After its release at the top of the first rise, a roller-coaster car moves freel

Question

After its release at the top of the first rise, a roller-coaster car moves freely with negligible friction. The roller coaster shown in Figure P8.26 has a circular loop of radius 19.3 m. The car barely makes it around the loop. At the top of the loop, the riders are upside down and feel weightless.

After its release at the top of the first rise, a roller-coaster car moves freely with negligible friction. The roller coaster shown in Figure P8.26 has a circular loop of radius 19.3 m. The car barely makes it around the loop. At the top of the loop, the riders are upside down and feel weightless.

Explanation / Answer

a)
At the top of the loop (point 3) if the riders are weightless, then the acceleration due to the car going round the loop exactly matches the acceleration due to gravity.

Acceleration due to going round the loop
= v^2 / 19.7 = 9.81
v^2 = 19.7 * 9.81
v = 13.90 m/s

b)
Potential energy at the top of the loop
= M * 19.7 * 2 * 9.81
= 386.514 * M J

Kinetic energy at the top of the loop
= 0.5 * M * v^2
= 0.5 * M * 13.90^2
= 96.6285 * M J

Total energy (potential & kinetic)
= 483.1425 * M J

At the bottom of the loop (point 1), the potential energy will be converted to kinetic energy
= 0.5 * M * v^2 = 483.1425 * M
v^2 = 2 * 483.1425
v = 31.085 m/s

c)
The potential energy at point 2
= M * 9.81 * 19.7
= 193.257 * M J

Kinetic energy at this point
= (483.1425 - 193.257) * M
= 289.8855 * M J
= 0.5 * M * v^2

v^2 = 289.8855 / 0.5
v = 24.078 m/s

d)
Kinetic energy at point 4
= 0.5 * M * 10.3^2
= 53.045 * M J

Kinetic energy at point 1
= 483.1425 * M J (answer b)

Potential energy at point 4
= (483.1425 - 53.045) * M
= 430.0975 * M J
= M * 9.81 * h
h = 430.0975 / 9.81
= 43.843 m

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