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Circular motion and momentum

  1. Mar 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Explain why in circular motion there is a force pointing to the center of the trajectory. Use the momentum principle.

    2. Relevant equations


    3. The attempt at a solution
    So I choose two arbitrary points in the circular path and know that momentum and velocity are tangent to the trajectory at any point.

    From ##\Delta \overrightarrow{p} = \overrightarrow{p_{f}} - \overrightarrow{p_{i}}## I know that picking up any two points in the trajectory, the change in momentum can never point outwards the trajectory, it always points inwards.

    From ##\Delta \overrightarrow{p} = F_{net} \Delta t \iff \overrightarrow{F}_{net} = \frac{\Delta \overrightarrow{p}}{\Delta t}## I can say that ##\overrightarrow{F}_{net}## must point in the same direction as ##\Delta \overrightarrow{p}##.

    Lastly, if I take ##\Delta t \rightarrow 0## I can see that ##\overrightarrow{F}_{net}## becomes closer and closer to ##\overrightarrow{F}_{net} \cdot \overrightarrow{v} = 0## or ##\overrightarrow{F}_{net} \cdot \overrightarrow{p} = 0##.

    Is that a good explanation?
     
  2. jcsd
  3. Mar 27, 2015 #2

    robphy

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    The last line sounds like a problem... you say that a vector "becomes closer and closer" to a scalar equation?

    Is the problem statement precise?
    Is the motion uniform circular motion (so that ##\vec v\cdot \vec v =constant## and the net-force is radially inward)?
    If not, then it is a general circular motion (allowing for speeding up and slowing down) (so the net-force need not point radially inward, although there will be a radially-inward component).

    In addition,
    I think you mean...
    ##\Delta \overrightarrow{p} = F_{net,average} \Delta t \iff \overrightarrow{F}_{net,average} = \frac{\Delta \overrightarrow{p}}{\Delta t}##

    As ##\Delta t\rightarrow 0##, then ##\vec F_{net,average} \rightarrow F_{net}##.

    (Sounds like the Matter and Interactions text. Is it?)
     
  4. Mar 27, 2015 #3
    Force is proportional to acceleration. The normal acceleration (in any movement) is ##\frac{v^2}{R}\vec{u}_n##. In a circular motion (uniform or not) ##R## is constant and not zero, so ##\vec{F}_n = m\vec{a}_n = m \frac{v^2}{R}\vec{u}_n## is not zero (assuming v is not zero).
     
  5. Mar 27, 2015 #4
    Matter and Interactions book.

    Last line was an attempt to say that if I make ##\Delta t## really close to zero, the force vector is closer to being ##\perp## in relation to momentum and velocity.
     
  6. Mar 27, 2015 #5

    haruspex

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    You are not asked to show they are orthogonal (and in general they won't be, as robphy pointed out). You are only asked to show there is a radial component of force.
     
  7. Apr 3, 2015 #6
    I missinterpreted the question.

    Forgot one important detail: coordinate system is cartesian and origin is at the center of the trajectory.

    Choosing any two points in the trajectory exposes one problem:
    if the two points are along the width of the circle, both momentum vectors are going to be parallel and the difference is a vector that doesn't points inwards but is tangent to the trajectory. In case the two points are the same, but one is placed "one lap" later, both vectors are going to be the same and the difference is going to be null.

    There is a slightly missconception in ##\Delta t \rightarrow 0##. I was using that to explain why the vectors are orthogonal. While it's true for circular motion along a circunference, it's not for the orbit of planets. That operation is more closely related to explaining why there is an assumption that the force is constant. Because with ##\Delta t \rightarrow 0##, there is "not enough time" for a change in the magnitude of the force to be measured.

    Third problem: I though that the question was the same as "Two students proposed two vectors to represent the force acting upon the object at one point of the trajectory. One is an arrow pointing towards the center of the circle, the other is an arrow pointing in the opposite direction of the former". As I see now, explaining why the force cannot point outwards it's slightly easier than explaining why the vector must point to the center of the trajectory, or, in case of a planet, why the force points to the center of mass of the other massive object its orbiting.

    The answer given by the teacher was conceptualy much simpler: Use ##\overrightarrow{p}_{f} = \overrightarrow{p}_{i} + \overrightarrow{F} \Delta t##. If the vector with coordinates pointing outwards the center of the circle is applied, the curve would be in the opposite direction of the actual direction of the motion (if object is turning right and moving clockwise, force would make it turn left, which is absurd). Therefore, the vector pointing to the center of the trajectory is right and the other is wrong.
     
    Last edited: Apr 3, 2015
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