Why does the friction force on a smoothly rolling wheel cause the centre of mass to accelerate?

Question

Suppose you have a wheel of mass $m$ rolling down an incline. Then the component of the weight along the incline acts on the centre of mass of the wheel, causing it to accelerate. The (static) friction force points up the incline. This friction force acts on the point of the wheel that is in contact with the incline and therefore a torque causes an angular acceleration, i.e. the wheel rotates while rolling down.

I have read that this friction force acts on the centre of mass of the wheel, too, causing the centre of mass of the wheel to accelerate more slowly than a block on a frictionless incline.

I don't understand how the friction force is transmitted from the bottom of the wheel to the centre of mass. I can only see that it acts as a torque on the bottom of the wheel, thus causing rotation. (I mean, a force cannot act on two different points at the same time, right?)

Answer 1

I don't understand how the friction force is transmitted from the bottom of the wheel to the centre of mass.

It's because the wheel is rigid and forces at any point on the object are distributed through to other portions of the object.

For linear momentum problems on rigid bodies, $F_{net} = ma$. The location where the force is applied does not matter.

The force does not act at two different points, but it can certainly have two different effects (changing the linear momentum of the object and simultaneously applying a torque to the object).

Answer 2

As you pointed out, the difference between the rolling wheel and one that slides, is the wheel’s angular acceleration. An object’s angular acceleration is related to its moment of inertia by the following formula:

$$\alpha = \frac{\tau}{I}$$

Where $\alpha$ is angular acceleration, $\tau$ is the torque applied to the object and $I$ is the object’s moment of inertia. For a solid wheel, the moment of inertia is given as:

$$I = \frac 12 mr^2$$

A bigger wheel (larger radius) will accelerate slower than a smaller wheel of the same mass and a lighter wheel will accelerate slower than a heavier wheel with the same radius. The location of the wheel’s mass is also important when determining the wheel’s acceleration. A “hoop” in which all the mass is concentrated around the perimeter will have a lower acceleration than a solid wheel of the same mass and radius:

$$I = mr^2$$

If you can imagine swinging a baseball bat with all of the weight at the far end vs one with the same mass that has the weight concentrated near your hands you can understand how the distribution of mass makes a difference.

The mass of the rolling wheel must be accelerated both linearly and angularly. The mass of the wheel that is sliding is accelerated by the same force but only linearly.

Answer 3

You are right on the point that friction cannot act on both centre of mass and point of contact at the same time.If we try to solve the problem by using newton's second law at the point of application of friction, the problem would turn out to be very complicated.

So we assume the mass to be concentrated at the centre of mass ,then apply force equation by assuming forces are acting at the centre of mass. By force equations I mean "F = ma" .(not τ=Iα).Note!but for torque we can't assume the force to be acting on the centre of mass,we should take into account the actual point of application.Otherwise it would make no sense.(Why should the body rotate if the force always acts on the centre of mass?).

It's our own way of simplifying the problem...

For example,you may have seen problems like the one below ,a lot of times.The tension due to the string is acting on the edge of the block ,but we assume the force to be acting on the center of mass of the block.It's the same analogy.

This way of solving problems can be used only for rigid bodies.For no-rigid bodies mass is not always concentrated at center of mass.