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Circular motion in gravity

  1. Nov 17, 2013 #1
    If a ball weighing 0.2 kg is fastened at the end of a 1.5m string and is released from the horizontal position, what will the velocity be when the pendulum is at the bottom of the circle.

    So I was wondering if I could use energy conservation for this question because that would make it a lot easier, if not how exactly would I start because this is not uniform circular motion.
     
  2. jcsd
  3. Nov 17, 2013 #2
    Energy conservation is perfectly applicable in case of non uniform circular motion.
     
  4. Nov 17, 2013 #3
    This ball has an unknown radius, so the initial energy would be Ei = mg(1.5 + x) and Ef = (1/2)mv^2 or does that x not matter?
     
  5. Nov 17, 2013 #4
    The ball is assumed to be a point particle.No need to consider 'x' .
     
  6. Nov 17, 2013 #5

    haruspex

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    Strictly speaking, yes, you need the vertical drop of the centre of mass, so you would need to add 1 radius.
     
  7. Nov 17, 2013 #6
    Well the full question asks for the velocity of the ball after it hits a block at rest, and the b) part is asking how much the centre of mass increases in the y direction if the collision is inelastic, so it isn't a point particle.
     
  8. Nov 17, 2013 #7
    Consider the ball as a point particle.Not only the ball,you have to assume the box to be a point particle as well.
     
  9. Nov 17, 2013 #8
    i = circle
    ii = block
    R = (MiRi + MiiRii) / (Mi + Mii)
    Since the change in height is all we want, lets say r = radius, and a = height of box, Riy = r, Riiy = a/2, then Ry = 0.4r + 0.3a, this doesn't seem right, what am I doing wrong?
     
  10. Nov 17, 2013 #9

    haruspex

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    You man, perhaps, how far the centre moves in the y direction? Would you mind posting the full text, word-for-word?
    As Tanya Sharma says, if you're not told the radius you pretty much have to take it to be a point particle. If you were to take the radius into account completely, you'd also have to consider the rotation of the ball as it swings down. Very messy.
     
  11. Nov 17, 2013 #10
    Here is the question
    2istlqu.png
     
  12. Nov 17, 2013 #11
    How can you get the height that the centre of mass rises if you don't apply the same logic from b) to a)?
     
  13. Nov 17, 2013 #12

    haruspex

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    OK, you previously left out the 'sticking together' bit.
    Clearly you can use energy conservation in a).
    For b), you must consider two events:
    - the collision itself, which is inelastic, so energy is not conserved. Instead, you know that the two objects move together afterwards, so you have an equation from that instead of the work equation. Find the velocity then.
    - after the collision, work is conserved during subsequent motion.
     
  14. Nov 17, 2013 #13
    oh wait so the b) part is saying, when they stick together and they continue on that path at the new velocity what is the maximum height it reaches?
     
  15. Nov 17, 2013 #14
    Also for the a) part I would like to check the logic of my answer. So first of all I got the initial v of the ball just before it hits the block with energy, then I made m1v1 = m1v1' + m2v2', but I had two unknowns so since it is an elastic collision I can use energy so then I did (0.5)m1v1^2 = (0.5)m1v1'^2 + (0.5)m2v2'^2, then when everything was subbed in I had to use the quadratic formula, and got -0.198 and 9.01 m/s, I think that -0.198 is the most applicable intuitively, right? Also I still don't understand b) is it looking for the change in height of centre of mass when they are stuck together versus when they are apart or what?
     
    Last edited: Nov 17, 2013
  16. Nov 17, 2013 #15

    haruspex

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    You can avoid quadratics. (1) m1v1 = m1v1' + m2v2'; (2) m1v1^2 = m1v1'^2 + m2v2'^2
    From (2): m1(v1^2 - v1'^2) = m2v2'^2 = m1(v1-v1')(v1+v1')
    From (1): m1(v1 - v1') = m2v2'; m1(v1 - v1')(v1+v1') = m2v2'(v1+v1')
    Combining: m2v2'^2 = m2v2'(v1+v1'); v2' = v1+v1'
    This is just Newton's Experimental Law with coefficient of restitution = 1.
    You know they have the same speed after impact in this case. Find that speed. What KE does that imply? If they continue together (still attached to the string), what will be their KE at the highest point of the swing?
     
  17. Nov 17, 2013 #16
    Yea now I understand b) its 1/2m(1+2)vf^2 = m(1+2)gh
     
    Last edited: Nov 17, 2013
  18. Nov 17, 2013 #17
    so then in that rearranged energy equation do we just get the momentum equation, and sub it in? It seems to me just by the equations you showed me that you can't isolate for one variable, I am probably wrong though. So after that I derived the equation for v1' = (m1v1 - m2v1) / (m1+m2), is that right?
     
    Last edited: Nov 17, 2013
  19. Nov 17, 2013 #18

    haruspex

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    Why the -m2v1? Pls post your derivation of that equation.
     
  20. Nov 17, 2013 #19
    m1v1 = m1v1' + m2v2'
    m1(v1 - v1') = m2v2'

    m1v1^2 = m1v1'^2 + m2v2'^2
    m1(v1-v1')(v1+v1') = m2v2'^2
    v2' = v1 + v1'

    m1(v1-v1')(v1+v1') = m2(v1+v1')(v1+v1')

    m1v1 - m1v1' = m2v1 + m2v1'
    m1v1 - m2v1 = m2v1' + m1v1'
    v1(m1-m2) = v1'(m2+m1)
    v1' = v1(m1-m2)/(m2+m1)
    There you go.
     
  21. Nov 17, 2013 #20

    haruspex

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    Oh, sorry, I thought your post #17 was on part b.
    Yes, this is right for part a.
     
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