A 2″-diameter polyurethane ball rides in the vee of the metal track in the photograph. When you release the ball from sufficiently high up the ramp (at left in the photograph), it will roll around the loop without leaving the track. Towards the top of the launch ramp, on the back, are three white marks. The lowest of these gives the minimum height from which you must launch the 2″-diameter polyurethane ball for it to travel around the loop without leaving the track. The middle mark (just slightly higher than the lowest mark) shows the minimum height from which you must release a 1-1/2″-diameter polyurethane ball, and the upper mark shows the minimum height for launching a 1″-diameter steel ball, for which the frictional losses are much greater than for the polyurethane balls. (The frictional losses for the two polyurethane balls are probably similar, but the geometry is somewhat different. (Vide infra.))

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.

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. (See demonstration 16.03 -- Ball on turntable.) 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.)

The ball, however, does not merely translate, but it rolls. So mgh = (1/2)mv² + (1/2)Iω², where I is the moment of inertia of the ball (2mr²/5, where m and r are the mass and radius of the ball, respectively) and ω is the angular velocity of the ball. Since ω = v/r, we have mgh = (1/2)mv² + (mr²/5)(v²/r²), and gh = v²[(1/2) + (1/5)] = v²(7/10). This gives v² = 10gh/7. Again setting v²/R = g gives 10gh/7R = g, and h = (7/10)R. So we see that because the ball also rolls, we must increase the height of the launch above the top of the loop by 40%. If we take the height from the bottom of the loop, this adds 2R to these values, and we have h = 2.5R without accounting for rotation, and h = 2.7R when we include rotation, for a difference of 8%.

With the 2″-diameter polyurethane ball, r = 0.025 m, and R = 0.165 m. So R/2 = 0.082 m, and (7/10)R = 0.116 m. In fact, the lower white mark on the launch ramp puts the ball about 0.175 m above the top of the loop, so slightly more than one radius above it. This is about 51% higher than our calculated value. Part of this difference is due to the fact that the contact point of the ball with the rail is not at the bottom, but a small distance above. The radius about which the ball rolls is about 3 mm shorter than the full radius of the ball, so it is about 0.022 m. To account for this, we must note that ω = v/r_c, where r_c is this shorter radius. If we do this, the energy equation above becomes mgh = (1/2)mv² + (mr²/5)(v²/r_c²), and v² = gh/[(1/2) + (1/5)(r²/r_c²)]. Now setting v² equal to g, we get h = R[(1/2) + (1/5)(r²/r_c²)]. If we take r²/r_c² = 0.025²/0.022², then h = R[(1/2) + (1/5)(1.29)], or 0.76R, about 8% higher than our previous calculation. For R = 0.165 m, this is 0.125 m. The additional 40% difference that 0.175 m represents (11% relative to the total height from the bottom of the loop) is due to frictional losses. Taken from the bottom of the loop, the height from which we must release the ball is 2R + 0.76R, or 2.76R. For R = 0.165 m, this is 0.455 m. This is about 2% larger than the 2.7R (=0.446 m) we obtain without the radius correction, which is in turn about 8% higher than the 2.5R (=0.412 m) we obtain when we do not account for the rotational kinetic energy. The 0.175 m height above the top of the loop represents a total height of 0.505 m, which is about 11% higher than the 0.455-m height we obtain when we account for rotational kinetic energy and correct for the contact radius of the ball, and about 13% higher than the 0.446-m height we obtain without the radius correction.

The 1-1/2″-diameter polyurethane ball sits a bit lower in the V than does the larger one, so for it, r²/r_c² is slightly greater than for the larger ball, so its mark is somewhat higher than the mark for the 2″-diameter ball. For the 1″-diameter steel ball, the contact points are farther from the bottom than for either polyurethane ball, so r²/r_c² is slightly greater still, and since it slides much more as it goes along the track than do the polyurethane balls, the loss to friction is significantly greater. Therefore, you must release it from the highest mark for it to stay on the track.