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Circular Motion - Banked Turns

  1. Sep 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Please could someone help explain the case of a friction-less banked race track. If you take an Olympic indoor banked cycle track and imagine for a moment that it was friction-less. The laws of circular motion tell us that without friction the banked geometry can enable a cyclist to hypothetically still race round the track. The equation that tells us this is v = sqrt(rgtantheta) where v is the velocity of the bike, r is the radius of the track, g is gravity and theta is the angle of the bank. It says in my book that if you solve for a value of v for a friction-less track then the bike must ride at this velocity and this velocity only. Any less and it will slide down and any more and it will slide up.

    My main question is, what keeps the bike from sliding on a friction-less surface if you maintain this speed? I understand that the vertical component of the reaction force counters the weight and the horizontal component provides the centripetal force. If you take the component of the weight acting down the slope then what counters this? I was thinking it might be the inertia of the bike but then inertia isn't strictly a force is it?

    2. Relevant equations

    v = sqrt(rgtantheta)

    3. The attempt at a solution
     
  2. jcsd
  3. Sep 27, 2014 #2

    Simon Bridge

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    Draw a free-body diagram for the object.
    To go in a circle, the vector sum of the forces has to point horizontally towards the center. Any other result is not a circle.

    This is not a static case - the object is in motion - so it is not so easy to think about inertialy.
    Basically there is no component of the net force which points along the surface of the track - therefore no acceleration up and down the track. You can have a go resolving the forces along the surface of the track like you would normally for an object on a slope, see what you get ;) You can also see what happens when the velocity is higher than that needed for circular motion.

    But if you take the frame of reference of the object, then it is easy to see that the normal force from the track balances the sum of the weight and the centrifugal pseudoforce.
     
  4. Sep 28, 2014 #3
    Thanks Simon. If you look at the weight force (still assuming the friction-less case) then surely I am entitled to apply the incline plane scenario where I can resolve the weight force to have a component pointing down the slope? I don't get what would cancel this component out if there were no friction. The only other force is the normal force and this has no component along the slope since it is orthogonal to the slope? Or is it simply the inertia of the bike wanting to fly upwards off the track that counters the weight wanting to push it down the slope?
     
  5. Sep 28, 2014 #4

    andrewkirk

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    It is a confusing situation. I think about it as follows:

    The sum of the forces applied to the bike, divided by its mass, must equal the sum of the accelerations experienced by the bike.

    On its own that's just a banal restatement of Newton's law, but we can write it out as:

    ##\vec{F_g}/m+\vec{F_n}/m=\vec{c}+\vec{r}##

    where ##\vec{F_g}, \vec{F_n}## are the forces from gravity and from the boards of the velodrome (normal force) on the bike, and ##\vec{c},\vec{r}## are respectively the centripetal acceleration necessary to follow the line on the boards, and the residual acceleration which will be parallel to the boards, either sliding directly up or down the surface.

    So we get ##\vec{r}=(\vec{g}-\vec{c})+\vec{n}## where we write ##\vec{g}=\vec{F_g}/m, \vec{n}=\vec{F_n}/m##.

    Note that ##\vec{g}## is fixed and ##\vec{c}## is fixed by the shape of the velodrome. The directions of ##\vec{n}## and ##\vec{r}## are fixed but not their magnitudes. So we can solve the problem by drawing the diagram of acceleration vectors as follows:

    Start at point O, draw the vector ##\vec{g}## downwards from O to A, then the vector ##-\vec{c}## (note the minus that reverses its direction) horizontally from A to B. In my drawing the top of the slope is on the left so ##\vec{c}## points right and so ##-\vec{c}## points left. From B draw ray ##l## in the direction of the upwards normal to the boards and drop a perpendicular from O to ##l##, crossing it at point C. Then ##\vec{BC}=\vec{n}## and ##\vec{OC}=\vec{r}## is the residual acceleration. If the the track is too steep (shallow) for the cyclist's speed and the track's curvature, ##\vec{r}## will point down (up) the boards and that's the way the cyclist will slide. If the track pitch is just right, we will have ##\vec{r}=0##.


    Approaching it this way, we can cover the inclined plane case as well. For that case ##\vec{c}=0## and the object slides down.

    If the diagram is drawn correctly, we can see that it comes down to a comparison of the directions of ##\vec{c}-\vec{g}## and ##\vec{n}##. If the former is more (less) vertical than the latter the bike slides down (up). The inclined plane case is where the former is absolutely vertical, so the object slides down.

    A final note - something I just realised. The bike would have to be propelled by rockets, jets or propellers as, if the surface is frictionless, it could not be propelled forward by turning the wheels!
     
    Last edited: Sep 28, 2014
  6. Sep 29, 2014 #5

    Simon Bridge

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    The acceleration is always in the direction of the total force, not just in the direction of the one force we happen to be looking at.

    The "unopposed" component of weight is contributing the component of the acceleration vector down the slope ... why does the bike, then, not slide down the slope? Well because there is also an unopposed net force in the normal direction.

    The sum of all the forces, remember, points horizontally towards the center; so, in the normal+down-slope coordinates, the resultant force will have two components.
     
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