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Centripetal Acceleration and velocity

  1. Sep 15, 2007 #1
    A ball on the end of a string is whirled around in a horizontal circle of radius .487 m. The plane of the circle is 1.06 m above the ground. The string breaks and the ball lands 2.26 m away from the point on the ground directly beneath the ball's location when the string breaks. The acceleration for gravity is 9.8 m/s^2. Find the centripetal acceleration of the ball during its circular motion. Answer in units of m/s^2.

    a[c] = (v^2)/r = (v^2)/.487

    I need to find the velocity. I'm imagining a disk (representing the circular path of the ball before the string is broken) 1.06 m off the ground, and the ball landing on the ground 2.26 m away from the point on the ground beneath the edge of this disk.

    I'm not sure if my train of thought is correct however, for the following calculations:

    I assume the initial y component of velocity is zero because this disk is completely horizontal. However, once the string breaks, free-fall acceleration acts upon it.

    So, y = v[yi]t + .5a(t^2) => -1.06 = (0)t -4.9(t^2)

    t = (1.06/4.9)^(.5)

    This would be the time it took for the ball to get where it is. Since I am considering the horizontal disk, however, I don't need y in my calculations. But, I know that 2.26 is a horizontal distance, and, if divided by t, will give me the x component of velocity. I get about 4.86 m/s. Then, squaring the term and dividing by the radius (.487), my result is 48.48 m/s^2.

    I am uncomfortable about these assumptions I am making, however, so an explanation as to why these things can be assumed would be greatly appreciated.

    Thank you!
  2. jcsd
  3. Sep 15, 2007 #2


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    Your work looks good to me. Can you describe which assumptions you feel uncomfortable with?
  4. Sep 15, 2007 #3
    Well, to find the velocity, I'm using the distance after the string breaks as well as the time. How can I use those numbers when I'm trying to find the acceleration as the ball is still in the circular path. The situation is different once the string breaks. Wouldn't the velocity as ball is spinning be different from the velocity as it is falling toward the ground? Or, does the fact that horizontal velocity (which I am using) is constant come into play here? When would there be a horizontal velocity that isn't constant (because it seems like the horizontal velocity here seems to wane, as the ball is kind of being shot outwards horizontally)? When would I need to consider the y component of velocity? If the situation does not include a perfectly horizontal circular path?
  5. Sep 15, 2007 #4


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    Yes... the moment the spring breaks... there is no more force acting in the horizontal direction... the only force acting on the ball is gravity... so from the moment the string breaks, the horizontal velocity of the ball remains constant until the ball hits the ground.

    As the ball is spinning, the horizontal velocity is changing, but the horizontal speed remains constant according to the problem.

    Horizontal velocity will remain constant as long as there is no horizontal force... force is what creates acceleraton and change in velocity... from the moment when the string is cut to when the ball falls to the ground... there is no horizontal force, so the horizontal component of velocity remains constant.
    Last edited: Sep 15, 2007
  6. Sep 15, 2007 #5


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    don't think of it like that... the horizontal force (the tension in the string) is taken away... so the ball keeps exactly the same horizontal velocity it had the instant before the string was cut...
  7. Sep 15, 2007 #6


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    That's the numbers you are given.
    Hence, you need to make that minimal set of idealizing assumptions for which those assumptions are sufficient.

    We idealize that the "snap" phase of the string is of zero duration, and that the velocity of the ball is unchanged just after the string snaps as it was just before it snapped.

    This is certainly true in the free-fall phase, since we idealize away air resistance.

    Its initial velocity of its free fall phase is set equal to its tangential velocity to its circular path just prior to the break of the string. Its horizontal trajectory is along that tangent.

    It is not "shot out" more than a tangent line to a circle shoots out from the circle. It just fails to curve within the horizontal plane, since the string is broken.
    In the free fall phase, its velocity is certainly constant, since only vertical gravity act upon it.
    This in contrast to the first phase, in which the horizontal velocity, albeit not speed, changes continuously.

  8. Sep 15, 2007 #7
    Thank you.

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