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Uniform Circular Motion of a ball

  1. Feb 1, 2005 #1
    i'm having trouble in solving this problem. i've tried several ways to solve this problem but still have not come up with the correct solution. the question is:

    A ball on the end of a string is revolved at a uniform rate in a verticle circle of radius 65 cm. If its speed is 4.25 m/s and its mass is .300 kg, calculate the tension in the string when the ball is (a) at the top of its path, and (b) at the bottom of its path.

    i tried using the equation: (EF)=ma....but i think you use that later on?
    there's another equation that i think i might have to use to solve this problem, however im not sure on how to find v1 and v2. this equation is:
    FT1 + mg = m * (v2,1/r).
  2. jcsd
  3. Feb 1, 2005 #2


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    ALLL problems in mechanics find their answer in this equation, and this one is no exception.

    The one equation that you will need besides F=ma is the expression of the acceleration of an object in uniform circular motion:

    [tex]a_r = v^2/R[/tex]

    You are basically asked to find what must the tension in the string be at these two points so that the sum of the forces on the ball always be equal to [itex]ma_r[/itex].

    P.S. In case you didn't knew, writting "EF" doesn't make much sense. I think you meant [itex]\Sigma \vec{F}[/itex]? This sign before F is the greek letter "capital sigma" and it usually stands in mathematical notation for "sum". So [itex]\Sigma \vec{F} = m \vec{a}[/itex] means "the sum of the forces (also called the 'net force') acting on a particle is equal to its mass times its acceleration".
  4. Feb 1, 2005 #3
    What's the E for?

    There are two forces acting on the ball: The tension of the string, which provides the centripetal force, and gravity. For an object in uniform circular motion, the centripetal acceleration is equal to [itex]v^2/r[/itex].

    We know that the net force acting on the ball is the sum of the two individual forces,
    Fnet = Fg + Ft
    Where Fg is the force of gravity, and Ft is the tension in the string.

    So the tension must be the difference between the net force and the force of gravity,
    Fnet - Fg = Ft

    In order to evaluate this expression for Ft, we need to know what Fnet and Fg are. The force of gravity is easy, it's just the product of the mass of the object and the acceleration due ot gravity. What about Fnet? Well, since we know the ball is moving in a uniform circle, we know the acceleration is [itex]v^2/r[/itex], and the problem gives us the mass. From F = ma, we just multiply those two things together to find the net force.

    You should decide which direction you want to make positive, and which direction you want to make negative. It's probably most intuitive to say that "up" is positive. If so, then when the ball is at the top of its swing, it has a net acceleration of [itex]-v^2/r[/itex], and when it is at the bottom of its swing, it has a net acceleration of [itex]+v^2/r[/itex]. Don't forget that this would mean Fg is a negative quantity, as well. The tension in the string should be greatest when the ball is at the bottom of its path, and least when it is at the top.

    Does this answer your question?
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