In physics textbooks, problems about rolling usually include the assumption that the rolling is without slipping, meaning that $v=R\omega$ throughout the motion, where $v$ is the translational velocity, $\omega$ is the angular velocity and $R$ is the radius. I do not understand the justification of this assumption. What is the mechanism that assures the equality $v=R\omega$, and under what conditions does it work?

Here is what I know. The velocity of the point of contact relative to the surface is $v-R\omega$ (here I assume that $v>0$ when the wheel moves to the right and $\omega>0$ when it rotates clockwise). Therefore,

  • If $v-R\omega>0$, then the point of contact slides right and then kinetic friction acts to the left, decreasing $v$ while increasing $\omega$.
  • If $v-R\omega<0$, similarly we have that $v$ increases while $\omega$ decreases.

This hints towards the intuition that a wheel rolling freely on a surface will tend to converge towards rolling without slipping. If the initial conditions do not match that of rolling without slipping, will we see the motion naturally converge to such? If so, how would that happen? Is it true? How does it look like?

More concretely, suppose that at time $t=0$ we release a wheel on a horizontal surface with velocities $v_0,\omega_0$. Assume that the friction coefficient is $\mu$. Can we give a quantitative description of $v(t),\omega(t)$?

  • 2
    $\begingroup$ It is a little clearer to visualize if you think of tank treads. They have a long flat surface on the bottom that doesn't slip. They have a long flat surface on top that moves at twice the tank's speed. $\endgroup$
    – mmesser314
    Commented Mar 25 at 23:48

3 Answers 3


"What is the mechanism that assures the equality v=Rω, and under what conditions does it work?"

The condition when the friction force at the interface between the wheel's edge and the ground never exceeds the the (maximum) static friction. Because if it does, that's when you get slip.

When you get no slip, that means the point of contact between the wheel and the ground has no relative motion. When that happens, if the ground is your frame of reference, the entire wheel rotates about that point. It behaves similar to an object that is continuously falling forward (on flat horizontal ground, anyways). That's where this diagram, which you have probably seen before comes from:

enter image description here


If the point of contact between the wheel and the ground cannot slide, it means that every point on the circumference of the wheel can be uniquely mapped to a point on the ground over a single rotation.

When deriving what a radian is from a unit circle (or any arbitrary circle, really) you come by the arc length:


Since you can map the circumference of the wheel over a single rotation to the ground you can use that to determine the distance traveled. If you differentiate with respect to time then you can get the speed:


which is the speed.

The reason passive rolling tends towards no slip is because the system is tending towards a lower energy state (or, more generally, least action as things in the universe seem to inexplicably do) with the the lowest energy state being the ultimate no-slip condition: zero speed. Slipping involves dynamic friction which causes energy loss while static friction does not. That means slipping will continue to dump energy until the system falls into an energy state low enough to have no-slip. From this, we can retro-actively conclude that slipping is a higher energy state than no slipping which may have been intuitively obvious, but rather difficult to explain beyond "something passively spinning really fast has more energy while also being much more likely to slip than something passively spinning slowly".

For active rolling there is no rule that says the system needs to tend towards no slip since we have continuous energy input that can offset the tendency to fall into an energy state low enough to have no slip. However, if something is actively rolling, it probably means a human is doing it and we strive to operate in no-slip conditions since slippage is inefficient if your goal is to move and therefore not desirable.

  • 1
    $\begingroup$ Thank you for the answer. I understand the mathematical derivation, my question is about the physical setting, which you address in the first part of your answer. Why is the friction force naturally "tuned" in such a way that the point of contact always has velocity 0? $\endgroup$
    – 35T41
    Commented Mar 25 at 23:48
  • $\begingroup$ @35T41 You might want to ask another question because what you're really asking doesn't have anything specifically to do with wheels. Asking about wheels just distracts from it. What you're really asking is about the mechanism that causes static friction to behave the way it does: why there is a force threshold known as maximum static friction where friction forces below that level have no sliding, but above that level have sliding when static friction give way to dynamic friction. That's not a question I can answer. $\endgroup$
    – DKNguyen
    Commented Mar 25 at 23:51
  • 1
    $\begingroup$ I can accept that this how friction works; the question is then why would we expect wheels to roll in such a way that it is static friction, not dynamic friction, that acts at the point of contact. That is, how slipping is prevented. I also tried to formulate a more well-defined question: if the initial conditions do not match that of rolling without slipping, will we see the motion naturally converge to such? If so, how would that happen? I believe an answer to this question will help me understand the situation better. $\endgroup$
    – 35T41
    Commented Mar 26 at 0:39
  • $\begingroup$ @35T41 Now that I can answer. I added the answer to the last part of my answer. $\endgroup$
    – DKNguyen
    Commented Mar 26 at 1:39

When an aircraft comes into land it wheels are essentially moving horizontally but have no rotation. When the wheels make contact with the runway there is obviously a torque acting to rapidly spin the wheels up to the required angular velocity and this can heard as a screeching noise while the wheels spin up.

A camera mounted on the aircraft would see the axles of the wheels as stationary and the runway moving horizontally. As soon as the wheel makes contact with the runway, there is a frictional force acting on the point of contact of $$F_f = \mu_k F_N \quad (Eq1)$$, where $\mu_k$ is the coefficient of kinetic friction, and $F_N$ is the force normal to the runway equal to $ m g$, where m is the mass of the aircraft and g is the acceleration of gravity. If the radius of the wheel is r then the torque on the wheel is $$\tau = \mu_k F_N r \quad (Eq 2)$$ and this will accelerate the rotation of the wheel until the wheel reaches an angular velocity such that the contact point is not moving relative to the runway. At this point the kinetic friction goes to zero, but there are now other frictional forces such as rolling friction and static friction that are still have to be allowed for.

Now if the aircraft engages reverse thrust to slow down the aircraft the wheel is momentarily rotating too fast relative to the runway, then there is a reverse torque due to the reappearance of kinetic friction that slows down the angular velocity of the wheel so that it once more has the correct angular velocity for rolling without slipping. In other words whenever the the wheel is slipping due to spinning too slow or too fast, there is a restoring torque that ensures the angular velocity $\omega$ of the wheel stabilises at $v/r$. This is providing there is sufficient kinetic friction to provide the restoring torque. For example if a car is braking hard on an icy road, the coefficient of friction of rubber on ice may be too low to restore the required matching angular velocity of the wheel for rolling motion and the car continues to slide.

More concretely, suppose that at time t=0 we release a wheel on a horizontal surface with velocities $v_0$, $ω_0$. Assume that the friction coefficient is μ. Can we give a quantitative description of v(t),ω(t)?

Edit: Originally I said the energy losses due to heat generated by the friction would be difficult to quantify, but the comments by Neil_UK have now convinced me (see addendum at the end of this answer) that the energy lost as heat etc during the spin up is exactly 50% of the input energy and I will label this factor as k =1/2.

If the initial angular velocity of the wheel is $\omega_0$ and the initial linear velocity is $v_0 =0$, then the initial stored angular kinetic energy in the wheel is $\frac12 I \omega_0^2$ where I is the moment of inertia of the wheel. If k = 1/2 of this initial energy is converted to the final linear and angular kinetic energy of the wheel, then the total final energy is: $$ k \left(\frac12 m v_f^2 + \frac12 I \omega_f^2 \right) = \frac 14 \left(v_f\left( m + \frac {I}{r^2} \right) \right) \quad (Eq 3)$$

To express the final velocity as a function of time, we can use the standard equation of motion $$v_f = at \quad (Eq 4)$$ where a is the horizontal acceleration. Using the formula for kinetic friction (Eq 1), we can restate this as $$v_f = \left( \frac {\mu_k F_N}{m} \right) t = \mu_k \ g \ t \quad (Eq 5)$$

ADDENDUM: Mechanical analogue of heat energy losses in an electrical RC circuit

Consider an initially stationary solid wooden block placed on a long belt sander, where the belt velocity is v. The block will accelerate due to the kinetic friction between the block and the belt. The final velocity of the block is equal to the velocity of the belt and kinetic energy it acquires is simply $1/2 mv^2$. If we now suddenly stop the belt, the block will continue to move due to its momentum but it will slow down due to the friction with the belt and eventually come to a stop. The kinetic energy it acquired has now all been dispersed as heat. The time and distance for the block to come to a stop due to friction is exactly the same as the same as the time and distance it took to accelerate the block using friction. This means by symmetry that $1/2 mv^2$ of energy was just wasted as heat during the acceleration and another $1/2 mv^2$ of energy was transferred to the block during the initial acceleration, so exactly 50% of the input energy was lost as heat/noise etc.

Kinetic friction is independent of the relative velocities of the objects and so the force and acceleration is constant, so that the time taken to accelerate or decelerate or the distance travelled to reach terminal velocity or stop can be found using the regular equations of motion.

  • $\begingroup$ I'm not sure sure you have to estimate the amount of energy lost spinning up a wheel. Like charging a capacitor through a resistor, isn't it always exactly half (making other assumptions like 'tearing rubber off the tyre' doesn't change its mass significantly). $\endgroup$
    – Neil_UK
    Commented Mar 26 at 7:29
  • $\begingroup$ You've missed my point, it's as simple as conservation laws, which are exactly the same for mechanics and RC's in electricity. Consider some massive things, like the runway and aircraft, and a small wheel, initially at rest under the aircraft. Once landed, and the wheel has spun up to aircraft speed, its energy is 0.5Iw2, its momentum is Iw. In transferring that momentum to the wheel, the runway exerted a total impulse of Iw, taking Iw2 from the plane's kinetic energy. The difference is exactly 50%, we can ignore the actual loss mechanism as mere detail, heat, sound, whatever. $\endgroup$
    – Neil_UK
    Commented Mar 26 at 11:08
  • $\begingroup$ Related: aviation.stackexchange.com/questions/3702/… $\endgroup$ Commented Mar 26 at 12:35
  • $\begingroup$ @Neil_UK I have given your comments some more thought and I have now convinced myself you are correct and edited my answer correspondingly. $\endgroup$
    – KDP
    Commented Apr 2 at 21:24

Imagine a rolling wheel. Rolling without slipping can occurs when the velocity $v$ is constant and the wheel of radius $r$ has exactly the correct angular velocity $\omega=v/r$. The mechanism is that the velocity at the contact point is zero.

  • $\begingroup$ Yes, there is a lot of physical hooplah surrounding rolling without slipping, but it all really boils down to meeting the non-holonomic constraint. $\endgroup$ Commented Apr 3 at 13:00

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