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Maximum velocity of motorbike

  1. May 10, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=58620&stc=1&d=1368173748.png


    2. Relevant equations



    3. The attempt at a solution
    I am not sure which equations to start with here.

    Let ##\mu## be the coefficient of friction.
    The maximum velocity is attained when it is just about to slip i.e
    [tex]\frac{mv_{max}^2}{R}=\mu mg[/tex]
    where m is the mass of motorbike.
    [tex]\Rightarrow v_{max}=\sqrt{\mu gR}[/tex]
    Next I think I have to use the energy conservation but I don't know the force which helps the bike to accelerate. :confused:
     

    Attached Files:

  2. jcsd
  3. May 10, 2013 #2
    I think you need to take into consideration that the cyclist is leaning inwards.
     
  4. May 10, 2013 #3
    I don't think that it would change anything being the circuit flat (at least there are no indications it is inclined). You may see the motorbike as a "point particle", or in any way gravity has no component in the horizontal direction so it doesn't seem to me it would change things.

    As for the rest, I think you can assume constant acceleration to reach total maximum speed, and simply use circular accelerated motion for the tangential part.
     
  5. May 10, 2013 #4
  6. May 10, 2013 #5
    I see what you mean, but I think that here it does not enter. It is right of course that the angle depends on the speed you are going, but this not to let the cyclist fall down. I don't think that this would permit a higher speed, as the forces are anyway the same. I mean the normal force is anyway only vertical and therefore has to be the same as weight, otherwise it would jump. Also the friction force is just given by the normal and therefore nothing changes. This is why I was thinking that in the problem in question you could consider the cyclist a point particle.

    I could be wrong of course, my two cents...
     
  7. May 10, 2013 #6
    Saying it perhaps better, to lean is the only way to be able to reach the maximal velocity allowed without killing yourself, but does not influence the value of this maximal velocity :smile:
     
  8. May 10, 2013 #7

    NascentOxygen

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    I see friction playing a rôle in two places here. Below a certain speed it stops the cyclist sliding off the track, but it also limits him to a certain acceleration which if he attempts to exceed he'll spin his back wheel.
     
  9. May 10, 2013 #8
    No, it is explicitly stated in the assumptions that the motorbike never slip as it accelerates
     
  10. May 10, 2013 #9

    TSny

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    It's the same force that lets you start walking from rest.
     
  11. May 10, 2013 #10
    The angle at which he leans will change from 0 degrees to a maximum as he accelerates around the track. This is because the speed around the track changes. One should therefore not assume that the centripetal acceleration is constant. It is the problem exercise 6 I am referring to in the link http://cnx.org/content/m42086/latest/?collection=col11406/1.7
     
  12. May 10, 2013 #11

    NascentOxygen

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    You are re-stating precisely what I said.
     
  13. May 10, 2013 #12
    You are completely right, but I do not see where it comes in in the problem. Here we are trying to find the maximum velocity it can reach... no doubt the centripetal will have to increase with velocity (by definition also, without speaking of angles which are just a consequence of that) but here we concentrate in the moment we actually reach this maximum velocity. To do so you impose to be at maximum friction allowable, which is ##F=\mu N##, and compute the velocity when this force is the centripetal force. And we are done in this way. Then of course we can consider that to manage not to fall, at this speed we will have to lean a certain angle, but this is not essential in finding the maximum speed.

    Then we want to find the space the man needs to reach such maximum velocity. We have initial velocity, final velocity (i.e. ##v_{max}##) and the time it needs, then also here (assuming constant TANGENTIAL acceleration and no sliding as stated in the problem) we have all the data... using uniformly accelerated circular motion we can find the space needed to reach ##v_{max}##. And the whole problem is solved.
     
  14. May 10, 2013 #13

    TSny

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    I don't think we want to assume constant tangential acceleration. Friction will allow only a certain maximum amount of net acceleration. As the motorbike speeds up, more and more of the acceleration will need to be centripetal, leaving less acceleration for tangential.
     
  15. May 10, 2013 #14
    I can agree with you if you said it need more and more power to speed up due to friction, but centripetal and tangential acceleration are not related, so I don't see why you can't have a constant tangential acceleration... also because if then tangential acceleration is not constant, then it would be loads more difficult to solve the problem without an expression of acceleration as function of time...
     
  16. May 10, 2013 #15

    TSny

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    Well, we'll see. I think we should wait for Pranav-Arora to return to the discussion.
     
  17. May 10, 2013 #16
    Yeah that's the best thing... and I could end out to be wrong, even if I am still convinced of what i was saying ;) anyway let's see what he says :)
     
  18. May 10, 2013 #17
    I am still lost on this one even after the replies.

    Should I consider that motorbike leans at a certain angle and make the equations?

    Friction? :confused:

    How both the accelerations are related? :confused:
     
  19. May 10, 2013 #18

    TSny

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    Yes, friction is the force. What is the magnitude of the maximum acceleration that the bike can have? How would you express this magnitude of acceleration in terms of tangential and centripetal components?
     
  20. May 11, 2013 #19
    Maximum acceleration can be ##\mu g##?

    It can be expressed as:
    [tex]\left(\frac{mv_{max}^2}{R}\right)^2+(ma_t)^2=(\mu g)^2[/tex]
    where ##a_t## is the tangential acceleration. Does this look correct?
     
  21. May 11, 2013 #20

    ehild

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    NO. Check.

    ehild
     
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