A roller-coaster car speeds down a hill past point A where R1 = 8.0 m and then rolls up a hill past point B where R2 = 14.0 m.
The car has a speed of 21.0 m/s at point A. if the track exerts a normal force on the car of 2.19 multiplied by 104 N at this point, what is the mass of the car?
I don't understand what R1 and R2 are. Are they heights?
fdh
R1 and R2 are both the radius of the circles at the bottom and top respectively. I don't know the answer, I'm looking into that as well. At least you know that it's a radius now! :D
To find the mass of the car, we can use the concept of centripetal force.
At point A, the only force acting on the car in the vertical direction is the normal force (since we assume no friction or air resistance). The centripetal force is provided by the gravitational force acting on the car.
The centripetal force, Fc, is given by the formula:
Fc = m * v^2 / R
Where:
- Fc is the centripetal force
- m is the mass of the car
- v is the velocity of the car
- R is the radius of the circular path
In this case, the centripetal force is equal to the gravitational force acting on the car, which can be expressed as:
Fc = m * g
Where:
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Since Fc = m * g, we can equate the two equations for Fc:
m * v^2 / R = m * g
Simplifying the equation and solving for mass:
m = (R * g) / v^2
Given that R1 = 8.0 m, v = 21.0 m/s, and g = 9.8 m/s^2, we can substitute these values into the equation:
m = (8.0 m * 9.8 m/s^2) / (21.0 m/s)^2
Evaluating the equation gives:
m = (78.4 N) / (441.0 m^2/s^2)
m ≈ 0.177 kg
Therefore, the mass of the car is approximately 0.177 kg.