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How does a motocylist move his center of mass?

  1. Jun 19, 2013 #1
    Hi all:


    we all know that a motorcyclist need to tilt his/her center of mass with the motorcycle when he is turning in order to balance the torque exerted on him.


    When he is tilted, gravity exert a torque on him that allows him to balance torque by centrifugal force about the point of contact with motorcycle and ground.

    or when he is tilted, normal force from the ground exert a torque on him balance the torque exerted by friction on his center of mass.



    but in order to be tilted in the first place, there must have been a net torque acted on him in order to move his center of mass perpendicular to his direction of travel. So did his torque come from pushing the air around him?


    thank you
     
    Last edited: Jun 19, 2013
  2. jcsd
  3. Jun 19, 2013 #2

    mfb

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    He uses the handle bars to start a curve. You do not have to, but that is the easiest way to lean towards one side: Rotate the handle bars (slighly) towards the opposite direction.

    I am sure an oar would work, too, but that is a bit... impractical.
     
  4. Jun 19, 2013 #3
    You can shift your center of mass, while angular momentum around the contact point of the wheels with the ground stays zero at all times, and there is no net torque on the rider+the motorcycle.. If you move your hips (and the motercycle) somewhat to the right, and your upper body to the left this will shift the center of mass to the right.

    if you move a mass m_1( the motorcycle + your lower body) to the right with speed v_1, at altitude h_1 from the ground and a mass m_2 (your upper body) to the left with speed v_2 at altitude h_2 from the ground. (we must have h_2>h_1)

    to conserve angular momentum you need

    [tex] m_1 h_1 v_1 = m_2 h_2 v_2 [/tex]

    The center of mass however moves [tex] \frac {m_1 v_1 - m_2 v_2}{m_{total}} [/tex] to the right, and since the conservation of angular momentum gives

    [tex] m_2 v_2 = m_1 v_1 \frac {h_1}{h_2} [/tex] and this is smaller than m_1 v_1 if h_2>h_1 so

    [tex] \frac {m_1 v_1 - m_2 v_2}{m_{total}} > 0 [/tex] and the center of mass moves to the right.


    tightrope walkers do the same,t they stick out a leg to the right and and arm to the left, this will move the center of mass to the right.
     
  5. Jun 19, 2013 #4

    rcgldr

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    It's called countersteering. Like mfb stated, the rider initially steers in the wrong direction, this results in a newton third pair law of forces, the tire outwards against the pavement, the pavement inwards against the tires. The force from the pavement results in centripetal acceleration of the motorcycle, and a torque that causes the motorcycle to lean. By initlally steering in the wrong direction, the bike leans "outwards" with respect to the wrong steering inputs, but this ends up being an "inwards" lean when the rider switches to steering in the correct direction.

    If you're wondering how people can steer without their hands on the handlebars, by body leaning inwards, the bike ends up leanind outwards, and the geometry of most bikes will tend to steer in the direction of the lean of the bike, so in this case since the bike is initially leaning outwards, it steers outwards. It's an indirect method of counter steering.

    There are debates about counter steering versus weight shifting for bicycles and motorcycles, but in the case of unicycles, there doesn't to be any debate and the community basically accepts the concept of counter steering.

    wiki article:

    http://en.wikipedia.org/wiki/Countersteering
     
  6. Jun 19, 2013 #5

    i was always under the impression that....you cannot change your center of mass using internal forces (in this case system is your body right?)


    if i extend my arm to the left, in the process i must have pushed the rest of my body to the right by a bit due to 3rd law, or perhaps I'm not understanding you completely?
     
  7. Jun 19, 2013 #6

    jbriggs444

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    That is true, of course.

    However, you are not necessarily using internal forces. Your feet (if standing) or the bottom of your wheels (if riding) are on the ground. A tight-rope walker can utilize this fact to push his waist one way while his head and feet go in the opposite direction. His feet on the wire provide the net thrust that moves his center of gravity in the desired direction.
     
  8. Jun 19, 2013 #7


    I think you are missing something here ( please tell me if I am wrong on this)


    when you say :


    [tex] m_1 h_1 v_1 = m_2 h_2 v_2 [/tex]


    , you are assuming that the force i exert on the motorcycle and the force it exert back on me is at our center of mass, but that cannot be the case when our individual center of mass were initially aligned vertically, so in order to exert a force that can cause change in velocity in the horizontal direction, the contact force between me and the motorbike must have exerted a torque on both me and the bike about each of our own center of mass.


    and i do believe the general angular momentum equation is :

    mVcm * r + moment of inertia * omega

    (parallel axis theorem)?

    please correct if I am wrong
     
    Last edited: Jun 19, 2013
  9. Jun 19, 2013 #8

    What I have problem with the most is the fact that center of mass of me and the bike is moving at all, this statement by itself already implies a net change in angular momentum. if we treat both me and bike as a system, the forces on the ground should be pushing my center of mass toward the opposite direction of turning (assuming we are turning to the left), yet now the COM of our system is moving to the left? this doesn't sound consistent.
     
  10. Jun 19, 2013 #9

    this makes perfect sense, thanks! (I initially thought you have to push against air)
     
  11. Jun 20, 2013 #10
    No, I'm not assuming that. I'm saying nothing about the forces needed to move your body or the motor cycle. what i'm assuming is that some mass at the level of your hips moves to the right, and that some other mass at a higher levels moves to the left.
    (this isn't a very accurate model, but only meant to show that movement of the center of mass, while the angular momentum is 0 is possible)


    A mass m moving at altidude h is at a distance h from the contact point of the wheels, so its angular momentum is m*h*v. (I'm considering momentum around an axis pointing in the direction of motion of the bike)
     
  12. Jun 20, 2013 #11

    mfb

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    I agree, and I would like to add that the required external forces come from the wheels. It is easier to see that with an extreme example: A small mass resting on the ground with a point-like contact, and a larger mass on top of it, without a fixed connection between the two masses. If you change the relative position between the two masses, you can get a strong sidewards force on the whole system, mediated by the (tilted) contact of the lower mass.
    The effect might be small, but it is self-accelerating, once a small tilt has been established gravity provides a net torque to increase the tilt.

    Another surprising fact: You can change the rotating angle of an object without any angular momentum. Cats do that when they are falling, for example.
     
  13. Jun 21, 2013 #12
    So are you saying that it is possible to cause your center of mass to shift TOWARD the direction of turning without counter steering and simply by pushing the bike to the other direction with your body?
     
  14. Jun 21, 2013 #13

    mfb

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    I would expect that, but I do not have an estimate about the size of the effect. Counter-steering is certainly more effective and easier to control (see post #2).
     
  15. Jun 21, 2013 #14

    jtbell

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    One can see this effect in the track left by a bicycle, either on smooth damp pavement or with wet wheels (from a puddle of water) on dry pavement . In e.g. a right turn, the track initially veers left slightly, by a few cm.
     
  16. Jun 21, 2013 #15


    do you think the shifting of center of mass can be validated if I analysis both me and the bike as one system?
     
    Last edited: Jun 22, 2013
  17. Jun 22, 2013 #16

    mfb

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    What do you mean with "as one system"? The whole system always contains the bike and you.
     
  18. Jun 22, 2013 #17

    rcgldr

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    Assuming the bike and rider are initially vertical, than any leaning is due to steering inputs. Once a steering input is applied, then the center of mass of the bike and rider will experience centripetal acceleration in the direction of steering, while at the same time leaning the wrong way (outwards relative to steering input). This is the countersteering phase of a turn. Once leaned over, then an additional torque exists due to gravity exerting a downwards force at the center of mass of bike and rider, and the pavement exerting an upwards force at the contact patch.

    On most bikes, the steering geometry (trail, caster) will tend to steer the front tire in the direction of lean, so once the rider stops or reduces the counter steering torque applied to the handlebars, the bike will then start to turn in the desired direction, with the goal of establishing a coodianted turn (no net torque about the roll / lean axis). On most bikes, there's a tendency to over correct for the lean angle and straighten up with no steering input, so a bit of counter-steering torque on the handlebars is required to maintain a steady lean in a steady turn.

    At high speeds like the ones experienced on racing motorcycles, the self correcting response related to lean transitions into a tendency to hold the current lean angle, and the rider needs to apply counter-steering inputs to both increase (outwards steering torque) and decrease lean angle (inwards steering torque).
     
  19. Jun 22, 2013 #18


    what i mean is is it possible to show that the angular momentum is conserved about the point of contact by considering me and the bike as one mass?

    if we consider the angular momentum of the system as a whole about the point of contact:

    it would be:


    angular momentum = mtotal * (center of mass velocity) * center of mass vertical location

    WALA : angular momentum = m1v1 *(1-(h1/h2)) * (m1h1 + m2h2 / mtotal)

    which doesn't equal to zero?
     
  20. Jun 22, 2013 #19

    mfb

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    As long as the center of mass is above the point (or line, in 3 dimensions) of contact, there is no torque and angular momentum around the road contact is conserved.
     
  21. Jul 12, 2013 #20
    what about angular momentum about the center of mass? it seems that the only force acting on the center of mass is friction toward the direction of turning.


    say the biker is moving out of the page, turning to the right, wouldn't this induce a moment about his center of mass counter clockwise?
     
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