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Central Forces

  1. Apr 26, 2010 #1

    bon

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    1. The problem statement, all variables and given/known data

    2 masses mass m connected by a light string length l. One moves on a smooth horizontal table in which there is a small hole. The string passes through the hole so that the second mass hangs vertically below. Asked to use conservation of energy and angular momentum to show that

    (where r' = r dot)

    r' ^2 = A - B/r^2 - gr

    where r(t) is the distance of the first particle from the hole...A and B constants

    2. Relevant equations



    3. The attempt at a solution

    Ok so im trying to use the central force eqn F(r) = -mg r hat

    But how do I use consv. of energy and mom?

    isnt consv of momentum here just r^2 theta dot = const? What about energy? Thanks
     
  2. jcsd
  3. Apr 27, 2010 #2

    ehild

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    That central force acts to the body on the table and it is equal to the tension of the string. ( Force of gravity - tension of the spring) is the resultant force on the suspended mass, but it accelerates, so the tension is not equal to mg.

    You have two bodies connected by the spring of length L. One moves on the table, at can have both radial and tangential velocity. The other body is suspended on the string and can move vertically downward. Its PE will change as it moves. It gains speed so its KE will also change. The string forces the mass on the table move inward with the same speed. What will be the sum of the KE an PE of both bodies?

    ehild
     
  4. Apr 27, 2010 #3

    bon

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    ok thanks i see now..but im still having trouble seeing how to arrive at their equation...

    what should i be looking at? Equations of motion individually? Changes in KE and PE for small displacements?

    Just not clear how to do this..


    THanks for your help
     
  5. Apr 27, 2010 #4

    ehild

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    Try to write out the expressions for KE and PE.

    ehild
     
  6. Apr 27, 2010 #5

    bon

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    Oh ok thanks i see..

    so i get for mass 2 PE = -mg(l-r) and KE = 1/2 m r' ^2

    for mass 1 PE = 0 KE = 1/2 mr' ^2 (where this r' is a vector)..

    So equating these and simplifying i get 1/2 r^2 theta dot ^2 = -gl + gr.. which isn't right?
     
  7. Apr 27, 2010 #6

    ehild

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    The energies are not equal, but their sum is constant!

    ehild
     
  8. Apr 27, 2010 #7

    bon

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    Thanks ehild..though im still a bit stuck. Decided to solve an easier problem first..please tell me where im going wrong:

    We want to put a satellite into orbit with apogee (Furthest distance from planet) of 5R/2 where R is the radius of the planet. The satellite is to be launched from the surface with a speed Vo at 30 degrees to the local vertical. If M is the mass of the planet, I'm trying to show that Vo^2 = 5GM/4R.

    Ok so using what you have said, I equate the sum of KE and PE for the satellite in the two positions 1) where it is just leaving the surface of the planet and 2) where it is at apogee..

    so surely 1) KE = 1/2 m Vo^2 PE = -GMm/R

    and at 2) KE = 25/8 m R^2 theta dot ^2 and PE = -GMm/(5R/2)

    So then I can find an expression for theta dot - theta dot = GM/r^3..?

    But it doesnt come out right?! Where am i supposed to use the 30 degrees from local normal piece of info? thanks ehild
     
  9. Apr 27, 2010 #8

    bon

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    ok i get what you are saying about the first one now..its just that the rdot cancels when you equate the sum of KE and PE of the two particules..why?
     
  10. Apr 27, 2010 #9

    bon

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    look:

    KE particle 1 = 1/2 m r'^2 where r' is the vector r'... PE = 0

    KE particle 2 = 1/2 m r'^2 where r' is the scalar r'...PE = -mg(l-r)

    but vector r' = r' r hat + r theta dot theta hat..

    so when you equate the sums, the r'^2 term disappears...which the term that appears in the final answer..! Why?
     
  11. Apr 27, 2010 #10

    ehild

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    That 30 degrees determines the initial radial and azimuthal velocity components. If v0 is the magnitude of the velocity at launch, the radial velocity is vr0=v0*cos(30°) and the azimuthal component, (r0*theta dot ) =va0= v0 *sin(30°).
    At the apogee, the radial velocity is zero (the satellite does not go farther), but still has an azimuthal component, vathat obeys the conservation of angular momentum L. The magnitude of L is the product of the radius with mass and azimuthal velocity:

    L=mv0sin(30°)*R=Mva5/2 R

    This equation, together with conservation of energy let you calculate both va and v0, or the angular velocities at launch and at apogee.

    The equation for the angular momentum is needed also in the solution of the previous problem.

    ehild
     
    Last edited: Apr 27, 2010
  12. Apr 27, 2010 #11

    bon

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    KE initial = 1/2mvo^2
    PE initial = -GMm/R

    KE apogee = 1/2m(r^2theta dot^2)
    PE apogee = -GMm/(5R/2)

    Is this right so far?

    So now all we need is an expression for theta dot..but i thought that initially the angular momentum = 0 as the body is moving directly away from the surface in a straight line (initially)

    I see how you can find components of vo knowing that it leaves at 30 degrees to vertical, but i dont see how you can use this in the conservation of angular momentum eqn..
     
  13. Apr 27, 2010 #12

    ehild

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    The angular momentum is not zero at the beginning. It is the vector product of the momentum with the radius vector. Read my previous post.

    ehild
     
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