A car that is initially at rest at the top of a roller-coaster track moves down the track without friction. What must be the maximm initial height from which the car descends so that it will not fly off the 10-m-high circular hill at the bottom?
To determine the maximum initial height from which the car can descend without flying off the circular hill, we need to consider the conservation of energy.
The car's initial energy can be split into two parts: potential energy (PE) at the top of the roller coaster track and kinetic energy (KE) at the bottom, just before ascending the circular hill. Assuming no friction or loss of energy, the total initial energy (Ei) is equal to the total energy at the bottom (Ef).
At the top of the track, the car has gravitational potential energy, given by the equation:
PE = m * g * h,
where m is the mass of the car, g is the acceleration due to gravity, and h is the height of the car from the reference point (usually chosen as the ground level). In this case, the height is the initial height from which the car descends.
As the car descends without friction, the potential energy is converted into kinetic energy when it reaches the bottom of the track. The kinetic energy is given by the equation:
KE = (1/2) * m * v^2,
where v is the velocity of the car at the bottom.
When the car reaches the top of the circular hill, it should not fly off, meaning the gravitational force must be sufficient to keep the car moving in a circular path. At the highest point of the circular hill, the net force acting on the car should be directed towards the center of the circle. This net force is provided by the downward gravitational force and the normal force pointing upwards.
So, we can use the centripetal acceleration formula to balance these forces:
m * g - m * (v^2 / r) = 0,
where r is the radius of the circular hill.
Simplifying this equation, we get:
g = v^2 / r.
Now, we need to express the velocity in terms of the total energy at the top of the track. The total energy can be written as:
Ei = PE + KE,
Ei = m * g * h + (1/2) * m * v^2.
Since Ei = Ef (conservation of energy), we have:
m * g * h = (1/2) * m * v^2 + m * g * (h - r),
where r = 10 m (height of the circular hill).
Simplifying and canceling m, we get:
g * h = (1/2) * v^2 + g * (h - r),
g * h = (1/2) * v^2 + g * h - g * r,
g * r = (1/2) * v^2.
Substituting g = 9.8 m/s^2 and r = 10 m, we can solve for v:
9.8 * 10 = (1/2) * v^2,
98 = (1/2) * v^2,
v^2 = 196,
v = 14 m/s.
Now that we know the velocity, we can calculate the maximum initial height (h) from which the car descends:
g * h = (1/2) * v^2,
9.8 * h = (1/2) * (14^2),
h = (1/2) * (14^2) / 9.8,
h ≈ 10.2 m.
Therefore, the maximum initial height from which the car can descend without flying off the 10-meter-high circular hill at the bottom is approximately 10.2 meters.