# Circular Motion of an auto

1. Oct 29, 2013

### Sylis

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

A 1500 kg car drives around a flat 200-m-diameter circular track at 25 m/s. What is the magnitude and direction of the net force on the car? What causes this force?

2. Relevant equations

F=mv2/r

3. The attempt at a solution

So I suppose the net force is the force that maintains the circular motion. Thus,

Fnet=(1500)(25)2/2100=9375N

However, how exactly would I come to the conclusion that the direction of Fnet is toward the center and how do I determine that the force that causes this force to be frictional?

I missed the class that we covered this entire chapter in, I'm doing my best to play catch up but am having trouble making the connection between linear and circular motion, so if the answers could be dumbed down a little for me that would be awesome. Thank you.

2. Oct 29, 2013

### hilbert2

Suppose we have an object that moves with constant speed on a circular trajectory that has radius $R$ and is centered at the origin. Now the position vector of the object as function of time is

$(x(t),y(t))=(Rcos\omega t, Rsin\omega t)$ ,

where $\omega$ is the angular speed. The acceleration of the object is the second time derivative of $(x(t),y(t))$:

$a=\frac{d^{2}}{dt^{2}}(x(t),y(t))=(R\frac{d^{2}}{dt^{2}}cos\omega t, R\frac{d^{2}}{dt^{2}}sin\omega t)=(-R\omega^{2}cos\omega t, -R\omega^{2}sin\omega t)$.

In other words, the acceleration vector has direction opposite to the direction of the position vector, and therefore points to the center of the circular trajectory. You can also calculate the norm of the acceleration vector to find that $|a|=R\omega^{2}=v^{2}/R$.

In your example, the only forces acting on the car are friction and gravity. As the car is at constant altitude, the net force must be horizontal and can only be caused by friction.

3. Oct 29, 2013

### johntcmb

You got the formula correct, but did the math wrong. It should be F=4687.5N. If you draw a free-body diagram of the vehicle, you will see that the only external forces acting on the car are friction, gravity and the normal force of the road against the tires. Since gravity and the normal force are perpendicular to the direction of the acceleration, the only force that could cause that acceleration is friction.

4. Oct 29, 2013

### nasu

Well, you are supposed to know that for uniform circular motion the acceleration is along the radius, pointing towards the center. (for this reason called centripetal acceleration).
This results from the fact that only the direction of velocity changes and not its magnitude.
This can be shown by different methods, depending on the level of your class.
Probably you missed this part but it should be in the book.
So if the acceleration is centripetal, the force should have the same direction - this follows from Newton's second law.

If the motion is not uniform (speed increases or decreases while moving in a circle) then the acceleration has a tangential component too so the direction is not towards the center anymore.

5. Oct 29, 2013

### haruspex

200m is the diameter, not the radius.

6. Oct 30, 2013

### johntcmb

Thanks, I need to read the problem more clearly.