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I'm writing a program to simulate a car driving. I'm wondering how I should find the radius when calculating the centripital force. If we let the car travel at a constant speed, then $F_{net} = mv^2/r$. Does each of the front wheels have their own radius? How do I find the radius?

My first thought was that to find the radius as in my illustration below. But at second thoughts I don't think that's correct.

enter image description here

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  • $\begingroup$ Take the radius as the distance from the center to the center of mass of the car. $\endgroup$ May 3 '15 at 8:44
  • $\begingroup$ @HritikNarayan What is the center though? $\endgroup$
    – lawls
    May 3 '15 at 8:45
  • $\begingroup$ The center should be the center of curvature of the turn! $\endgroup$ May 3 '15 at 8:47
  • $\begingroup$ @HritikNarayan Obviously, but that is the problem. Say the car is driving in a parking lot. All information I have is how much the wheels are turned. $\endgroup$
    – lawls
    May 3 '15 at 8:50
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    $\begingroup$ How is this information given? Would you be able to obtain an arc like trajectory of sorts, from the given information? That'd help you find the radius. $\endgroup$ May 3 '15 at 8:52
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The issue here is that your front wheels are turned/steered by the same angle.

When you try to find the instantaneous centre of curvature, you may first want to assume the wheels won't slip from side to side, like you may get if you drive around a corner on a slippy road. As there is no slip, the velocity of each wheel must occur in the direction the wheel, as you'd expect.

Then, for all four wheels, draw lines perpendicular to the velocities. All four lines must meet at the same point, otherwise the assumption that there is no slip is false.

So, your front wheels are steered by the same angle, and this is not good as this would mean the perpendicular lines you drew would never meet. So, there would have to be slip. If there is slip, this would become a dynamics problem rather than just kinematics (i.e. have to use Newton's 2nd Law). So the front wheels have to turn by different amounts (unless the front wheels are going straight forward).

This is shown in the following diagram: enter image description here

Notice how all the lines connecting the instantaneous centre and the wheels are perpendicular to the velocities. In this system, there may be no slip (provided the magnitudes of velocity are calculated appropriately!).

It's important to note the inst centre always lies on the back wheels' axis line. Note the odd case when all wheels face forward: it seems the 4 perpendicular lines never meet. However, this is fine as the inst centre actually lies at infinity, and parallel lines 'meet at infinity'. An inst centre at infinity implies the car moves in a straight line.

So, to find the inst centre, it would be ideal to assume no slip. To do so, provided the back wheels don't steer, arrange one of the front wheels as you want, then determine the inst centre, then arrange the fourth wheel accordingly. If you include slip, velocities may not necessarily align with the direction of the wheel.

Now that you know the inst centre, you need to find the speed of the centre of mass of the car. If you don't know it, then you need to know the speed of one of the wheels. To do so, note the formula:

$$v_{wheel} = \omega r_{wheel}$$

Where $v_{wheel}$ is the speed of a particular wheel, $r_{wheel}$ is the distance of that wheel from the inst centre, and $\omega$ is the angular velocity of the car about the inst centre. Similarly, the centre of mass follows the formula:

$$v_{g} = \omega r_{g}$$

Where $v_g$ is the speed of the centre of mass, and $r_g$ is the distance of the centre of mass from the inst centre.

Finally, the centripetal acceleration can be determined by:

$$a_c = \omega^2 r_g = \frac{v_g^2}{r_g}$$

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  • $\begingroup$ Thank you for a very thorough answer! I will work a bit on my own and try out your ideas and possibly return with a follow-up question if I get stuck. $\endgroup$
    – lawls
    May 4 '15 at 6:25

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