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Ball speed at the top of the loop

  1. Sep 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A compressed spring is used to propel a ball-bearing along a track which contains a circular loop of radius 0.10 m in a vertical plane. The spring obeys Hooke's law and requires a force of 0.20 N to compress it to 1.00 mm.

    327781daa759.jpg

    (a) The spring is compressed by 30 mm. Calculate the energy stored in the spring.
    (b) A ball-bearing of mass 0.025 kg is placed against the end of the spring which is then released. Calculate (i) the speed with which the ball-bearing leaves the spring, (ii) the speed of the ball at the top of the loop, (iii) the force exerted on the ball by the track at the top of the loop.

    Assume that the effects of friction can be ignored.

    Answers: (a) 9 * 10-2 J, (b) (i) 2.7 m s-1, (ii) 1.8 m s-1, (iii) 0.55 N

    2. The attempt at a solution
    (a) E = 0.5 k x2 → E = 0.5 * [(0.2 * 30) / 0.03] * 0.032 = 0.09 J

    (b) (i) KE = 0.5 m v2 → v = √(2 KE) / m = √(2 * 0.09) / 0.025 = 2.7 m s-1

    (b) (ii) v = √2 g h = √2 * 10 * 0.2 = 2 m s-1 (in this case I took h as 0.1 + 0.1 -- twice the radius)
    or v = √g r = √10 * 0.1 = 1 m s-1
    or v = √2 g h = √2 * 10 * 0.1 = 1.4 m s-1
    or PE = m g h = 0.025 * 10 * 0.1 = 0.025 J → v = √(2 KE) / m = √(2 * 0.025) / 0.025 = 1.4 m s-1
    or PE = m g h = 0.025 * 10 * 0.2 = 0.05 J → v = √(2 KE) / m = √(2 * 0.05) / 0.025 = 2 m s-1
    So part (b) (ii) is wrong and I can't see my mistake.

    (b) (iii) I took v = 1.8 m s-1 from the answer. ∑ F = N + m g = (m v2) / r → N = [(m v2) / r] - m g = [(0.025 * 1.82) / 0.1] - 0.025 * 10 = 0.56 N

    As far as I can see only (b) (ii) is wrong. Any suggestions please on where is the mistake? Thanks in advance.
     
  2. jcsd
  3. Sep 18, 2016 #2

    kuruman

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    Homework Helper
    Gold Member

    What is the meaning of all these ors? Why is v = √2 g h ? Write the conservation of mechanical energy equation correctly. KEbottom + PEbottom = KEtop + PEtop. If you can, use symbols and put the numbers in at the end. Then it will be easier to see where you went wrong.
     
  4. Sep 18, 2016 #3

    gneill

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    Staff: Mentor

    You forgot to conserve energy. The ball-bearing enters the loop at the bottom with some KE.
     
  5. Sep 18, 2016 #4
    when you are using conservation of energy of a body you must use Initial E (KE +PE) = Final E (KE + PE)
    i think you are missing the initial part of energy.
     
  6. Sep 18, 2016 #5
    So we have: KEBottom + PEBottom = KETop + PETop
    0.5 m v2 + m g h = 0.5 m v2 + m g h
    m is included everywhere, so simplify: 0.5 v2 + g h = 0.5 v2 + g h
    PEBottom is equal to 0 since h = 0, so: 0.5 v2 = 0.5 v2 + g h
    0.5 * 2.72 = 0.5 v2 + 10 * 0.2 [2.7 m s-1 is the speed with which the body leaves the spring and 0.2 is the height = radius 0.1 *2]
    3.645 = 0.5 v2 + 2
    1.645 = 0.5 v2
    3.29 = v2
    v = 1.8 m s-1

    This should be it, right?
     
  7. Sep 18, 2016 #6

    gneill

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    Staff: Mentor

    Looks good.
     
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