A roller coaster works by gravitational energy. The coaster car is pulled up to a high point and then released, rolling downwards on the track through all manners of curves and loops. I have a short roller coaster car that I pull up to the top of a hill of height H. The coaster car is released from this height and must go around a perfectly circular vertical loop with a radius of 20 m (and the bottom of the loop is on the ground). If I don't want the coaster car to fall off the loop at any point, what should be the minimum value of H in meters?
50 meters
thank you
To determine the minimum value of H necessary for the coaster car to stay on the loop, we need to consider the forces acting on the car at different points of the loop.
At the top of the loop, there are two primary forces acting on the car: the gravitational force (mg) and the normal force (N). The normal force exerted by the track must be sufficient to supply the centripetal force required to keep the car moving in a circular path.
To find the minimum value of H, we can equate the centripetal force to the sum of the gravitational force and the normal force:
Centripetal force = gravitational force + normal force
mv^2/R = mg + N
where:
m = mass of the car (assumed to be constant)
v = velocity of the car at the highest point of the loop
R = radius of the circular loop
g = acceleration due to gravity (approximately 9.8 m/s^2)
Since the velocity at the topmost point of the loop is zero (the car momentarily comes to rest), we can set v = 0. Therefore, the equation simplifies to:
0 = mg + N
Since the weight of the car (mg) is downward and the normal force (N) is upward, their magnitudes will be equal but opposite in direction. So, we can rewrite the equation as:
mg = N
Substituting this equation into the earlier centripetal force equation, we get:
mv^2/R = 2mg
Canceling out the mass, the equation becomes:
v^2/R = 2g
Rearranging the equation to solve for v^2:
v^2 = 2gR
Now, we can substitute the given values:
v^2 = 2 * 9.8 m/s^2 * 20 m
v^2 = 392 m^2/s^2
Taking the square root of both sides:
v = √(392) m/s
Finally, to determine the minimum value of H, we can use the equation for kinetic energy:
KE = 0.5mv^2
where KE is the kinetic energy and m is the mass of the car.
At the top of the loop, all of the initial potential energy (mgh) is converted to kinetic energy (KE). Therefore, we can equate the potential energy to the kinetic energy:
mgh = 0.5mv^2
Canceling out the mass:
gh = 0.5v^2
Solving for H:
H = (0.5v^2)/g
Substituting the values:
H = (0.5 * (392 m^2/s^2))/9.8 m/s^2
H = 10 m
So, the minimum value of H necessary for the coaster car to stay on the loop is 10 meters.
To find the minimum value of H to prevent the coaster car from falling off the loop, we need to consider the forces acting on the car at the topmost point of the loop. At this point, there are two significant forces: the gravitational force and the normal force.
The gravitational force always acts vertically downward and can be calculated using the equation:
F_gravity = m * g,
where m is the mass of the coaster car and g is the acceleration due to gravity (approximately 9.8 m/s^2).
The normal force, denoted by N, is the force exerted by the track perpendicular to the track's surface. At the top of the loop, the normal force acts downward to counterbalance the gravitational force and prevent the car from falling off.
To determine the normal force, we need to consider the net force acting on the car. At the topmost point of the loop, the net force should provide the necessary centripetal force to keep the car moving in a circular path.
The net force is given by the equation:
F_net = N - F_gravity.
The centripetal force can be calculated as:
F_centripetal = m * (v^2 / r),
where v is the velocity of the car at the topmost point and r is the radius of the loop.
Since the car is moving in a perfect circle, the net force must equal the centripetal force:
N - F_gravity = m * (v^2 / r).
Now, let's consider the situation where the normal force is at its minimum value, which occurs when the car just barely maintains contact with the track. At this point, the normal force becomes zero, resulting in the car losing contact with the track and falling off.
To find the minimum value of H, we assume the normal force is zero and solve for H. At the topmost point, the gravitational force provides the necessary centripetal force to keep the car moving in a circular path:
m * g = m * (v^2 / r).
The mass cancels out:
g = v^2 / r.
Rearranging the equation to solve for v^2:
v^2 = g * r.
Now, we can use the conservation of energy to relate the velocity at the topmost point to the height H. At the topmost point, all the gravitational potential energy is converted to kinetic energy:
m * g * H = (1/2) * m * v^2,
where m cancels out.
Substituting the expression for v^2:
g * H = (1/2) * g * r.
The gravitational acceleration, g, cancels out:
H = (1/2) * r.
Plugging in the given radius of the loop (r = 20 m):
H = (1/2) * 20 m = 10 m.
Therefore, the minimum value of H should be 10 meters to prevent the coaster car from falling off the loop at any point.