Loop-the-loop: energy margin and the meaning of a light-gate speed
A student is testing a loop-the-loop track. A trolley of mass 0.85 kg is launched from the bottom by a spring and must stay in contact with the track at the top of a vertical circular loop of radius 0.42 m. A light gate at the launch point records 5.8 m/s. The track is not perfectly smooth: the student estimates 0.18 J is lost to friction and air resistance between the launch point and the top of the loop. The trolley's length is 0.11 m, which is not negligible compared with the loop radius. The student's partner claims that because the speed at the bottom is more than double the speed needed at the top, the trolley will certainly complete the loop. The student is unsure whether to trust the energy calculation or to add a safety factor. They also notice that the light gate reports the time for a 0.050 m card to pass, not the instantaneous speed. The track is clamped to a bench that vibrates slightly when the spring is released.
1. Using the recorded speed, determine whether the trolley's centre of mass has enough energy to reach the top with the minimum speed required for contact. State any assumptions and comment on whether the answer is marginal.
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Minimum speed at the top for contact: the normal force can fall to zero, so mg = mv_top²/R, giving v_top,min = sqrt(gR) = sqrt(9.81 × 0.42) = 2.03 m/s.
Energy at the bottom: KE = ½mv² = 0.5 × 0.85 × 5.8² = 14.3 J.
Energy needed at the top: PE gain = mg(2R) = 0.85 × 9.81 × 0.84 = 7.00 J. KE at minimum speed = ½ × 0.85 × 2.03² = 1.75 J. Total required from the bottom = 7.00 + 1.75 = 8.75 J, plus 0.18 J friction = 8.93 J.
Available energy is 14.3 J, so there is about 5.4 J spare. The trolley should complete the loop comfortably, even allowing for the finite trolley size. A more careful centre-of-mass calculation would use a slightly smaller path radius, which lowers both the required height gain and the minimum speed; the conclusion is unchanged. The answer is not marginal. Assumptions: the 0.18 J loss is reliable, the spring has finished pushing before the light gate, and the track is rigid enough that vibration does not remove contact.
2. The partner says the speed at the bottom is 'more than double' the minimum top speed, so it must work. Critically evaluate this reasoning and identify the physics error.
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The ratio is 5.8/2.03 = 2.86, but comparing speeds directly is not a valid energy argument. Gravitational potential energy must be supplied, and kinetic energy depends on v², not v. The correct test is an energy balance, not a speed ratio.
A quick check without friction gives v_top = sqrt(v_bottom² - 4gR) = sqrt(5.8² - 4 × 9.81 × 0.42) = sqrt(33.64 - 16.48) = 4.14 m/s. With 0.18 J friction, v_top = sqrt((14.3 - 7.00 - 0.18) × 2/0.85) = sqrt(16.75) = 4.09 m/s. Both exceed 2.03 m/s.
The conclusion happens to be correct, but the reasoning is not. If the loop radius were larger, the same bottom speed could still be 'more than double' the minimum top speed yet fail because the required PE gain would be larger. The partner has treated speed as if it were conserved energy.
3. The light gate measures the time for a 0.050 m card. If the card takes 8.6 ms, calculate the recorded speed and discuss whether this is the instantaneous speed at the launch point. Suggest one improvement.
Read the worked answer
Recorded speed = card length/time = 0.050 m / 8.6 × 10⁻³ s = 5.81 m/s, consistent with the quoted 5.8 m/s.
Strictly, this is the average speed over the card's passage, not the instantaneous speed at a single point. If the trolley is accelerating or decelerating over the 0.050 m card, the average lies between the entry and exit speeds. Near the launch point the acceleration is probably small once the spring has released, so the approximation is reasonable, but the card is not negligible compared with the trolley length (0.11 m).
Improvement: use a much shorter card or two light gates a measured distance apart to obtain speeds at two positions and calculate acceleration; alternatively use video tracking to extract position-time data. Repeating the launch several times would also show whether the 5.8 m/s value is reproducible.