Force applied to wheel in pure rolling motion at contact point with road

Suppose a wheel with radius $R$ is resting on a non-inclined surface. A torque $\tau$ is applied to the wheel center. In an attempt to prevent wheel from spinning, the ground applies a static friction force to the wheel at the contact point (parallel to the surface), then the wheel starts rolling without spinning. The same friction force also acts as a torque on the wheel around its axis. This scenario is depicted below:

I'm trying to find the magnitude of the force the ground applies to the wheel - that is, the force which drives the wheel forward (which should be the same in magnitude as the force the wheel applies to the ground at the contact point).

That's how I'm doing it:

The relation between linear acceleration $a$ and angular acceleration $\alpha$ for a pure rolling movement is given by $$\tag{1} a = \alpha \centerdot R,$$

The relation between torque $\tau$ and angular acceleration $\alpha$ is

$$\tau = I \centerdot \alpha$$

where $I$ is the moment of inertia of the wheel around its axis.

The relation between torque $\tau$, force $f$ and lever arm $R$ is:

$$\tau = f \centerdot R$$

Being the engine torque $\tau_e$, the friction force $f$, the counter torque due to friction force $\tau_f$ and the moment of inertia of the wheel $I$ around its axis given by $\frac{1}{2}mR^2$:

The linear acceleration of the wheel is due to the friction force only:

$$f = ma$$ $$a = \frac{f}{m}$$

This is the counter torque the ground applies on the wheel's edge (negative, because it pointing in the opposite direction of $\tau_e$):

$$\tau_f = -f \centerdot R$$

The net torque causes angular acceleration on the wheel:

$$\tau = \tau_e + \tau_f$$ $$\tau = \tau_e - f \centerdot R$$ $$\tau = I \centerdot \alpha$$ $$\alpha = \frac{\tau}{I}$$ $$\alpha = \frac{\tau_e - f \centerdot R}{I}$$ $$\alpha = \frac{\tau_e - f \centerdot R}{\frac{1}{2}mR^2}$$

Substituting $\alpha$ and $a$ in $(1)$ gives:

$$\frac{f}{m} = \frac{\tau_e - f \centerdot R}{\frac{1}{2}mR^2}R$$

Rearranging:

$$f = \frac{2}{3}\frac{\tau_e}{R}$$

And that is the force $f$ from static friction which pulls the wheel forwards without making it spin or slip, and consequently, the force the wheel applies to the road surface at the contact point.

And it says: "The torque on the rear axle can be converted to a force of the wheel on the road surface by dividing by the wheel radius. (Force is torque divided by distance)."

That statement doesn't match the approach I used above. If the force of the wheel on the ground was simply engine torque divided by radius (negative, since it pointing in the opposing direction):

$$f = -\frac{\tau_e}{R}$$

then the counter torque applied to the wheel would be

$$\tau_f = -f \centerdot R$$

that implies that $\tau_f = -\tau_e$.

That means the net torque would be zero, and the wheel would just slip without rotating at all.

Am I doing something wrong on my calculations?

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a) It is very hard to apply a pure torque, and b) there is no friction unless there is normal force though the contact. –  ja72 May 5 '14 at 16:37

The trick here is that at the contact point the velocity of the wheel relative to the ground (with no slip) is just zero. Furthermore the acceleration is centripetal so the forces tangent to the wheel are zero.. So the frictional and torque forces must sum to zero. And therefore the ground force is just the opposite sign and the same magnitude as the force (torque/radius) exerted through the wheel by the engine.

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Thanks! But - The zero velocity of the contact point is in fact the instantaneous velocity, but even this point has non zero angular velocity, and non zero angular acceleration in this case - otherwise the wheel would not be rotating nor accelerating its rotation. Then, in order to have angular acceleration, there must be a non-zero net torque acting on this point. Furthermore, an engine exerting clockwise torque on the wheel and a ground force acting on the edge of the wheel (causing a torque of same intensity, but counter-clockwise) should cancel each other. Agree? –  ri_ri Dec 28 '13 at 1:46
And, assuming your line of thought, can you point me which step in the derivation I did something that led to a different ground force magnitude? –  ri_ri Dec 28 '13 at 2:02
Torque (and force) are vector quantities. Notice that I am resolving into horizontal and vertical forces. Since the acceleration is entirely vertical (a condition for rotation) the force from the wheel center is vertical. I am claiming no net horizontal force on the contact point. There is a net force that is accelerating that point in an upward direction. An instant later, it will lift up, away from the ground. Note also that I am focussing on acceleration and NOT on either velocity or angular velocity since they are not directly related to force. –  DWin Dec 28 '13 at 2:23
Sorry for the insistence - I have no problems in understanding that the velocity at the point of contact is zero, and that an centripetal acceleration maintains the rotational motion of every point on the wheel. But I'm still not able to see how it solves the contradiction I presented - if the magnitude of the ground force is (torque/radius). Once there is a CW torque T from engine, and a CCW torque Tf (equals -T) that the ground delivers to the wheel, then what would be the net torque acting on the whole wheel? To me it seems clear that it would be zero, but this means no rotation. –  ri_ri Dec 28 '13 at 15:05
If you are looking for other points to think about then look at central point of the wheel and the points at the edge of the wheel at the same height as the axis and also the point opposite the contact point. Then you can think about integrating over the entire surface if you want "the net torque acting on the whole wheel". (That's not a quantity that I know exactly how to deploy. I would have thought that the no-slip condition imposed a constraint on the system dynamics.) –  DWin Dec 28 '13 at 17:47

I think your analysis is good and correct. The statement you've quoted doesn't contradict what you've done, it probably suggests a different way to look at the problem.

I think the statement simply states a method in mechanics- transferring a torque to another point (coordinate axes). In this case, from the centre of the wheel to the point of contact.

Mathematically, a torque can be replaced by a pair of forces, with one force at the centre and the other a point $P$ at a tangential distance of $r$. The magnitude of the force would be:

$F = \tau_e / r$

This $F$ is not the friction force.

The "counter torque" term doesn't make sense anymore since we have replaced the torque by a pair of forces. You can instead say the "torque" applied by the ground would be:

$\tau = F \cdot r - f \cdot r$

which is

= $\tau_e -f \cdot r$

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The problem is in assuming the force is forwards at a contact point right underneath the axis of the wheel, together with the no-slip condition. This leads to zero friction force, since their is no friction when their is no slipping.

A solution can only be found when deformation of the wheel is taken into account, allowing the force to act on a different position of the wheel. See for example: http://www.phy.davidson.edu/fachome/dmb/PY430/Friction/rolling.html

A similar problem is when the wheel is rolling and is decelerated by "friction" with the surface. When you make these same assumptions, the force must be zero, since it would otherwise either make the wheel spin faster while slowing down its forward movement, or make the wheel turn slower and accelerate its forward movement. Neither is possible.

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