4.0 x 10 m
10^what ???? left exponents out twice?
anyway I will call the initial height H
and I will call the max loop height (the diameter) D
a)
(1/2) m V^2 = m g H
V = sqrt (2 g H)
b) radius = R = D/2
v = velocity at top of loop = sqrt [2 g* (H-D)] = sqrt [2 g* (H- 2R)]
centripetal acceleration = v^2/ R
centripetal acceleration = g or it will fall
2 g* (H- 2R) /R = g
2 H - 4 R = R
2 H = 5 R
R = (2/5) H
cool :)
A roller coaster car is at the top of a 4.0 x 10 m hill, above the ground. It starts from rest. It travels down the hill to ground level and then goes into a loop. Assuming there is this is an ideal roller coaster and that there is no friction between the car and track and no air resistance. Determine:
a) What the speed is at ground level after the first hill. [2 marks]
b) The minimum speed required to clear a loop, is 1.0 x 10 m/s at the top of the loop. Determine the maximum height the loop can be for the car to safely complete the loop. [3 marks]
1 answer
1 answer