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Car speeding up from rest around a circular track

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Homework Statement



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?

Homework Equations



a_rad = 4*pi^2*R/T^2

F=ma

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.
 

Answers and Replies

  • #2
gneill
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Are you sure that you've identified all of the forces acting on the car?
 
  • #3
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Well, there is gravity, but doesn't the normal force cancel it out?
 
  • #4
gneill
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Well, there is gravity, but doesn't the normal force cancel it out?
Yes. What else? Hint: it's a circular track.
 
  • #5
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Is it centripetal acceleration?
 
  • #6
gneill
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Is it centripetal acceleration?
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
rcgldr
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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
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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
gneill
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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
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?
 
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  • #10
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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
gneill
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But I haven't got which forces you mean. There is static friction that accelerates the car, right ? I mean the road-tire interaction.
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.
 
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  • #12
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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 circle's radius = 1.5
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
gneill
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A new question should go in a new thread. Please start a new thread for this.
 
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  • #14
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Right.
 
  • #15
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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
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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))
 

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