Steel Ball on a Loop Track: Energy Loss and Contact
A physics club is testing a table-top roller coaster. A steel ball is released from rest at a height of 0.850 m above the lowest point of the track and enters a vertical circular loop of radius 0.180 m. The ball is small enough to be treated as a particle, and rotational energy is ignored. A light gate at the top of the loop records a speed of 2.10 m/s. The club knows that the ball loses some mechanical energy on the track, but they want to know whether it still completes the loop and how large the loss is. Use g = 9.81 m/s². The ball has mass 12.0 g. The track is smooth at the lowest point but has a rough patch before the loop. One member claims that because the ball is light, it should not lose much energy; another claims that a heavier ball would certainly fall off the loop.
1. Calculate the speed at the top predicted by conservation of energy with no loss, and the fraction of initial mechanical energy lost.
Read the worked answer
Take the lowest point as zero potential energy. Initial energy: E_i = mgh = 0.0120 × 9.81 × 0.850 = 0.100 J. At the top the height is 2r = 0.360 m. With no loss, E_i = mg(2r) + ½mv², so v² = 2g(h − 2r) = 2 × 9.81 × 0.490 = 9.61, v = 3.10 m/s.
Measured v = 2.10 m/s. Measured energies at top: PE = 0.0120 × 9.81 × 0.360 = 0.0424 J; KE = 0.5 × 0.0120 × 2.10² = 0.0265 J; total = 0.0689 J. Loss = 0.100 − 0.0689 = 0.0311 J, or 31% of the initial mechanical energy. The ball still reaches the top, but it has lost nearly a third of its initial energy before the top.
2. Show whether the ball stays in contact at the top. What minimum speed is needed, and how does your answer change if the speed were 1.00 m/s?
Read the worked answer
At the top, forces toward the centre are the normal force N (downward) and weight mg (downward): mg + N = mv²/r. Thus N = m(v²/r − g). For v = 2.10 m/s, v²/r = 4.41/0.180 = 24.5 m/s², so N = 0.0120(24.5 − 9.81) = 0.176 N > 0; the track pushes down, so contact is maintained.
The minimum speed for contact is where N = 0: v_min² = gr = 9.81 × 0.180 = 1.77, v_min = 1.33 m/s. At 1.00 m/s, v²/r = 5.56 m/s² < g, so gravity alone is stronger than the required centripetal force; the ball cannot follow the loop and would lose contact before the top. The measured 2.10 m/s is above the minimum, so the loop is completed.
3. Evaluate the two members' claims: (a) a lighter ball loses less energy; (b) a heavier ball would certainly fall off the loop.
Read the worked answer
(a) Not generally true. In the no-loss limit, gravitational potential energy and kinetic energy both scale with mass, so the speed at any height is independent of mass. Friction loss is F_f × distance; for sliding friction F_f = μN, and N often scales with weight, so the loss also scales with mass, leaving the fractional loss roughly mass-independent. If the rough patch gives a nearly constant resisting force, a lighter ball would lose a larger fraction, not a smaller one.
(b) Wrong. The contact condition v² ≥ gr contains no mass. A heavier ball has the same speed at the top, if the track is the same and rotational energy is ignored, and the same minimum speed. It would not certainly fall off. Mass affects the normal force and friction, but not the speed requirement in this model.