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Centripetal acceleration car problem

  1. Jan 4, 2007 #1
    1. The problem statement, all variables and given/known data
    A race car starts from rest on a circular track of radius 400m. The car's speed increases at the constant rate of .5m/s^2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, determine a) the speed of the race car, b) the distance traveled, and c) the elapsed time.

    2. Relevant equations
    centripetial accel. =rw^2
    tangential accel.= r(alpha)

    3. The attempt at a solution
    Hmm.. I'm a little bit confused at how to find the point where the two accelerations are the same. Well first I made the equations equal to each other and I tried solving for (t) but it seems like everything just cancels out except theta...

    Another question, have you every been in a situation where you never understood anything that your class learned? How did you over come that? I mean I've listened to my teacher, read books and internet stuff, but I can't seem to get some facts straight no matter what. What did you do when you were totally stuck? I mean you didn't understand anything? :rolleyes:

    Thanks! :smile:
  2. jcsd
  3. Jan 4, 2007 #2

    Why make the accelerations equal to each other when the problem tells you what one of them is?

    I think most of us are in that situation from time to time. I know I am. Just keep plugging away, ask questions, eventually it seems to sort itself out.

  4. Jan 4, 2007 #3
    you know that the linear speed of the car is constant, so from


    you will have a function of [itex]\theta[/itex] in terms of time.
  5. Jan 5, 2007 #4


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    I don't think it is; (perhaps you meant the linear acceleration...)

    Or am I missing something?
  6. Jan 5, 2007 #5

    Doc Al

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    OK, but:
    (1) The tangential acceleration is given, so use it.
    (2) You can also write the centripetal acceleration in terms of tangential speed (instead of angular speed), which you can figure out from the tangential acceleration.

    Show exactly what you did.
  7. Jan 5, 2007 #6
    Okay here is what I did lol:

    Since a_c=rw^2 and a_t=r(alpha)


    --Divide the radius from both sides



    Kept simplifying for some reason and ended up with t=(delta)theta

    I wanted to solve for a variable,but how would you know what theta is? Or do you not use theta at all?
  8. Jan 5, 2007 #7
    Ok, so how do you know what is tangential acceleration when objects are moving in a circle? Do you look at the units (m/s^2) ? Centripetal acceleration is measured in rad/s^2 right?

    If .5m/s^2 is the tangential acceleration, I think I can solve this. Let me try. =]
  9. Jan 5, 2007 #8
    Alright I've found both the angular speed and tangential velocity:

    Since a_t=0.5m/s^2



    w=0.0354 rad/s^2 <---- Do you need to include (rad)? I've heard that it is a pure unit and can be counted as NOTHING. Is this true? If so, can I just write is as 0.0354 1/s^2 or should I include the (rad)?

    Ok, a) says: determine the speed of the race car. Should I always assume that they are talking about the tangential speed and not the angular speed??
    How do you know when they are talking about either one?

    If they want the v, v=rw

    v=(400m)(0.0354 rad/s^2)

    v= 14.14m/s

    Are my steps correctly done? Thanks for your help!
  10. Jan 5, 2007 #9
    Oops I think I made a mistake. =/
  11. Jan 5, 2007 #10
    Wait nvm. Since 0.5=a_t and a_t=r(alpha)

    and the magnitudes of a_c and a_t are equal, then a_c=a_t right?

    So, rw^2=.05m/s^2 correct?
  12. Jan 5, 2007 #11

    Doc Al

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    The units of w are rad/s, not rad/s^2. Radians (and other angle measures) are an unusual unit, since angles have no real dimension: so give your answer for angular speed in terms of rad/s. But when you use w to find the tangential speed, you'll drop the radians.

    You have to understand the context. If someone asked you "how fast" a car was going around a track, would you be tempted to give the angular speed? Of course not--they mean linear speed.
    When they ask for speed, it generally means linear speed--as measured in m/s; but if they ask for angular speed, then that's what you give them.

    Looks good to me.

    You could save a bit of work by expressing the centripetal acceleration directly in terms of speed:
    [tex]a_c = \omega^2 r = v^2/r[/tex]
  13. Jan 5, 2007 #12
    Oh, ok yeah I don't know why I put ^2 with speed. lol. Ok, I get the (rad) units thing and when to use them now. I should remember that other equation, it could have been a lot faster like you said.

    By the way, what is the relationship between v^2/r and rw^2? I know that a_c is the vector that is pointed towards the circular rotational motion and that it is can be caculated from v_f-v_i/t right? I don't see how you can change (omega) into tangetial velocity though.

    I've done both b) and c) now using the speed I found. Here's my work:

    Well I found time using the tangetial speed. + tangential accel




    b) asks for the distance traveled
    I used x=1/2at^2


    x= 200.223m

    c) asks for the elapsed time and I think I'v already found it!

    It would be 28.3s right? Thanks for all your help doc al and everyone else!
  14. Jan 5, 2007 #13


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    [tex]\omega = \frac{v^2}{r}[/tex]
  15. Jan 5, 2007 #14

    Doc Al

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    What do you mean you don't see how? You did it yourself in solving the first part! :wink:
    [tex]v_t = \omega r[/tex]
  16. Jan 5, 2007 #15
    I thought v=rw so how can w = v^2/r? Shouldn't it be w=v/r

    EDIT: wait let me solve this since a_c=rw^2
  17. Jan 5, 2007 #16

    Doc Al

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    Looks good to me!
  18. Jan 5, 2007 #17

    Doc Al

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    Hoot made a typo. He is allowed one mistake per year--that's it for 2007.:rofl:
  19. Jan 5, 2007 #18


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    My mistake, brain fart so to speak.

    [tex]\alpha = \frac{v^2}{r}[/tex]

    [tex]\omega = \frac{v}{r}[/tex]

    As you correctly say.

    The pressure is on now :rofl:
  20. Jan 5, 2007 #19
    Oh ok I got it now! =D a_c=rw^2 and v=rw so v/r=w, w^2=v^2/r^2

    substitute that in a_c=rw^2 and you get a_c=r(v^2/r^2) =v^2/r
  21. Jan 5, 2007 #20
    lol. I'm sorry Hoot. :frown: Okay, thanks to both of you especially and everyone else that helped me! :smile:
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