Roller coaster circular motion normal forces

[wegman14]

Homework Statement

A roller-coaster car has a mass of 1200 kg when fully loaded with passengers. As the car passes over the top of a circular hill of radius 17 m, its speed is not changing. (a) At the top of the hill, what is the normal force (using the negative sign for the downward direction) FN on the car from the track if the car's speed is v = 8.2 m/s? (b) What is FN if v = 17 m/s?

Homework Equations

-Fn= m (-v^2 / R)

The Attempt at a Solution

A) 1200(- 8.2^2/ 17)
-Fn= -4746.352941

B) 1200(- 17^2/ 17)
-Fn= -20,400 N
do not believe these answers are correct, but cannt determine any other way to solve, thanks for the help

 

[rock.freak667]

Consider the forces acting on the car at the top of the loop. The normal reaction,R, upwards and the weight,W. The resultant of theses 2 provide the centripetal force needed to keep the car on the track. So that resultant force=Weight-Normal Reaction.

 

[Moetasim]

I would first check the equations and units used in the attempt at a solution. The equation given, -Fn= m (-v^2 / R), is correct for calculating the normal force in circular motion. However, the units used for mass (kg) and velocity (m/s) do not match, which could lead to incorrect results. The mass should be in kilograms (kg) and the velocity in meters per second (m/s). Also, the radius (R) is given in meters, so it should be kept in the same unit when plugging into the equation.

Using the correct units, the solutions would be:

A) -Fn= (1200 kg)(-8.2 m/s)^2 / 17 m
-Fn= -4,746.4 N

B) -Fn= (1200 kg)(-17 m/s)^2 / 17 m
-Fn= -20,400 N

These answers seem reasonable, as the normal force would decrease as the speed increases due to the centripetal acceleration becoming greater. However, it is always important to double check the calculations and units to ensure accuracy.