at). Try looking up Linear Acceleration in rigid body rotation in your text book. atan=dv/dt = (r)(dw/dt)=ra

    Tan^-1(2.40/2.00) =50.19 deg
  9. Nov 8, 2015 #8
    Oh wow, that makes a lot of sense. Thank you so much!!
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