Vertical loop: contact, normal force, and mass
An amusement park engineer is checking a vertical loop on a new coaster. The car and riders have a combined mass of 420 kg, and the loop has radius 8.5 m. Instruments show the car enters the bottom of the loop at 22.0 m/s. Between the bottom and the top, friction and air resistance remove 2.4 kJ of mechanical energy. The track is modelled as a particle moving in a circle, with no rotational energy of the wheels. A safety inspector claims that because the car is still moving at the top, the riders will not feel weightless; another says that the normal force, not the speed, decides whether contact is lost. The design must also work when the car is full or nearly empty, so the team tests whether the minimum entry speed depends on mass. Use g = 9.81 m/s².
1. Calculate the speed of the car at the top of the loop and the normal force on the car there. State whether contact is maintained.
Read the worked answer
Energy at the bottom:
KE_bottom = 0.5 × 420 × 22.0² = 1.016 × 10^5 J
Gain in potential energy:
ΔPE = m g (2R) = 420 × 9.81 × 17.0 = 7.004 × 10^4 J
Kinetic energy at the top:
KE_top = 1.016 × 10^5 − 7.004 × 10^4 − 2.4 × 10^3 = 2.920 × 10^4 J
v_top = sqrt(2 × 2.920 × 10^4 / 420) = 11.8 m/s
Normal force at the top: at the top both weight and normal force act downward toward the centre, so
N + mg = m v² / R
N = 420 × (11.8² / 8.5 − 9.81) = 2.75 × 10^3 N
N is positive, so the track still pushes on the car: contact is maintained. The riders feel lighter than normal but not weightless.
2. Determine the minimum entry speed that would just keep the car in contact at the top, including the stated 2.4 kJ energy loss. Compare this with the actual 22.0 m/s.
Read the worked answer
Contact is just maintained when N = 0, so at the top
mg = m v_top² / R → v_top² = gR = 9.81 × 8.5 = 83.4 m²/s²
For the minimum entry case, the energy equation from bottom to top is
0.5 m v_entry² = mg(2R) + E_loss + 0.5 m v_top²
Divide by m and substitute:
0.5 v_entry² = 2gR + E_loss/m + 0.5gR
v_entry² = 5gR + 2E_loss/m
v_entry² = 5 × 9.81 × 8.5 + 2 × 2400 / 420 = 417 + 11.4 = 428
v_entry = 20.7 m/s
The actual entry speed is 22.0 m/s, above the 20.7 m/s minimum, so there is a positive normal force at the top. The margin is about 1.3 m/s, so the design is safe under the stated assumptions.
3. A fuller car has a larger mass but the same entry speed and the same total energy loss. Decide whether it is more or less likely to lose contact at the top, and whether the minimum entry speed is independent of mass. Justify your answer.
Read the worked answer
Using the same energy equation, the top speed satisfies
v_top² = v_entry² − 4gR − 2E_loss/m
If the entry speed and the absolute energy loss are fixed, increasing m makes 2E_loss/m smaller, so v_top² is larger. The normal force is
N = m(v_top²/R − g)
so the heavier car has both a larger bracket and a larger mass factor. It is therefore less likely to lose contact at the same entry speed, assuming the energy loss stays at 2.4 kJ.
The minimum entry speed is
v_entry,min = sqrt(5gR + 2E_loss/m)
This is not independent of mass when the energy loss is a fixed absolute amount: a heavier car needs a slightly lower entry speed. Mass would cancel only if the energy loss were proportional to mass, for example if friction loss per kilogram were constant.