<p>You have a supporting point and a ball connected by a string of length 100cm. There is a peg at a distance d below the supporting point. You release the ball from rest at an instance when the ball is totally horizontal to the point, it swings along an arc. When the string catches on the peg, the ball swings upward but does not complete a circular path around the peg. Instead, the abll leaves its circular path, and strikes the peg.</p>
<p>Find the distance d below the point of support at which the peg should be placed for this to happen.</p>
<p>If you get the answer to this problem, I commend you. of course, I would like to know how you got the answer.</p>
<p>any physicists out here?</p>
<p>Ok, so the point and then the ball are horizontal, so the ball is going to fall with a velocity caused by gravity. Its been a while since Ive done something like this and its late, so basically you find the rate at which the ball is dropping in m/s. I thing it is sqrt(2gh) and then you say that when it hits a peg it makes it to the top, but at that point it has only potential energy and no kinetic so that it just drops. So you find the speed at which it hits the peg based on the variable h, and then treat 100cm-h as the radius of the new vertical circle that the ball fails to make it around.</p>
<p>Wow, this is a nice question, makes the stuff we did in school look rather basic... Here is how I did it,</p>
<p>At the top of the motion, E[k] = 0 and E[p] = mgh (J)
At the bottom of the bottom after a quarter rotation, E[k] = .5mv^2 (J)and E[p] = 0</p>
<p>Therefore,</p>
<p>mgh = .5mv^2
v = SQUARE ROOT (2gh)
= SQRT (19.6)</p>
<p>Now, at this point the ball enters a new circular motion, with radius (1-d)m. Again, remembering that because a[c] = 0, v= 0 at the peak of the circular motion, then simply working with Potential Energy and Kinetic Energy we get,</p>
<p>.5v^2 = 2g(1-d)
d = .5m = 50cm</p>
<p>Hope thats correct...</p>
<p>Whoah...I got 50 cm too. It has to have zero velocity at the top of the loop, so KE = 0 too. Therefore all of the energy must be gravitational potential E. The GPE has to equal the original GPE, or mgh. Therefore the height has to be the same as the drop height. The only way to do that is to make the peg half-way down, or 50 cm away.</p>
<p>was this a MC or OR q'</p>
<p>mattd1688. it was long answer. anyway, i'll check on the .50 m answer. somehow, i got one equation with 2 variables. Anyway, there was a second part to the question. You have to find a function d(theta) at which the peg needs to be placed for this to happen.</p>
<p>It was midnight so I wasnt thinking all that good. But the .5m answer is right. I was thinking kinematics and energy all mixed up. Potential = mg(1m) above the ground. For a ball to fall down, gaining kinetic and then losing it as it comes back up until it reaches 0, w/o friction, its original and final potential energies would have to be the same. Which would mean that it would have to reach 1m again. THe only way for this to happen is by having a peg 1/2 way down the string.
d(theta) = 0 = mg(h(inital)-2(h(intial)-h(final)))
I am in a B class so I dont do equations but that would be my guess.</p>
<p>"Therefore,</p>
<p>mgh = .5mv^2
v = SQUARE ROOT (2gh)
= SQRT (19.6)</p>
<p>Now, at this point the ball enters a new circular motion, with radius (1-d)m. Again, remembering that because a[c] = 0, v= 0 at the peak of the circular motion, then simply working with Potential Energy and Kinetic Energy we get,</p>
<p>.5v^2 = 2g(1-d)
d = .5m = 50cm"</p>
<p>at the peak of the circular motion, it is not directly above the peg, so shouldnt the potential energy be 2g (1-d)cos(theta) with theta unknown?</p>
<p>i was thinking more along the lines of finding the height when tension is 0.</p>
<p>"It has to have zero velocity at the top of the loop, so KE = 0 too"</p>
<p>it doesn't just DROP at the top of the loop, there is still a velocity. in fact, it loses contact with the circular motion BEFORE it reaches the top of the loop. </p>
<p>I actually found an equation for it. unfortunately, it involves 3 unknown variables...</p>
<p>someone please answer my questions?</p>
<p>where are you guys???</p>
<p>i just got confirmation that the answer is .47???</p>