A roller coaster car of mass 960 kg starts at a distance of H = 20< above the bottom of a loop 12 m in diameter. If the friction in negligible, find the normal force of the rails on the car when i it is a) pside down the top of the loop and b) at the bottom of hte loop
thank you, i would really apreciate a clear explenation...
thank you, i would really apreciate a clear explenation...
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Energy at the top of the loop = m . g . ( 20 – 12 ) = 960 . 9.8 . 8 = 75264 J
This is all Kinetic energy so the velocity must be
v = √ 2 . E / m = 12.52 m/s
The reaction of the track ( Normal force ) = mv^2/R – mg = 960 ( 26.1 – 9.8 ) = 15680 N
b) Track reaction = mg + mv^2 / R
v = √ 2 . g . h = √ 40 . 9.8 = 19.80 m/s
N = 960 ( 9.8 + 65.3 ) = 72128 N
This is all Kinetic energy so the velocity must be
v = √ 2 . E / m = 12.52 m/s
The reaction of the track ( Normal force ) = mv^2/R – mg = 960 ( 26.1 – 9.8 ) = 15680 N
b) Track reaction = mg + mv^2 / R
v = √ 2 . g . h = √ 40 . 9.8 = 19.80 m/s
N = 960 ( 9.8 + 65.3 ) = 72128 N