2
$\begingroup$

Let's say we have a car with four wheels and the center of the car has a speed of 50km/h. There is no friction or air resistance. If the car is moving straight each wheel will also have a speed of 50km/h, since that's the speed of the center of the car.

If the car's front wheels would turn a little bit, then we would have uniform circular motion (since there is nothing to slow down the speed). Each wheel would have its own speed (since their radius to the center of the curve would be different for each wheel). What I wonder is if the speed of the center of the car is still 50km/h?

UPDATE

I realize that there need to be friction to turn. What I meant was that there is no friction that will slow down the speed. When the car turns there will be a centripetal force perpendicular to the car's velocity. Which should indicate that the car's speed will remain the same, but it's direction will change.

NEW UPDATE

Here's a simulation I made: https://vid.me/HYz9. The red line is the car's velocity, and the green line is the front wheel's velocity. You will notice that the red line only changes direction, whilst the green line (the front wheel's velocity) also gets a higher speed as the wheel is turned. Is this simulation accurate?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Technically if there is no friction you could argue that even if the wheels turned the car would not turn. Or actually the car would end up in a ever ending spin as the result of the mass movement. Probably not what you meant but I could not resist :) $\endgroup$
    – Waxhead
    Commented May 16, 2015 at 12:00
  • $\begingroup$ You can't turn without friction. If there's no friction, the wheels will still continue to spin exactly as they did before you turned them and the car would still have the same velocity vector and thus not experience any acceleration. $\endgroup$
    – Madde Anerson
    Commented May 16, 2015 at 13:46

2 Answers 2

Reset to default
1
$\begingroup$

You are correct that centripetal force does not affect speed. So let us consider tangential speed, which is the speed the turning car would have if centripetal force were removed. Linear speed = tangential speed = distance / time = (2 * pi * radius) / time. (To simplify, assume the car goes through one full circle at uniform speed.) The linear speed of the center of the car is proportional to the radius of the circle. Likewise the linear speed of each wheel is proportional to the radius of the respective circles they traverse.

When the front wheels turn, they place not only themselves, but also the center of the car, on the circumferences of circles of differing radii.

Each wheel and the center of the car will be moving at different linear speeds, and none of them will necessarily be at the same speed they had when they were on a straight line. Also, the rear wheels will be on different radii from the front wheels, from the center, and from each other.

If there were some way for you to measure the speed of the car's center independently of the speed of each wheel, you could intentionally keep it at 50 km/h by adjusting the rotation of the wheels. But if the wheels connected to the power train kept turning at the original number of revolutions, the center of the car would not continue at an instantaneous speed of 50 km/h, UNLESS you carefully chose to place it on the radius of a circle with speed component of the velocity vector equal to 50 km/h.

Alternatively, you could equate linear average speed with circumferential average speed: v = r * w, where "v" is the linear speed, "r" is the radius of a circle, and "w" is angular velocity.

Incidentally, without friction there would be no centripetal force and neither the front wheels nor the center of the car could move in circles, as warhead pointed out.

Improve this answer
$\endgroup$
8
  • $\begingroup$ I realize that friction is needed to turn, what I meant to say was that there are no forces that affect the speed of the car. Correct me if I'm wrong, the car's speed should not change just because it's turning, since the centripetal force is perpendicular to the motion (velocity). Which should indicate that the car's speed should still be 50 km/h. $\endgroup$
    – Mark
    Commented May 16, 2015 at 14:04
  • $\begingroup$ @Mark: You're right, centripetal force changes the car's direction, not its speed. But if the power train wheels keep turning at the original rpm, you need to choose a radius that will keep the center of the car at 50 km/h. In other words, 50 km/h won't automatically be maintained regardless of the radius. $\endgroup$
    – Ernie
    Commented May 16, 2015 at 15:05
  • $\begingroup$ But the wheels don't turn at the original rpm during a turn, do they? Each wheel need to change their speed since they all have different radiuses during the turn, right? $\endgroup$
    – Mark
    Commented May 16, 2015 at 15:26
  • $\begingroup$ @Mark: True, each front wheel will seek its own rpm, and the power train wheels (assuming rear power) will be adjusted by the differential gears to rotate at differing rpm in order to maintain different speeds. What I'm trying to say is that the speed at the center of the car will depend on the radius, so you need to adjust the accelerator to maintain 50 km/hr in a turn. In other words, your speed in the turn will depend on the radius of the turn, other things being equal. $\endgroup$
    – Ernie
    Commented May 16, 2015 at 16:25
  • $\begingroup$ How can the radius change the speed? Obviously the angular speed changes, but the translational speed should remain the same? Did you see my video I added to my question? Would you say that's an inaccurate simulation? $\endgroup$
    – Mark
    Commented May 16, 2015 at 16:53
0
$\begingroup$

I would say that indeed the center of the car still move at 50km/h.

The angular velocity is $\omega = v/r$, with v the tangantial velocity and r the radius.

Therefore the wheels on the inside will go slower (smaller v) and the wheel on the outside will go faster (larger v) to keep the same angular velocity $\omega$.

Best,

Samuel

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.