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Circular Motion of a ball Problem

  1. Oct 20, 2007 #1
    1. The problem statement, all variables and given/known data
    A 60 g ball is tied to the end of a 50-cm-long string and swung in a vertical circle. The center of the circle, as shown in Figure, is 150 cm above the floor. The ball is swung at the minimum speed necessary to make it over the top without the string going slack.

    If the string is released at the instant the ball is at the top of the loop, where does the ball hit the ground?

    http://session.masteringphysics.com/problemAsset/1000600/5/knight_Figure_07_62.jpg

    2. Relevant equations

    Critical velocity = [tex]V{c}[/tex] = [tex]\sqrt{r*g}[/tex]
    [tex]\Delta[/tex]d=V{i}*t + (1/2)(a)(t^2)

    3. The attempt at a solution
    I attempted this problem as a projectile motion problem, and divided the calculations into two parts: Horizontal and Verticle

    Verticle Part

    So I used this equation: [tex]\Delta[/tex]d=V{i}*t + (1/2)(a)(t^2) and I know the verticle initial is 0 so I can eliminate V{i}*t of the equation and get: [tex]\Delta[/tex]d=(1/2)(a)(t^2)

    2 = (1/2)*(9.8)*(t^2)
    2 = 4.9*(t^2)
    t^2 = 2 / 49
    t = 0.64 seconds

    Horizontal Part

    The horizontal parts only has 3 variables: V, d, and t (which I know from the verticle part). and V = V{c} = [tex]\sqrt{r*g}[/tex] = 0.59 m/s.

    and d = v*t = 0.59*0.64 = 0.3776 m

    This doesnt look right. I was wondering if someone can point me in the right direction. Any help is greatly appreciated. Thanks.
     
    Last edited: Oct 21, 2007
  2. jcsd
  3. Oct 20, 2007 #2

    PhanthomJay

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    Looks ok except for a math error in your critical speed calc.
     
  4. Oct 21, 2007 #3
    yea I see what you mean. I accidently used the mass instead of the radius.

    so it would be v_{c}=[tex]\sqrt{r*g}[/tex] = [tex]\sqrt{0.5*9.8}[/tex] = 6.9

    therefore:

    d = v*t = (6.9) * (0.64) = 4.4 m/s?

    Does this seem right?

    Edit: Still not the right answer. Help?
     
  5. Oct 21, 2007 #4

    Doc Al

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    What are you asked to find? (I don't see a question in the problem statement.)
     
  6. Oct 21, 2007 #5
    oops.Sorry about that.

    If the string is released at the instant the ball is at the top of the loop, where does the ball hit the ground?
     
  7. Oct 21, 2007 #6

    PhanthomJay

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    you've made anoteher math error,
    [tex]\sqrt{r*g}[/tex] = [tex]\sqrt{0.5*9.8}= \sqrt4.9 = 2.2m/s[/tex] And then calculate the distance d which wil be in meters.
     
  8. Oct 21, 2007 #7
    yep

    d = v*t = 2.2*0.64 = 1.4 m

    Thanks a lot!

    still can't figure out why I put down 6.9 instead of 4.9....
     
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