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Front flip of a bike. Conserving Angular Momentum.

  1. Apr 18, 2015 #1
    1. The problem statement, all variables and given/known data
    We model a dirt bike as a rod of mass (M) 100 kg, and length (L) 2 m to which are attached two wheels of mass (m)=40 kg kg, and radius R=0.5 m. The bike goes off a spine ramp with angle ##\phi=1deg.## to the vertical on either side (see diagram below).

    Our question is, at what time, ##t_{pull}## (in seconds), after leaving the ramp, should the rider pull the brakes (stopping both wheels instantaneously) if they want to do a front flip with minimum takeoff speed?

    • The answer to this physics question is a real number.
    • On landing, the bike is parallel to the landing ramp, both wheels are in contact with the landing surface, and the back wheel contacts the very top of the ramp.
    • The wheels are rings, i.e. the spokes are massless and all the mass is concentrated at the radius and ##I=mR^{2}##.
    • For simplicity, assume that both wheels lose contact with the ramp at the same angular velocity, ##v/R## , where ##v## is the velocity of the motorcycle upon losing contact with the ramp.
    • The bike coasts off the ramp, i.e. there is no power from the engine putting a torque on the wheel.
    • The bike lands on the opposite side and does a full front flip, I.e. 540 - 2= 538 degrees
    Mecc.jpg

    2. Relevant equations


    3. The attempt at a solution

    I first tried to analyse the problem. When the brakes are applied then the wheels stop spinning and as these forces are internal forces, if we consider bike as a system, so they can not brought deviation in the trajectory of the CoM of the system.

    As these are internal force so we can conserve angular momentum of the system about its CoM. This will give the magnitude of the angular velocity after the brakes are applied. We can also conserve ttotal energy of the system.

    Am I right??

    When brakes were not applied till then the bike was not rotating about the its CoM. So the orientation of the bikes remains same.

    ssss.png


    As the wheels were not rotating about the CoM of the bike before the application of the brakes. So the velocity of the CoM of the wheels wrt the CoM of the system is zero.

    Now the angular momentum of the wheels about the CoM( let it be point O) of the frame is equal to
    Angular momentum of the CoM of the wheels about point O+Angular momentum of wheels about its CoM.


    Now the 1st term is zero be cause the wheels were not rotation about point O. So

    Initial angular momentum is ## 2I_{w}\omega##.

    When the brakes are applied then whole bike would start rotating about the CoM of the bike. Let the angular velocity be ## \omega_{0}##.

    So Final Angular momentum is ##I_{system}\omega_{0}##.

    So ##I_{system}\omega_{0}=2I_{w}\omega##.

    After that I would get a constraint relation that the bike have to rotate 538 deg. in the time laft to land on the ground. I mostly make mistakes in conserving Angular momentum. :confused:

    Have I done correctly this time?:oops:

    EDIT: I have taken Counter Clockwise as +ve while calculating ##L##.



     
  2. jcsd
  3. Apr 18, 2015 #2

    haruspex

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    All looks right so far, except that work will not be conserved when the brakes are applied. But you don't seem to have used that anyway.
     
  4. Apr 18, 2015 #3
    So,

    ##I_{system}=\frac{ML^2}{12}+2 \left( m{ R }^{ 2 }+m{ \left( \frac { l }{ 2 } \right) }^{ 2 } \right) ##

    ##I_{w}=mR^{2}##

    So on putting these I got

    ##\omega_{0}=0.3v##

    Now time period of projectile is

    ##\frac { 2v\cos { \phi } }{ g } ##

    Time requires by wheels to rotate 538 deg. =## \frac { 269\pi }{ 90(0.3v) } ##

    So Answer is ##\frac { 2v\cos { \phi } }{ g }- \frac { 269\pi }{ 90(0.3v) } ##.

    I think I need the value of 'v'. How should I get it?
     
  5. Apr 18, 2015 #4

    haruspex

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    You mean, the time required for the whole bike to rotate, yes?
    It isn't made clear, but I assume the start point is when the front wheel leaves the ramp. That makes it symmetric with the landing position.
    Consider the horizontal morion while airborne.
     
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