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Racecar on a circular track

  1. Oct 13, 2011 #1
    The problem statement
    A race-car is driving on a circular track, and above a particular speed, the race-car may flip over towards the outside track (inward wheels lift).
    Frictional force is present. The distance of the center of mass from the ground is L, the distance between the wheels is 2d, and the distance from the projection of the center of mass on the floor to the center of the circular track is R.
    The question is to find the normal reactions on each wheel.

    Here's a diagram I drew:
    [PLAIN]http://s4.postimage.org/jnhgqkhkr/Untitled.png [Broken]

    Relevant equations
    While I know how to arrive at the solution with these 3 equations:
    1. [itex] N_1+N_2=Mg [/itex] --------(ƩFy)
    2. [itex] f_1+f_2=\frac{Mv^2}{R}[/itex]----------(ƩFx)
    3. [itex](f_1+f_2)L+N_1d-N_2d=0[/itex]---------- (τ about CM)
    Solving by eliminating frictional force from the eq and I get the required solution.

    My question however, is this:
    Since I am assuming for now that the car does not flip over, why am I restricted to taking torque about the center of mass.
    The point (P) directly below the center of mass is also stationary wrt the CM, so If I assume that point to be my origin and calculate torque about this point, I can eliminate both frictional forces and weight of the body since they pass through P.

    However, I get N1=N2, which is obviously wrong.
    To avoid this conflict it would imply that at this point, the net torque is NOT zero. But as far as I can see, there is no angular motion about P either.

    What changes do I have to make to my equations if I choose (if I am allowed to) to analyze torque about the point P.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 13, 2011 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The problem with using point P as your origin is that it is accelerating. The CM has the special property that the torque about it equals the rate of change of angular momentum in all cases. Not so for point P.

    You can get around that by essentially viewing things from an accelerating frame which introduces a pseudoforce at the CM. You should get the same answer then, using P as your origin.
     
  4. Oct 13, 2011 #3
    It works for me. Normal force is same equation. Moments don't have friction term.

    N1+N2=mg
    (mv^2)*L/R + N1d - N2d = 0
    etc
     
  5. Oct 13, 2011 #4
    Doc, How is point P accelerating any different than the CM is ?
    From what i can see, the CM is moving with the same acceleration as point P at all times during this motion. And if there is no need to get into an accelerating reference frame when analyzing the CM motion, why is there a need to get into one which has the same velocity with respect to the CM ?

    Of course, I do see how moving to the accelerating reference frame would help..
    Now f1 and f2 are balanced by the centrifugal force C which points in the [itex]\hat{r}[/itex], but again, where is C acting ?
    Wouldn't it act on the center of mass of the body ? In that case, again, analyzing about P, torque would still not factor C (and therefore f1 and f2)

    What am I missing? Could you please elaborate on where the fictitious forces act ?
     
  6. Oct 13, 2011 #5

    Doc Al

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    Staff: Mentor

    Yes, both point P and the CM accelerate. Without going into a detailed derivation, I'll just say that the CM is a special case. When you use an accelerating point as your origin, there's an extra term in your equation (the pseudoforce); that extra term goes to zero if the accelerating point is the CM. (Check out any intermediate Mechanics book for a derivation.)
    The centrifugal force acts at the CM and thus exerts a torque about P. So if you take P as your origin, you'll get the equation given by LawrenceC above.
     
  7. Oct 14, 2011 #6
    Ah, that's right. I kept thinking that C was acting downwards even though I mentioned that it was in [itex]\hat{r}[/itex] in my previous post.

    Indeed, it works out fine. Thanks a lot.
     
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