# Physics behind Wheel Slipping

Lets say that I'm in a car and I apply full acceleration suddenly. Now, the wheels would slip and hence the car doesn't displace much.

But If I start with some constant acceleration, slipping doesn't appear and the car moves normally. I think that its related to some friction mechanism.

But I don't understand why the wheel slips at high speeds and not at low speeds. It's like, when the speed is high, rules are changing.

Also, In each step F(s) (friction) should be equal to F (force in other direction). Isn't it? Any physical explanations?

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The law of inertia. – Sponge Bob Oct 2 '12 at 3:58
when you say "high speeds" and "low speeds" do you mean of the vehicle, or of the wheels? In videos where the cars spin their tires fro a second or two before taking off they are applying the brakes to the unpowered wheels to hold them in place. Otherwise the friction is usually only slightly reduced by slipping (due to the difference in static vs. kinetic friction) and the car should displace at only a slightly lower rate than usual. – Rick Jul 13 '15 at 15:12

It's hard to make the wheels spin at high speeds because you're in a higher gear, so the torque at the wheels is less. So I assume you are only asking about wheel spin in first gear i.e. it's quite easy to spin the wheels when pulling away in first gear but much harder if e.g. you're travelling at 10 mph in first gear.

The reason is that if you're stationary and drop the clutch the angular momentum of the engine contributes to the torque. That is, the torque at the wheels is the torque from the engine plus the torque from angular momentum stored in the flywheel, crankshaft etc. This happens because the engine is spinning faster that it would if the clutch were engaged, so engaging the clutch slows the engine speed. The extra torque is given by:

$$\tau = I\frac{d\omega}{dt}$$

where $I$ is the moment of inertia of the spinning bits of the engine and $\omega$ is the engine speed, so $d\omega/dt$ is the rate of change of engine speed. If you drop the clutch the engine speed changes rapidly so $d\omega/dt$ is large and the extra torque is large. If you ease the clutch out $d\omega/dt$ is small so the extra torque is small and the wheels won't spin.

When you're driving at (e.g.) a steady 10 mph the engine speed matches the wheel speed, so if you now suddenly stamp on the accelerator it's only the torque from the engine that's available to spin the wheels. You don't get the contribution from $d\omega/dt$.

To see this try driving at 5 mph, then disengage the clutch, rev the engine and drop the clutch. As the clutch bites the wheels will spin just as they do when the car is stationary.

It's worth noting that a powerful car can spin the wheels in first gear even without playing with the clutch. In fact an old sports car I had many years ago would spin the wheels in second gear in the dry and in third gear if the road was wet!

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In your answer you state why the tires don't slip when the car is moving quickly , but that's the opposite of the OP's claim: the wheel slips at high speeds and not at low speeds – Rick Jul 13 '15 at 17:48

Traction (The friction between a moving body relative to the surface) plays an important role here, 'cause one or more tires in the car lose traction and lead to Wheel Spin (i.e.) The car remains slipping until it attains some stable Traction. This is best explained via Starting Tractive Effort. It is an important factor that is given some higher priority in Railway Engineering. They use "Locomotive Wheel-slip" instead of our "Wheel-spin"..!

This is because the weight of a car is too much to pull immediately at a given period of time (i.e.) Power-to-weight ratio should also be noticed. But, it's less for vehicles and great for Locomotives and it's calculated using the Curb Weight of vehicles.

Surface conditions: This slipping is more common in winters 'cause the coefficient of friction is too low for lubricants like water, oil grease, mud, etc. Hence, the colder water in between the road and tires prevent them from sticking to the roads. In a more specific manner, the differentials provide enough torque for the wheels to spin. Similar thing is applicable to the puck in Ice Hockey..!

Inertia of motion also plays here, 'cause the Inertia of the engine and the regulator wheel (Flywheel) is at a higher RPM than the gear that tries to bring the shaft of transmission of the heavy vehicle to the same velocity, starting at rest..! (which makes the situation more complicated...)

You could see this in most common Drag races and it's called a Burnout where those racers release the clutch and accelerate while holding the brakes. They even use reserved wet tracks as Burnout boxes for proving their freestyle. But, the only difference is that those guys are doing it in a purpose, while here - It happens when you have no experience regarding this..!

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The car's engine can only control the angular acceleration provided to the wheels. The more you press on gas, the more is the angular acceleration of the wheels.

It is friction's responsibility to convert the angular acceleration of the wheels into linear acceleration of the car.

Now consider a wheel of radius $r$, rotating with an angular acceleration $\alpha$. If there is no slippage, that means the wheel is moving forward with a linear acceleration of $\alpha r$. And if the car's mass is $m$, that would mean there must be friction of amount $m\alpha r$ acting on the car. However, if this quantity is larger than what the ground can support, the wheels will slip.

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The whole question of myne is shrinked to your last ( unexcplaind) sentence if this quantity is larger than what the ground can support – Royi Namir Sep 30 '12 at 15:33
Your answer is actually explaining OP's question... You just explained how the vehicle moves..! – Waffle's Crazy Peanut Sep 30 '12 at 17:39
I'm sorry, perhaps I misunderstood the question. OP seems to be confused about why the rules change with speed. I just pointed out that the rules are the same for all speeds. The ground does not really behave differently when the car is moving fast. It's just that it is not able to provide the amount of friction needed to support a high acceleration. Anyway, I apologise if this is not what OP was asking. – Vinayak Pathak Oct 1 '12 at 0:47

Slipping happens when the force applied to move the car is larger than the friction between the wheel and the ground can resist. Friction is needed to stop the wheel from slipping.

There is a difference between static and kinetic friction. As long as the wheels are not slipping, the contact point is stationary with respect to the ground and static friction applies. Once the wheels slip, the contact point actually moves on the ground, i.e. kinetic friction. While the contact point is stationary, the applied force exactly matches the friction force - up to the point where the force is greater than the maximum friction possible between tyre and ground.

Kinetic friction is lower than static friction so, once the force applied is sufficient to overcome the friction, the required force drops, which makes it even easier for the wheels to spin. At that point you need to reduce the force significantly to stop the slipping.

If you reduce the friction(e.g. on ice) slipping is much easier but the same rules still apply. The same rules also apply in the reverse situation, when you are braking. If you apply too much force the wheels will lock up, and you're into kinetic friction. At that point you need to release the brakes and try again. This is what ABS systems do automatically - hence the "rattling" when you brake hard.