# Car speeding up from rest around a circular track

1. Jul 13, 2013

### postfan

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

A car of mass m takes off from rest around a circular track of radius r . Speeding up at a constant rate, the car takes t seconds to go around the track once. What is the magnitude of the net force acting on the car at the end of the trip?

2. Relevant equations

F=ma

3. The attempt at a solution

I simply used the acceleration formula and plugged it into N2L to get (4*pi^2*r*m)/t^2, but it's not right. Could someone please tell me what I am doing wrong and how to fix it.

2. Jul 13, 2013

### Staff: Mentor

Are you sure that you've identified all of the forces acting on the car?

3. Jul 14, 2013

### postfan

Well, there is gravity, but doesn't the normal force cancel it out?

4. Jul 14, 2013

### Staff: Mentor

Yes. What else? Hint: it's a circular track.

5. Jul 14, 2013

### postfan

Is it centripetal acceleration?

6. Jul 14, 2013

### Staff: Mentor

Yes indeed. Be sure to note the directions of the forces. After all, they are vector quantities and will need to be added accordingly.

7. Jul 14, 2013

### rcgldr

Does the car continue to speed up at a constant rate after the first lap? If so, you'll need to take that into account as well.

8. Feb 22, 2015

### Poetria

Could you tell me what is wrong here:

a = sqrt(((12.56*r)/(t^2))^2+((((2*3.14*r)/t)^2)/r)^2))

Tangential acceleration: (12.56*r)/(t^2)

Centripetal accelration: (((2*3.14*r)/t)^2)/r

Net force of course = m*a

9. Feb 22, 2015

### Staff: Mentor

It would be more clear if you would leave constants like $\pi$ in their symbolic form. It can be tricky to trace the provenance of decimal numbers in a formula.

Thinking about the question some more, imagining a free body diagram for the car at the end of one lap, I realized that several of the forces acting on the vehicle come in pairs that cancel each other. You yourself identified the gravitational and normal force pair. So their contribution to the NET force will be nil.

Can you think of another such pair of forces?

10. Feb 22, 2015

### Poetria

I figured out the answer. I confused velocity with average velocity. Phew. Many thanks.

velocity = (2*2*pi*r)/t

Force net = m*sqrt((157.754*r^2)/t^4+(256*pi^4*r^2)/t^4)

But I haven't got which forces you mean. There is static friction that accelerates the car, right ? I mean the road-tire interaction.

11. Feb 22, 2015

### Staff: Mentor

Right. Part of that friction force causes the car to accelerate ahead (tangentially), and part prevents the car from sliding sideways, providing the centripetal force that keeps the car moving in a circle.

You can ignore where I was going with the force pairs argument, it's not necessary here. You have arrived at a correct result.

12. Feb 23, 2015

### Poetria

Ok, many thanks. :) It was interesting though. :)

By the way I have to find a tutorial how to compute average velocity as a vector in a circular motion. I screwed this up irreparebly in a different exercice. You can't just divide the sum of velocities by two because there is acceleration.
This was a three-quarter path: http://www.mathgoodies.com/lessons/fractions/circles/circle_three_fourths_blue.gif

The bicyclist's speed: 2.5
Acceleration at the point A: (-25/6,0)
Velocity at the point A: (0, 2.5)
Velocity at the point B (the final point of the journey): (2.5,0)
And the average velocity (-0.53, -0.53)

I have computed everything correctly except for average velocity and I have no idea how to get the right angle. I guessed there should be minus signs (-1.25, -1.25) but of course magnitudes are wrong.

13. Feb 23, 2015

### Staff: Mentor

14. Feb 23, 2015

### Poetria

Right.

15. Feb 24, 2015

### dean barry

i went this way:

Basic equation for centripetal force:

f = m * ( v ^2 / r )

But: v = a * t

Splice to get: f = m * ( ( (a * t ) ^2 ) / r )

16. Feb 24, 2015

### Poetria

If I solve it in this way I get an imaginary number:

a=sqrt(r^2/(2 t^4)-(3 i sqrt(43751) r^2)/(50 t^4))