# Racecar travelling a banked curve

1. Jul 7, 2010

### ninetyfour

1. The problem statement, all variables and given/known data

A racecar is traveling at a speed of 120km/h on a banked curve (no friction). What is the radius of the rotation if the turn is banked at an angle of 25 degrees to the horizontal?

2. Relevant equations

v = 120 km/h
angle = 25 degrees
r = ?

3. The attempt at a solution

I have no idea how to approach this problem. Help? Ideas? Hints? :(

Last edited: Jul 8, 2010
2. Jul 7, 2010

### jgm340

Here's the hint: If the car were sitting on tilted track completely still, only the friction of the wheels would keep it from sliding inwards (towards the center of the circular path). What force is acting on the car in this case?

It ought to be that this force acting on the car equals the force required to keep it on the circular path.

The equation for centripetal force (the force required to keep an object on a circular path) is this:

where F = the force required to keep the object on a circular path, v is the velocity of the object, and r is the radius of the circular path.

3. Jul 7, 2010

### ninetyfour

Okay, so I should use the force equation. What should I substitute for mass, as that is not given?

4. Jul 7, 2010

### jgm340

There is another equation, which it is up to you to find.

This is another equation for force will also involve mass (allowing you to cancel it out). It is an equation for the force which would cause a ball to roll to the side of a tilted track.

Last edited: Jul 7, 2010
5. Jul 7, 2010

### ninetyfour

Alright, here's what I have done. I didn't know how to incorporate the centripetal force formula though...

I split Fn into two components, the vertical and the horizontal.
Fy = force of gravity = m*g
Fx = some force = m*a

theta = 25 degrees

tan25 = m*a / m*g
- the masses cancel out
tan25 = a / g
tan25(g) = a
a = 4.57 m/s^2

a = v^2 / r
4.57 m/s^2 = (33.3333 m/s)^2 / r
r = 243m

Is this right? D: