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2 sphere system, Conservation of Energy and Momentum

  1. Sep 13, 2015 #1
    1. The problem statement, all variables and given/known data

    A sphere of radius $1m$ and mass $25 Kg$ is put on another sphere of radius $5 m$ and $7 Kg$ which is placed on a smooth ground. Now the upper sphere is pushed very slightly from it's equilibrium position and it begins to fall.

    Now when the line joining the centre of the 2 spheres makes an angle $\theta$ with the vertical the upper sphere loses contact with lower sphere. Then find $\theta$ in degrees.

    Note that there is no Friction between the 2 spheres.


    2. Relevant equations

    $$g=9.8 ms^{-2}$$
    $$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$
    $$U_{initial} + K_{initial}=U_{final}+K_{final}$$

    3. The attempt at a solution

    Unfortunately, I haven't been able to get anywhere with the solution.

    I took the small sphere (of radius r) as $m$ and the large sphere (of radius R) as $M$. The velocity of the small sphere is $v_m$ and the velocity of the large sphere is $v_M$.


    When the small sphere will fall off, the Normal Reaction would be 0. Using this relation,
    at the time of falling off (taking radial components),

    $$\dfrac{mv_m^2}{R+r}=mg\cos\theta$$

    I was told that the next step would be to conserve linear momentum in the horizontal direction since there was no external force in that direction. Lastly I was to apply the Conservation of Mechanical Energy for the system sphere+sphere+Earth.

    Also, I took the 0 Potential Energy reference level as the common horizontal tangent to both spheres.



    Unfortunately I'm unable to form the Equations for Conservation of Linear Momentum and Mechanical Energy.

    I would be grateful for any help with this question. Many thanks in advance!

    PS. My (incorrect?) equation for Conservation of Energy is as follows: Taking 0 Potential Energy at the top of the sphere, $$-MgR+mgr=\dfrac{1}{2}MV_M^2+\dfrac{1}{2}mv_m^2-MgR-mgR(1-\cos\theta)$$
     
    Last edited: Sep 13, 2015
  2. jcsd
  3. Sep 13, 2015 #2

    mfb

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    The big sphere will move as well.
    For the initial positions?
    I guess the floor would be more intuitive.

    If there is friction, the spheres will rotate as well, but the problem statement is a bit unclear in several aspects.
     
  4. Sep 13, 2015 #3
    Sir, I'm really sorry. I had meant to write that there is no friction. Is the problem more clear now?
    Sir, why would the floor be more intuitive, please?

    Also, could you please show how you arrived at $$
    \vec{F} = a^2 \cos(\theta) \cdot \{ \sin(\theta) \cos(\varphi), \sin(\theta) \sin(\varphi), \cos(\theta)^2 \}?$$
     
  5. Sep 13, 2015 #4

    mfb

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    It is the usual way to define "zero height".

    That equation in my post was a copy&paste error, sorry. Still had a formula from an unrelated thread in the clipboard. I wanted to copy this formula:
    The big sphere will move as well.
     
  6. Sep 13, 2015 #5
    Sir, I realised that it would move. This would be because of the Normal Force between the 2 spheres. However, I still cannot understand how to apply the law of Conservation of Momentum.
     
  7. Sep 13, 2015 #6

    mfb

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    Then you should take this motion into account.

    How can you describe the horizontal momentum in terms of the velocities of the spheres?
    What do you know about the horizontal position of the center of mass?
     
  8. Sep 13, 2015 #7
    I'm sorry Sir, I don't know the answer to the first question. Had there been 2 separate spheres with velocities v1 and v2, the momentum (initial) of them would be $m1v1$ and $m2v2$. However, the confusion is arising since these 2 spheres are in physical contact and it thus seems that the motion of the bottom sphere will influence the motion of the top sphere. Thus, the velocity of the top sphere would not simply be $v_m$. This is what is bothering me, Sir.

    Regarding the second question, I didn't understand. Wouldn't the center of mass of each sphere always be at the center of each sphere respectively?
     
  9. Sep 13, 2015 #8

    mfb

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    The contact does not matter.
    That is true, but it does not change for formulas for their momentum.
    You can define the velocity to be v_m. By the way, for LaTeX you can use two $ or two # here: ##v_m##.
    The center of mass of the spheres yes, but I meant the center of mass of the whole system.
     
  10. Sep 13, 2015 #9
    Sir, regarding the center of mass, in the horizontal direction, it would be at the same level (ie the ##Y## coordinate would be the same) since there is no external force acting in the horizontal direction.
    Sir, please could you explain your first 2 points? I couldn't understand why the formula for momentum of the upper sphere wouldn't be affected if the velocity of the upper sphere is influenced by the lower sphere. As an example, suppose a block is placed on a surface moving towards the left with ##v_1## while the block moves on the surface with ##v_2##. Wouldn't the momentum of the block in the ground frame be ##m_{block}v_2# and in the frame of the surface, ##m_{block}(v_2-v_1)##?
    Also Sir, could you please explain what you meant by ''define the velocity'? Did you mean wrt a specific frame?
     
  11. Sep 13, 2015 #10

    mfb

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    Right. You can write that down as formula. It helps to find a relation between velocities.

    The momentum of an object is always the mass multiplied with the velocity. It does not depend on which other objects it touches.
    The velocity will depend on the contact, of course, but that is a different part of the problem.

    What is vm? You introduced the variable, so you have to choose what it means. If you say "vm is the velocity of the small sphere", then it will be the velocity of the small sphere. My comment was on your question if vm can be the velocity of the small sphere. Sure it can, if you define it to be.
     
  12. Sep 13, 2015 #11
    I couldn't understand why the formula for momentum of the upper sphere wouldn't be affected if the velocity of the upper sphere is influenced by the lower sphere. As an example, suppose a block is placed on a surface moving towards the left with ##v_1## while the block moves on the surface with ##v_2##. Wouldn't the momentum of the block in the ground frame be ##m_{block}v_2## and in the frame of the surface, ##m_{block}(v_2-v_1)##?

    Sir, I'm assuming that originally, the center of both spheres lie on the Y axis. Hence, the ##X## coordinate of the center of mass will always be 0 since there are no external forces acting in the horizontal direction. Hence, ##-mx_1## where ##x_1## is the displacement of the upper sphere's center of mass, is always equal to ##Mx_2## where ##x_2## is the displacement of the lower sphere's center of mass.
    ##v_m## is the velocity of the upper sphere wrt the ground frame.
    Could you please tell me how to proceed next?
     
    Last edited: Sep 13, 2015
  13. Sep 13, 2015 #12

    mfb

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    I am traveling at 20 km/h to a town 20 kilometers away. How long do I need to go there? To find the answer, do you have to know if I run there, take a bike, a car or a train?
    It simply does not matter where the velocity comes from. Momentum is always* mass multiplied by velocity.

    The reference frame here is the ground. Sure, momentum does depend on the reference frame, but every other choice blows up the equations and does not help.

    *ignoring relativity
     
  14. Sep 13, 2015 #13
    Sir, I got that. I had meant to ask how we account for the velocity of the upper sphere.
     
  15. Sep 13, 2015 #14

    mfb

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    With a variable describing this. Like vm. Better: with its horizontal component.
     
  16. Sep 13, 2015 #15
    So Sir, at the instant when the upper sphere loses contact, the total momentum of the system is $$-Mv_M+mv_m\cos\theta?$$ But doesn't this assume that the velocities of the respective spheres do not change with time, ie they remain the same as they initially were? How then would these velocities account for the effect of the respective spheres on each others' velocity?
     
  17. Sep 13, 2015 #16

    mfb

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    The angle of the motion won't agree with the contact angle at this point.
    Why? One sphere will start moving to the left, the other one to the right. The total momentum stays constant, which allows to find the ratio of these (horizontal) velocities.
     
  18. Sep 13, 2015 #17
    Sir, I couldn't understand your point regarding the angle of motion. Did you mean that the upper sphere would (at the instant of losing contact) be moving tangentially to the lower sphere? In that case, resolving the velocity into 2 components, I still get the horizontal component of ##v_m## as ##v_m\cos\theta##. However, with respect to the ground, the actual value of the velocity of the upper sphere should be ##v_m\cos\theta-v_M## (hopefully this is right since the upper sphere is being carried on the lower sphere which in turn is moving with ##v_M## towards the opposite side). Thus, my (modified) equation for the total momentum of the system at the time the upper sphere loses contact is $$-Mv_M+m(v_m\cos\theta-v_M)$$ It would have been really useful if we could conserve momentum from the initial moment ##t=0## when the 2 spheres were just kept on each other and had no velocity. However, at ##t+\Delta t##, an external force is applied on the upper sphere. Thus, we would have to conserve momentum from ##after## this first external force has been applied right?

    But Sir, at ##t+\Delta t## at the instant the force is applied, would the 2 spheres have any velocity? In that case, could the initial momentum of the system (at ##t+\Delta t##) be ##0##?
     
    Last edited: Sep 14, 2015
  19. Sep 14, 2015 #18
    Please could you ignore the part I've included in this quote? I' not sure about this; it could (and probably is) wrong.
     
  20. Sep 14, 2015 #19
    Assuming ##v_m## represents speed of the upper sphere with respect to the lower one and ##\theta## is the angle which the line joining the the two centers make with the vertical , the above relation looks correct .
     
  21. Sep 14, 2015 #20
    Sir, actually ##v_M## is the velocity of the lower sphere wrt the ground while ##v_m## is the velocity of the upper sphere wrt the ground. I know that the velocity of the lower sphere will affect the velocity of the upper sphere, but am unable to understand how to use this while writing the equation representing the conservation of linear momentu in the horizontal direction.
     
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