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]
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 :)
To solve both parts of this problem, we can make use of the law of conservation of mechanical energy. In an ideal roller coaster, where there is no friction or air resistance, the total mechanical energy of the system remains constant. The mechanical energy is the sum of the potential energy and the kinetic energy of the roller coaster car.
a) To determine the speed at ground level after the first hill, we need to find the potential energy at the top of the hill and equate it to the kinetic energy at ground level.
The potential energy at the top of the hill is given by the equation:
Potential energy = m * g * h
where m is the mass of the roller coaster car, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the hill (4.0 x 10 m).
The kinetic energy at ground level is given by the equation:
Kinetic energy = (1/2) * m * v^2
where v is the speed at ground level that we need to find.
Since the mechanical energy is conserved, we can equate the potential energy to the kinetic energy:
m * g * h = (1/2) * m * v^2
Simplifying the equation, we can cancel out the mass of the roller coaster car:
g * h = (1/2) * v^2
Solving for v, we find:
v = √(2 * g * h)
v = √(2 * 9.8 * 40)
v ≈ 19.8 m/s
Therefore, the speed at ground level after the first hill is approximately 19.8 m/s.
b) To determine the maximum height of the loop for the car to safely complete it with a minimum speed of 10 m/s at the top of the loop, we need to consider the total mechanical energy at the top of the loop.
At the top of the loop, the roller coaster car has maximum potential energy and minimum kinetic energy. The minimum speed of 10 m/s corresponds to the minimum kinetic energy required to maintain contact with the loop.
Since the mechanical energy is conserved, we can equate the potential energy at the top of the loop to the kinetic energy at the top of the loop:
m * g * h_loop = (1/2) * m * v_loop^2
Cancelling out the mass of the roller coaster car, we find:
g * h_loop = (1/2) * v_loop^2
Rearranging the equation to solve for h_loop, we have:
h_loop = (1/2) * v_loop^2 / g
Substituting the given values, we find:
h_loop = (1/2) * (10^2) / 9.8
h_loop ≈ 5.1 m
Therefore, the maximum height the loop can be for the car to safely complete it with a minimum speed of 10 m/s at the top of the loop is approximately 5.1 meters.