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This is actually a college level question.

  1. Nov 2, 2003 #1
    I'm having trouble with two of my Physics Problems. It goes a little something like this:
    A bowling ball weighing w is attached to the ceiling by a rope of length L. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is v.

    What is the acceleration of the bowling ball, in magnitude and direction, at this instant? Take the free fall acceleration to be g.

    And then the last:
    A bob of mass m is suspended from a fixed point with a massless string of length L (i.e., it is a pendulum). You are to investigate the motion in which the string moves in a cone with half-angle theta.

    What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle theta from the vertical? Express the speed in terms of m, g, L, and/or theta.

    I'm having a particularly hard time with the first one. THanks if you decide to help a brother out!!!
  2. jcsd
  3. Nov 2, 2003 #2
    How about this

    Original Roational Velocity = Final rotational velocity + g * time

    The rotational velocity will take into account the length

  4. Nov 3, 2003 #3


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    1. the question seems elementary...i'm not sure i'm reading it right.

    the resultant acceleration of the ball is the vector sum of the tangential components and centripetal components of acceleration. however, since the ball is at the midpoint of the swing, there is a zero tangential component, because there are no forces acting on the ball tangentially to its path. the only 2 forces acting are the tension in the rope and the weight of the ball, both of which are perpendicular to its path.

    therefore, the resultant acceleration of the ball is simply the centripetal component, which is directed towards the centre of rotation/swing (exactly upwards).

    the circular motion equation applies here:

    F = mv^2 / L
    ma = mv^2 / L
    a = v^2/L

    that seems to be about it.

    2. let A = theta. now obtain a general expression relating the required variables. first draw a diagram and resolve the forces into vertical and horizontal components:

    (eqn 1) TsinA = mv^2 / r (using eqn of circular motion)


    (eqn 2) TcosA = mg (since the ball's height in the air doesn't change)

    r is the radius of rotation, and is equal to LsinA. T is the tension in the string.

    dividing eqn (1) by (2),

    tanA = v^2 / rg

    substituting r = LsinA into eqn,

    tan A = v^2 / gLsinA
    v^2 = gLtanAsinA
    v = sqrt [ gLtanAsinA ]

    and that's it.
  5. Nov 3, 2003 #4


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    Science Advisor

    One problem we have is that you have not given us an idea of what you have to work with. Do you have formulas that are given to you or are you to set up a "force diagram"?

    The way I would do this, in general, is set up a force diagram with force -mg downward (gravity) and the break that into two components, one tangent to the path of motion (a circle) and the other normal to it. When the pendulum (i.e. rope) makes angle θ with the vertical, then tangential force is give by
    FT= -mg sin(θ) and the normal force is given by
    FN= -mg cos(θ).

    Of course, "As the rope swings through the vertical", θ= 0 so FT= 0 and FN= -mg just as joc said. The normal force is, of course, offset by the equal tension in the rope: the acceleration vector is the 0 vector, exactly as if the bowling ball were simply hanging there!
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