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Homework Help: Linear speed of car coasting on two tracks

  1. Apr 25, 2014 #1
    This is another one of those concept problems where if I think about it one way, it makes sense, but if I think about it another way I don't reach the same conclusion...again, any clarifications in explaining this would be much appreciated.

    A streetcar is freely coasting (no friction) around a large circular track. It is then switched
    to a small circular track. When coasting on the smaller circle its linear speed is...


    2. Relevant equations: kinetic energy = 1/2mv^2

    3. Attempt at solution

    If I take the system to be the two tracks, earth and the car, then I can consider this as a closed system, where energy is conserved. There is no change in potential energy in the system, and we'll assume that there is no friction/heat loss. Since in a closed system, no energy enters or leaves the system, and the mass of the car does not change, then we can conclude that the kinetic energy of the car must stay the same, and that the velocity must then be the same.

    BUT...what if I try to look at this from a angular momentum approach? It is a closed system and there are no external torques. So then couldn't you liken this to a revolving ice skater problem? When the ice skater brings his/her arms in, the linear/rotational speed increases. So likewise, when the coasting car transitions to the smaller track, why doesn't it speed up like the ice skater but just stays at the same speed instead?
  2. jcsd
  3. Apr 25, 2014 #2
    In order to transfer the car from a large to a small circle, is any energy input required?
  4. Apr 25, 2014 #3
    No energy input is required. The big/small tracks are tangential to each other. So the car is coasting on the bigger track, then at that connection point, naturally transitions into the smaller track. Somewhat similar to a figure 8.
  5. Apr 25, 2014 #4
    OK, so I think you're right about the energy.

    But the angular momentum is a product of 2 variables (mass is constant). Could angular momentum still be conserved if the velocity is the same?
  6. Apr 25, 2014 #5
    When you spiral in, the force from the track, which is perpendicular to your speed, is not in the direction of the middle of the spiral, so there is a torque that reduces your angular momentum.
  7. Apr 25, 2014 #6
    It is in the direction of the center of the small circle, so I dont see where the torque comes in. The frictionless tracks only cause a radial acceleration.
  8. Apr 25, 2014 #7
    That's only true if the track is a circle. If the track gets closer to the center of the circle, then the direction of the acceleration, wich is perpendicular to the track is not in the direction of the center, so there is a torque present.
  9. Apr 25, 2014 #8
    I thought the tracks were both a circle.

    A torque must change the tangential velocity. I think the OP correctly reasoned that the tangential velocity remains unchanged.
  10. Apr 25, 2014 #9
    But how are you ever going from the larger to the smaller circle? There must be a section of track between them that's not a circle.
  11. Apr 25, 2014 #10


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    If the circles are concentric, then you have to spiral in.

    On the other hand...
    If you suddenly are on a track on a different circle by way of a shared tangent, such as in a figure 8 , then you will likely be determining angular momentum about a completely different point. While the train travels around the first circle, its angular momentum w.r.t. the center of the second circle is not constant. Likewise, while the train travels around the second circle, its angular momentum w.r.t. the center of the first circle is not constant.
  12. Apr 28, 2014 #11
    The solution states that energy is conserved, so the answer is that, yes, tangential velocity does remain unchanged.

    There is no external torque in the system, so I see that the rule of conservation of angular momentum it breaks must be not being about the same point (center of the first circle). But what is the specific reason that the point must be the same in order to say that momentum is conserved? Because that is implicitly our reference frame/point as well? But isn't it just the distance between the point of rotation and train (treated as a point mass in this case) that we're concerned about in the equation L = r * mv ?
  13. Apr 28, 2014 #12


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    Define angular momentum of a point mass.
  14. Apr 28, 2014 #13
    For an object with a fixed mass that is rotating about a fixed symmetry axis, the angular momentum is expressed as the product of the moment of inertia of the object and its angular velocity vector:

    L = I w

    Ok, so fixed axis. I guess I was overthinking it =)

    Thanks SammyS!
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