2
$\begingroup$

If you have a stone tied to a string, and it is given a certain velocity at the lower most point such that it completes a quarter circle but then the string becomes slack, and it undergoes projectile motion.

The problem is, is it necessary that this parabola will be symmetric about the verticle? If so, is it impossible that the stone passes through the center of this circle during downward motion?

Please give a detailed solution

Thanks in advance

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ I don't understand, what has this got to do with homework? $\endgroup$
    – Saurabh Raje
    Sep 12, 2013 at 12:39

1 Answer 1

Reset to default
1
$\begingroup$

if the stone is released at exactly quarter circle, the stone goes perfectly vertical--not a parabolic trajectory. for the stone to return to the centre of the circle, it has to be released slightly after a quarter circle is reached. this is so the stone has some horizontal velocity (to go anywhere other than just vertical). essentially this gives the stone a velocity angled to the horizontal "theta". since the the magnitude of the velocity (Vo) is known (by how fast you were swinging it about), and range "d" is the radius of your original circle, then you can calculate at which angle the stone must be released for its trajectory to pass through the center of the original circle.

enter image description here

there are a few projectile motion equations out there which are simple to understand. this one here answers your question directly.

Improve this answer
$\endgroup$
4
  • $\begingroup$ you say that it will go vertical, how is it even possible? the string hasnt broken... $\endgroup$
    – Saurabh Raje
    Sep 12, 2013 at 15:02
  • 1
    $\begingroup$ @SaurabhRaje, gregsan probably assumed that the stone will not be influenced by the string after a quarter circle, since that it the situation you stated in your question. But if this can only happen when the string becomes slack, then either the velocity of the stone would have to be zero at a quarter circle or the angle at which the string becomes slack should be greater then a quarter circle, just as gregsan stated. $\endgroup$
    – fibonatic
    Sep 15, 2013 at 14:27
  • $\begingroup$ that is fine, @fibonatic, but my doubt was, how will it go vertically upward for a measurable/substantial time? moreover, if the velocity is zero at quarter circle, then it will fall down in the same circular path. My main problem was, is the parabola symmetric about the vertical? If no, then how will you prove it? $\endgroup$
    – Saurabh Raje
    Sep 17, 2013 at 7:13
  • $\begingroup$ What do you mean by symmetric about the vertical? That the parabola will be symmetric around the vertical axis going though the center of the circle? Or that the path of the stone will be vertical symmetric around the highest point of the parabola? The answer to this would be no to the first and yes to the second, if friction can be neglected, the mass of the rope is much smaller than that of the stone and the stone will not collide with anything. $\endgroup$
    – fibonatic
    Sep 17, 2013 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.