The Roller Coaster Loop and the Missing Energy
A 500 kg cart is released from rest at the top of a 20 m hill on a frictionless track. It descends and enters a circular loop of radius 8.0 m. At the top of the loop, an onboard sensor reads the cart's speed as 9.0 m/s. The designer expected the cart to just barely maintain contact (normal force → 0) at the loop top, and is alarmed that the measured speed is 'too high' for that condition. Use g = 10 m/s².
1. Using energy conservation, predict the speed at the top of the loop if the track is truly frictionless, and compare to the sensor reading.
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Height at loop top = 2R = 16 m above the bottom; the cart starts 20 m above the bottom.
- Energy: mgh_start = ½mv_top² + mg(2R).
- 500(10)(20) = ½(500)v² + 500(10)(16)
- 100000 = 250v² + 80000 → 250v² = 20000 → v² = 80 → v ≈ 8.9 m/s.
Predicted ≈ 8.9 m/s vs. measured 9.0 m/s - essentially agreement within rounding. The sensor is not anomalously high; the designer's alarm is misplaced.
2. The designer claims the cart is 'too fast to stay on the track' at the loop top. Evaluate the contact condition quantitatively.
Read the worked answer
At the loop top, gravity and normal force both point toward the center: N + mg = mv²/R → N = m(v²/R − g). With v = 9.0 m/s, R = 8.0 m: v²/R = 81/8 = 10.1 m/s². So N = 500(10.1 − 10) = 500(0.1) = 50 N, positive → contact maintained (barely).
Minimum speed to stay on: v_min = √(gR) = √80 ≈ 8.9 m/s. The cart at 9.0 m/s is just above the threshold. Higher speed means more normal force required, so 'too fast to stay on' is backwards: too slow loses contact. The judgment error is inverting the condition.
3. If instead the measured speed were 7.0 m/s, what would you conclude, and what does that imply about the track?
Read the worked answer
v_min = 8.9 m/s; 7.0 m/s < v_min means the cart cannot maintain contact - it would leave the track before the top. Since the cart was observed at the top, a speed of 7.0 m/s is inconsistent with a frictionless track and the stated release height. Energy accounting: required KE at top for 7.0 m/s is ½(500)(49) = 12250 J, but available is 100000 − 80000 = 20000 J. The 7750 J deficit implies energy was removed - friction/air resistance - or the release height/sensor is wrong. The judgment: a low measured speed signals dissipation, not a violation of physics.