A roller-coaster car speeds down a hill past point A where R1 = 8.2 m and then rolls up a hill past point B where R2 = 15.4 m, as shown below.
(a) The car has a speed of 21.0 m/s at point A. if the track exerts a normal force on the car of 2.30 ✕ 104 N at this point, what is the mass of the car?
kg
(b) What is the maximum speed the car can have at point B for the gravitational force to hold it on the track?
(a) To find the mass of the car, we can use the centripetal force equation:
F = mv² / R
Where:
F is the normal force exerted on the car at point A,
m is the mass of the car,
v is the speed of the car, and
R is the radius of the circular path (in this case, R1).
Rearranging the equation, we get:
m = F * R / v²
Plugging in the given values:
F = 2.30 ✕ 10^4 N
R = 8.2 m
v = 21.0 m/s
m = (2.30 ✕ 10^4 N) * (8.2 m) / (21.0 m/s)²
m ≈ 743 kg
Therefore, the mass of the car is approximately 743 kg.
(b) To find the maximum speed the car can have at point B, we need to consider the gravitational force acting on the car. At the top of the hill (point B), the normal force and the gravitational force must be equal for the car to remain on the track. The gravitational force is given by:
F_gravity = mg
Where:
m is the mass of the car, and
g is the acceleration due to gravity (9.8 m/s²).
Therefore, we have:
F_gravity = mg = F_normal
Using this information, we can find the maximum speed:
F_normal = mv² / R
Rearranging the equation, we get:
v² = R * g
Plugging in the given value for R2:
R = 15.4 m
v² = (15.4 m) * (9.8 m/s²)
v ≈ 11.6 m/s
Therefore, the maximum speed the car can have at point B for the gravitational force to hold it on the track is approximately 11.6 m/s.
To solve this problem, we need to use the concept of centripetal force, which is the force acting on an object moving in a circular path. In this case, the centripetal force is provided by the normal force exerted by the track.
(a) To find the mass of the car at point A, we can use the formula for centripetal force:
F = (m * v^2) / R
Where:
- F is the centripetal force (2.30 x 10^4 N)
- m is the mass of the car (unknown)
- v is the speed of the car (21.0 m/s)
- R is the radius of the circular path (8.2 m)
Rearranging the formula, we can solve for the mass (m):
m = (F * R) / v^2
Substituting the given values:
m = (2.30 x 10^4 N * 8.2 m) / (21.0 m/s)^2
Calculating this expression will give you the mass of the car at point A in kilograms.
(b) To find the maximum speed the car can have at point B, we need to consider the gravitational force acting on the car. At point B, the gravitational force acting downwards is balanced by the normal force exerted by the track, preventing the car from losing contact with the track.
The gravitational force is given by:
F_gravity = m * g
Where:
- F_gravity is the gravitational force (unknown)
- m is the mass of the car (the same value we found in part (a))
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Since the gravitational force provides the centripetal force at point B, we can set it equal to the centripetal force formula:
F_gravity = (m * v^2) / R
Simplifying and rearranging the formula, we can solve for the maximum speed (v) at point B:
v = √(g * R)
Substituting the given values:
v = √(9.8 m/s^2 * 15.4 m)
Calculating this expression will give you the maximum speed the car can have at point B in meters per second (m/s).