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?
Thank you!
To determine the height above the bottom of the loop from which the roller coaster must descend, we need to use the concept of centripetal force.
The centripetal force acting on the rider at the top of the loop is the sum of the gravitational force (weight) and the normal force exerted by the seat. In this case, we want the magnitude of the normal force to be equal to the weight of the rider, which means that the centripetal force should be twice the weight.
We can calculate the centripetal force using the formula:
F = m * v² / r
where:
- F is the centripetal force,
- m is the mass of the rider,
- v is the velocity of the roller coaster, and
- r is the radius of the loop.
In this case, the centripetal force is equal to twice the weight:
2 * m * g = m * v² / r
Where:
- g is the acceleration due to gravity (approximately 9.81 m/s²), and
- we assume the mass of the rider (m) cancels out on both sides of the equation.
Now we can rearrange the equation to solve for v:
v² = 2 * g * r
Next, we can use the conservation of mechanical energy to determine the initial height (h) above the bottom of the loop from which the roller coaster must descend. At the top of the loop, the total mechanical energy is the sum of the potential energy (mgh) and the kinetic energy (1/2mv²):
2 * m * g * h = 1/2 * m * v²
We can substitute the value of v² from the previous equation:
2 * m * g * h = 1/2 * m * 2 * g * r
The mass of the rider cancels out, and we can simplify the equation further:
2 * g * h = g * r
Finally, we can solve for h:
h = r / 2
Therefore, the roller coaster must descend from a height equal to half the radius of the loop. In this case, the loop has a diameter of 30.0 m, so the radius is half of that, which is 15.0 m.
Therefore, the roller coaster must descend from a height of 15.0 meters above the bottom of the loop to satisfy the requirement.
To determine the height above the bottom of the loop from which the roller coaster must descend to satisfy the requirement, we can use the principle of conservation of mechanical energy.
The roller coaster will have gravitational potential energy at the initial height, and it will also have kinetic energy as it reaches the top of the loop. At the top of the loop, the centripetal force must be equal to the gravitational force acting on the rider to provide the necessary force of the seat on the rider.
Let's assume the mass of the rider is "m," and the acceleration due to gravity is "g."
First, let's calculate the speed of the roller coaster at the top of the loop:
The centripetal force is given by the formula:
F_c = m * (v^2 / R)
where F_c is the centripetal force, m is the mass of the rider, v is the speed, and R is the radius of the loop.
Since the centripetal force must be equal to the weight of the rider, we have:
F_c = m * g
Setting these two equations equal to each other, we get:
m * (v^2 / R) = m * g
Canceling out the mass on both sides, we find:
v^2 = g * R
Taking the square root of both sides:
v = √(g * R)
Now, at the top of the loop, the kinetic energy (KE) is given by:
KE = (1/2) * m * v^2
This kinetic energy must be equal to the gravitational potential energy (PE) at the initial height:
PE = m * g * h
Setting these two equations equal to each other, we get:
(1/2) * m * v^2 = m * g * h
Canceling out the mass on both sides:
(1/2) * v^2 = g * h
Simplifying further, we find:
h = (1/2) * (v^2 / g)
Substituting the expression for v we found earlier, we have:
h = (1/2) * ((√(g * R))^2 / g)
Simplifying further, we get:
h = (1/2) * (g * R / g)
Canceling out the "g" terms, we find:
h = (1/2) * R
Therefore, the roller coaster must descend from a height equal to half the radius of the loop, which in this case is half of 30.0 m.
Thus, the roller coaster must descend from a height of 15.0 m above the bottom of the loop to satisfy this requirement.
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