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Motion in a vertical circle

  1. Nov 23, 2012 #1
    A 0.8 kg ball is whirled on a string r = 0.4 meters long in a vertical circular path. At the bottom of the circle, the ball is h = 0.45 meters from the ground. At the top of the circle, the ball has a speed of 3 m/s. Assume that the total energy of the ball is kept constant.

    a. Calculate total energy of the ball (ground is zero potential energy)
    I chose the top most point of the circle to do ths since PE is maximum there.
    ME = KE + PE
    Ke = mv^2/2 = (.8kg)*(3m/s)^2/2 = 3.6 J
    PE = mgh = (.8kg)*(10m/s^2)*(1.25m) =10 J
    Total energy = 13.6 J

    Im not sure if i calculated the PE correctly. I know that PE at the top of the circle is mg2r but then i have to consider the height from the ground to the top of the circle. So i figured i should add the distance from the ground up and multiply by mg. is that right?

    Next part of the question asks me to calculate velocity at the bottom of the circle. Here KE is larger than PE.
    ME = KE + PE
    13.6 J = mv^2/2 + mgh
    13.6 J = (.8kg * v^2)/2 + (.8kg)*(10m/s^2)*(.45m)
    When i solve for velocity i get 5m/s. how can this be? Shouldn't it be 3m/s as it is on the top? What am i doing wrong?
     
  2. jcsd
  3. Nov 23, 2012 #2

    K^2

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    Science Advisor

    Everything is correct. Gravity pulls down on the ball, so it speeds up going down, and slows down going up. That's why it's going 3m/s at the top and 5m/s at the bottom.

    And yes, you compute potential energy correctly for both cases, because problem explicitly states that 0 is to be taken at ground level. So the height of the ball is h at the bottom and h+2r at the top. That gives you mgh and mg(h+2r) respectively for potential energy.
     
  4. Nov 23, 2012 #3
    Ok, thanks a lot for your help!
     
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