Play with the loop the loop demonstration which involves a ball and a track that
ID: 1580153 • Letter: P
Question
Play with the loop the loop demonstration which involves a ball and a track that looks like a roller coaster. Point A is on the high side of the track and point B is on the low side. (a) Predict the minimum height from which the ball should be released on side A to be able to reach the end of the track on side B. What was the reasoning used for your prediction? Now release the ball from that height to verify your prediction. (b) What is the minimum height from which the ball should be released so as to complete a loop without losing contact with the track? If the loss of mechanical energy can be ignored, can the ball be released from side B from exactly the same height as the height of the loop (Hint: What would it mean about the speed of the ball at the highest point of the loop and what will it mean for the centripetal acceleration necessary for making the ball go in a circle?)? (c) What are the possible sources of loss of total mechanical energy? (d) Describe an interesting feature of this demonstration. (e) Describe four other situations from your every day experience where mechanical energy is approximately conserved because work done by the non-conservative forces is ignorable. Don't forget to mention which non-conservative force is doing negligible work in each case you cite. 1/10/05 Page 1 of 1Explanation / Answer
The ball starts with potential energy of mgh, where m is its mass, g is the acceleration of gravity and h is its height above the bottom of the track, and zero kinetic energy. As it rolls down the ramp, gravity converts its potential energy into kinetic energy, so that when the ball reaches the bottom of the loop, its potential energy is zero, and its kinetic energy ((1/2)mv2) is a maximum. As the ball continues around the loop, its kinetic energy decreases while its potential energy increases. At the top of the loop, if the ball has sufficient kinetic energy, it can continue around the loop without leaving the track. If not, it falls away from the track as it approaches the top of the loop.
Part-b For the ball to remain on the track at the top of the loop, the acceleration necessary to keep it moving in a circle must equal or exceed the acceleration of gravity, or v2/R must equal or exceed g, where v is the ball’s velocity, R is the radius of the loop in which the ball travels, and g is the acceleration of gravity. If we ignore the rolling motion of the ball, then if h is the height of the launch ramp above the top of the loop, mgh = (1/2)mv2, or gh = (1/2)v2, and v2 = 2gh. Our initial constraint (v2/R g) gives 2gh/R = g, and h = R/2. So we must launch the ball from a height that is at least half the radius of the loop above the top of the loop, or a total height of (5/2)R from the bottom of the loop. (All of this is measured with respect to the center of mass of the ball.)