A roller coaster has a vertical drop of height 66 m and then is followed by a oop. It starts as a steep slope 75 degrees to the horizontal and then enters into a circular path of radius R. In order to stay below 4 tims the force of gravity at the bottom of the bend, what is the minimum radius of the bottom section of the loop.
After the bottom of the dip of radius R th track gradually transitions to the inverted part of the loop. What is the force of the rider at the top of the loop if the height of the loop is 33 m? Woud the change fall out of the riders pockets?
To find the minimum radius of the bottom section of the loop, we need to consider the centripetal force and gravitational force acting on the roller coaster at that point.
1. Let's start by finding the velocity of the roller coaster at the bottom of the dip. We can use the conservation of energy, assuming no energy losses due to friction or air resistance. The initial potential energy at the top of the roller coaster is converted into kinetic energy at the bottom of the dip.
Gravitational potential energy at the top = Kinetic energy at the bottom
mgh = (1/2)mv^2
Where m is the mass of the roller coaster, g is the acceleration due to gravity (approximately 9.8 m/s^2), h is the height of the vertical drop (66 m), and v is the velocity at the bottom.
Simplifying the equation, we can cancel out the mass:
gh = (1/2)v^2
2. Now, let's calculate the centripetal force required to stay below 4 times the force of gravity at the bottom of the bend. The centripetal force is provided by the normal force acting on the roller coaster at the bottom of the loop.
Centripetal force = m * (v^2 / R)
Where R is the radius of the loop.
3. To stay below 4 times the force of gravity, the centripetal force should be less than 4 times the gravitational force:
m * (v^2 / R) < 4 * mg
Canceling out the mass again, we have:
(v^2 / R) < 4g
Substituting the value of g, we get:
(v^2 / R) < 4 * 9.8
4. Since we know the height of the loop is 33 m, we can use this information to find the velocity at the bottom of the dip. Using the conservation of energy once again, assuming no energy losses:
mgh = (1/2)mv^2
h = (1/2)v^2 / g
Rearranging the equation:
v^2 = 2gh
Substituting the values of g and h, we get:
v^2 = 2 * 9.8 * 33
v^2 = 646.8
v ≈ 25.46 m/s
5. Now we can substitute the values of v and g back into the inequality:
(25.46^2 / R) < 4 * 9.8
651.31 / R < 39.2
Solving for R:
R > 651.31 / 39.2
R > 16.61 m
So, the minimum radius of the bottom section of the loop should be greater than approximately 16.61 m.
Regarding the force of the rider at the top of the loop, we need to consider the net force acting on the rider. At the top of the loop, the net force is the difference between the gravitational force and the centrifugal force.
1. The gravitational force is given by:
F_gravity = mg
Where m is the mass of the rider and g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The centrifugal force is given by:
F_centrifugal = m * (v^2 / R)
Where v is the velocity of the rider at the top of the loop and R is the radius of the loop.
The net force is:
F_net = F_gravity - F_centrifugal
3. The force of the rider at the top of the loop will be the net force acting on the rider.
Regarding whether the change would fall out of the rider's pockets, it depends on the magnitude and direction of the force. If the net force is greater than zero and directed upward, then the change would remain inside the pockets. If the net force is negative or directed downward, then the change could fall out.