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So here is the standard problem setup of a car turning on an unbanked road:

A 1000 kg car is going around a curve with radius 30 meters. If the coefficient of friction between the car's tires and the road is 0.5, what is the maximum speed at which the car can make the turn?

You setup $$f_K = \mu_k mg = m \frac{v^2}{r}$$

and solve for $v$ to get the maximum possible velocity with which you can still make the turn.

However, what about if you go lower than this velocity. Clearly you will still be able to make the turn, however the above equation no longer works, even though the equation should hold.

While your centripetal acceleration decreases, the force from kinetic friction remains at the same value it was before, and now both sides of the equation are not equal.

What changes in regards to the frictional force that compensates for the decrease in acceleration?

Essentially, my question is that while it is "clear by logic" that going lower than the maximum velocity allows you to make the turn, how would you show that using the above equations as per newton's second law?

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    $\begingroup$ As the tire is (presumably) not slipping on the road, I think you would want to use $\mu_s$ instead of $\mu_k$. $\endgroup$
    – BowlOfRed
    Commented Jun 22, 2015 at 7:40
  • $\begingroup$ @BowlOfRed Could you please elaborate further, perhaps posting as an answer. If you indeed use $\mu_s$ then the dependence on velocity is obvious, but I do not see why you should use $\mu_s$ instead of $\mu_k$ -- isn't the car in motion? $\endgroup$
    – 1110101001
    Commented Jun 22, 2015 at 9:02
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    $\begingroup$ Because the bottom of the tire is not in motion with respect to the road. Rolling is not skidding. $\endgroup$
    – BowlOfRed
    Commented Jun 22, 2015 at 9:03
  • $\begingroup$ @BowlOfRed Ok, that makes sense; post it as an answer and I'll accept it. As a tangentially related question, does the force of kinetic friction depend on speed like static friction does? $\endgroup$
    – 1110101001
    Commented Jun 22, 2015 at 9:08
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    $\begingroup$ Static friction always has relative $v=0$, so it doesn't depend on speed. It depends on net force. $\endgroup$
    – BowlOfRed
    Commented Jun 22, 2015 at 9:10

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The coefficient of static friction and the normal force together allow you to calculate the maximum force that friction can apply, not the actual force.

A block sitting on a shelf with no external horizontal forces will have a friction force of zero. If you apply a force from the side that is less than the maximum friction force, then the actual friction force will rise so that it is equal to the external force.

The same thing happens with the car in this case. When the velocity is less than the maximum, the tires are able supply a centripetal force equal to the net (centripetal) force.

$$f_s \leqslant \mu_s F_n$$

when you travel in the car at a slower speed then, what is the force that is being lessened on the bottom of the tire to account for the decrease in static friction?

Inside the car, the force being decreased is the (fictitious) centrifugal force. It's fine to think of it that way in the frame of the car.

Otherwise, let's think of it just a bit differently. To turn the car means that you are changing the velocity vector. This is therefore an acceleration. An acceleration that changes the car from $1m/s$ pointing N to $1m/s$ pointing NE is less than an acceleration that changes the car from $5m/s$ N to $5m/s$ NE. That lesser acceleration in the first case requires a lesser force.

So when the car is moving slowly, the wheel is being asked to provide only a certain amount of force to change the velocity vector.

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  • $\begingroup$ Your edit makes sense in the case when the car is turning. What about when we are moving in a straight line though. If we move at a slower constant velocity that should lead to an decrease in static friction right? But in this case we are not accelerating in any direction, so where does the decrease in static frictional force come from? $\endgroup$
    – 1110101001
    Commented Jun 22, 2015 at 9:31
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    $\begingroup$ If the car is not turning, then the tendency to skid no longer depends on velocity. The force would be be attributable simply to braking or power acceleration. Neither one is easier or harder to do at different speeds (except for the case that the engine can develop less torque at higher speed). $\endgroup$
    – BowlOfRed
    Commented Jun 22, 2015 at 9:36
  • $\begingroup$ I see -- but static friction between the tire and road would still exist even when the car is not turning right? $\endgroup$
    – 1110101001
    Commented Jun 22, 2015 at 9:40
  • $\begingroup$ We consider static friction is zero if there is no net force between the tire and the ground. If your engine is turning the same speed and you don't turn the wheel, then static friction will be arbitrarily low. Imagine an ice sheet. What conditions would keep the car from skidding? Because the ice will supply (almost) no force, it's the same conditions that would let no (frictional) force appear on a normal roadway. $\endgroup$
    – BowlOfRed
    Commented Jun 22, 2015 at 9:58
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Direction of frictional force is perpendicular to the motion. Coefficient of friction cannot exceed certain value. If the centripetal force normal to motion exceeds the permissible value, car will skid. The equation you are using will measure speed for maximum frictional force. You are assuming that it does not change, which is not the case. Coefficient of friction does not remain constant for all speeds. When you decrease the speed, coefficient of friction also decreases. e.g. zero for v = 0.

Consider the situation, force is applied to move the real mass block on the rough surface. It will not move until you apply force more than static friction force which has certain value, let say 20 N. But this does not mean that frictional force is always 20 N, independent of force. Should the block move when you apply force as 5 N ? No, it shouldn't as frictional force will not be 20 N but 5 N.

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  • $\begingroup$ You stated that the force from kinetic friction depends on speed, but according to physics.stackexchange.com/questions/2408/… and physics.stackexchange.com/questions/48534/… kinetic force should not depend on speed so long as you are in motion $\endgroup$
    – 1110101001
    Commented Jun 22, 2015 at 8:59
  • $\begingroup$ I accept it, I wrote it in wrong. Coefficient of friction does not change with speed. What I meant to say was, as speed decreases net frictional force decreases to balance centripetal force. $\endgroup$
    – Mad Max
    Commented Jun 22, 2015 at 12:36
  • $\begingroup$ No, the coefficient of friction does NOT depend on speed (in the first order). It depends on the two surfaces. Static friction is limited by $\mu_sF_N$, but does NOT relate directly to a coefficient as the last sentence of the first paragraph implies. $\endgroup$
    – Bill N
    Commented Jun 22, 2015 at 22:28
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The thing is that you cannot measure friction between the tires of the car and the road and then calculate its speed.There are two kind of friction. The first is the static friction and the other is kinetic friction. the maximum static friction is measured by

$f_s \le f_{smax}\left( = \mu_s mg\right) $

In order for the car not to slip and make the turn,

$f_s \le \mu_s mg $

The friction has the role of a centripetal force so

$f_s = (mv^2)/r$

So: $ (mv^2)/r \le \mu_s mg $

In other words, with this equation you can calculate the maximum speed as you correctly observed. However this equation is used for the calculation of friction when the speed is not maximum, since the friction cannot be measured otherwise.

For example if $ v = U/2 $ where U = Vmax then $f_s = (mU ^2)/4 = (\mu_s mg)/4 $ which be measured.

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  • $\begingroup$ Few clarifications: You said that the maximum static friction is measured by mu_k mg -- I believe you meant maximum kinetic friction, right? As since the car is in motion kinetic friction is what provides the centripetal force. Second, isn't the force from kinetic friction always a constant value? According to physics.stackexchange.com/questions/2408/… the force of kinetic friction remains the same $\endgroup$
    – 1110101001
    Commented Jun 22, 2015 at 8:55
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    $\begingroup$ No. The road and the tire (at least the part touching the road) are not in motion with respect to each other. So static friction is correct here. If the tire were skidding, then you would consider kinetic friction. $\endgroup$
    – BowlOfRed
    Commented Jun 22, 2015 at 9:00
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    $\begingroup$ Your first equation uses $\mu_k$, but is described as measuring maximum static friction. Do you mean to use $\mu_s$ instead? $\endgroup$
    – BowlOfRed
    Commented Jun 22, 2015 at 9:06
  • $\begingroup$ You still mention kinetic friction ($f_k$) in your answer. There is no kinetic friction unless the tires are skidding! $\endgroup$
    – Bill N
    Commented Jun 22, 2015 at 22:31
  • $\begingroup$ I am from Greece i am not familiar with the signs and stuff... $\endgroup$
    – George Smyridis
    Commented Jun 22, 2015 at 23:21
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In order for an object to travel on a curved path of any type (circular or otherwise), it must experience a component of acceleration perpendicular to the path. The magnitude of that sideways acceleration must be, instantaneously, $$a_{\perp}=\frac{v^2}{r},$$ where $r$ is the instantaneous radius of curvature of a circular path, and $v$ is the instantaneous speed.

If you travel on a non-circular curve, the radius of curvature and the center of curvature are changing. For example, Halley's Comet orbits the Sun in an extremely elliptical orbit. At every point (to first order), it experiences a sideways components acceleration (due to the Sun's gravity) which is $\frac{v^2}{r}$ at every moment, but $v$ is not constant, and $r$ is not the distance to the Sun. Both parameters have complicated relationships to where on the ellipse the comet is located.

On the other hand, if you want to travel in a circle, $r$ and the center will be constant. If you want uniform circular motion, then both $r$ and $v$ will be constant.

This sideways acceleration is called a centripetal acceleration because the acceleration vector points toward the instantaneous center of curvature of the path. The formula tells us how much sideways acceleration is required to be on that curved path and how much sideways acceleration we actually have.

Now, let's think about the force. There is no force named the centripetal force. Colloquially (and I believe, unfortunately), a lot of people use that term. ``Centripetal'' merely refers to the direction and/or consequence of the force, not its fundamental interaction. A force which relates to a centripetal acceleration must be a real force such as friction, tension, a spring, gravity, electromagnetic attraction or repulsion, etc.

So, now, Newton's 2nd Law says for the sideways component $$F_{\perp}=ma_{\perp} = m\frac{v^2}{r}.$$

For an object to travel at speed $v$ on a curved path of radius $r$, there must a a component of some real force acting sideways on the object. The magnitude of that force component must (and will) be numerically equal to $m\frac{v^2}{r}$. If it's not, then the path must change.

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Clearly by logic if we will go lower than the velocity obtained we can make an easy turn.....So we need to find the max velocity ..

If u will increase Ur velocity by this max velocity then the two tires will lift from the ground and then Ur car will get damage.

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  • $\begingroup$ No, I got that but the equation $\mu_k mg = m \frac{v^2}{r}$ should still hold when you go at less that maximum velocity right? But if that is the case when you plug in values you find that they are not equal. Or essentially, my question is that while it is "clear by logic" that going lower than the maximum velocity allows you to make the turn, how would you show that using the above equations as per newton's second law? $\endgroup$
    – 1110101001
    Commented Jun 22, 2015 at 7:33

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