Asked by Mel
Part of a roller-coaster ride involves coasting down an incline and entering a loop 30.0 m in diameter. For safety considerations, the roller coasters speed at the top of the loop must be such that the force of the seat on a rider is equal in magnitude to the rider’s weight. From what height above the bottom of the loop must the roller coaster descend to satisfy this requirement?
Answers
Answered by
Damon
R = 15 meters
Ac = w^2 R = v^2/R
Ac = 2 g to balance weight plus to give additional mg force
9.8 * 2 = v^2/15
v^2 = 294 m^2/s^2
v at top of loop = 17.1 m/s
potential energy at top of loop = m g h
= m (9.8)(30) = 294 m
kinetic energy at top of loop = (1/2) m (294) = 147 m
total energy at top of loop = 441 m
we need 441 m Joules of potential energy at start to make it
m g h = 441 m
9.8 h = 441
h = 45 meters high
Ac = w^2 R = v^2/R
Ac = 2 g to balance weight plus to give additional mg force
9.8 * 2 = v^2/15
v^2 = 294 m^2/s^2
v at top of loop = 17.1 m/s
potential energy at top of loop = m g h
= m (9.8)(30) = 294 m
kinetic energy at top of loop = (1/2) m (294) = 147 m
total energy at top of loop = 441 m
we need 441 m Joules of potential energy at start to make it
m g h = 441 m
9.8 h = 441
h = 45 meters high
Answered by
Anonymous
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