Force exerted on ground from a slipping wheel

Question

I am trying to do some car calculations for a game.

I have a tyre (radius $r$, inertia $I$) rotating around it's axle with an angular velocity $\omega$ faster than the center of mass is moving across the surface. So the tyre is not rolling along the ground, but slipping. The surface has a friction coefficient $\mu$. The tyre is pressed down by the car with a weight $N$

I need to calculate two things.

1. The force the tyre exerts on the ground.
2. The resulting (de)acceleration of the tyre.

Imagine a car doing a wheelspin. I need to find the force the excess speed of the tyre transfers to the ground, giving an acceleration to the car, and thereby also de deacceleration of the tyre.

I have calculations for the external forces (engine, brakes, drag, lateral slipping) and this is the final piece to complete the puzzle. So help is much appreciated, as i have been trying to solve this in a myriad of ways, with no success.

//Edited - It is the wheels connected to the enigne i am trying to calculate. However i am in this case not interested in the torque coming from the engine, only actions from the extra energy stored in wheel from building up the spin.

The engine provides a torque $\tau$ on the tyre

Answer 1

To answer the first part of your question, I'm going to assume that the center of mass of the car is at the center of the car itself. This is usually not true of real cars - the engine block is quite heavy and is usually positioned well forward of the center of the car. In the absence of any information about how heavy the engine is and where it is, I'm going to assume a symmetrical car.

The force that the tyre exerts on the ground is equal to the weight of the tyre itself plus one-fourth the weight of the car. Any other force that the tyre might exert on the ground would result in a car that jumps up off the ground or digs into the ground, which is clearly not the case.

There is not enough information to answer the second part of your questions. Once you break the rolling-without-slipping assumption, we need to have some information about the torque on the tyre.

To clarify why we need the torque on the tyre, consider this: for all we know the tyre has a torque of say one billion N-m on it, and will begin accelerating much faster than the speed of sound or anything else. Once the rolling-without-slipping condition is discarded, no assumptions can be made about the rotational rate of the tyre.

Answer 2

Assumptions : 1. car is moving straight in the entire time period.

2. Centre of mass of the car is at centre, so that the reaction force on each wheel is $\frac{N}{4}$

Clarification : The excess speed of wheel does not give any acceleration to the vehicle. the wheel would just continue spinning. The forward frictional force would slowly reduce the speed of the wheel so that finally condition for rolling is achieved. From then on, the vehicle starts moving

Answer :

force tyre exerts on the ground :

Tyre exerts two forces. One tangential frictional force and one weight of the car. Weight of car on each wheel is $\frac{N}{4}$ by assumption. While slipping, the frictional. force is the maximum possible, ie : $\mu\frac{N}{4}$

deceleration of tyre

Since the wheels are slipping, the speed or torque generated on the wheels does not matter. The vehicle would decelerate uniformly, under the frictional force from all 4 wheels. The net backward force on vehicle is $4$ x $\mu\frac{N}{4} = \mu N$ deceleration : $ a = \frac{(\mu N)}{M}$ If the vehicle's current speed is $v_{cm}$, and after some time it reduced to $v'$, $v'= v_{cm}-a.t$, then as long as $v'$ is less than $R\omega$ the tyre continues to slip and after that the car starts rolling.

after rolling starts, deceleration of the car would be the deceleration of the wheels(due to less throttle) and proportional to $\frac{d\omega}{dt}$ (which is off topic)

Removing assumptions:

Assumption 1 is fair enough.. slight deviations occur during start due to angular momentum of spinning wheels

Assumption 2 can be removed (but it does not matter much). Let $x_f$ and $x_b$ be the distance of front and rear wheels from centre of mass. Then balancing torque about centre of mass or by simple intuition,

$N_f = Mg(\frac{x_b}{x_b + x_f})$ (ie: normal reaction is inversely propotional to distance from centre of mass). (Here $N_f and N_b are total forward and backward reactions... by symmetry, reaction on wheels would be half of it since there are two wheels in front)

This does not change the net frictional force, but since the wheels closer to centre of mass have more load and hence more friction, they slow down faster and more quickly reach rolling condition. Thus, the vehicle starts to roll in less time than a vehicle with com at centre. Usually it is the rear wheels which starts rolling first.