# Which force makes a wheel roll down the hill? What causes friction?

A wheel rolling down a hill has two axis of rotation. One is where the center or mass is and the other is the point of contact with the surface which acts as a fulcrum. I was trying to understand how this happens, how it rotates down the hill. What causes friction and torque? Please mind me over this simple issue, but I just want to further understand how things work. My ideas are as follow:

• F1, the component of gravity, pulls the wheel from the center of mass (CM). This pull will create friction f. Friction then creates a torque about its center of mass. The object is at the same time rotating around the point where it touches the surface, like a fulcrum, the torque is created with lever-arm L and force W.

I don't know if this is right or not. Maybe one should not consider the component of gravity at all acting on the wheel. If so, what is rotation originated? Is it because the position of the CM? Maybe someone can explain better.

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if gravity wasn't involved, the wheel wouldn't even be on the hill to roll down, it would be floating ;) and it will only rotate when someone/something rotates it (an external push or force)! – rps Oct 7 '13 at 6:17
Gravity is the main cause indeed. But, how exactly? Is it because it makes the wheel rotate around the support point, or is it because of the component of gravity that makes the wheel to try to slide and then creates friction, then torque? – user990692 Oct 7 '13 at 12:39

First things first, Torque is always relative to a given point. You can calculate the torque about the centre of mass or about the point of contact.
Next, Any object, moving howsoever, can be analysed in one definite way, by considering the pure translational motion of the Centre of mass and the pure rotation of the body about the centre of mass. (This is true because in the COM reference frame, the pseudo-forces due to the rotating create torques which are balanced about the axis passing through centre of mass.) Therefore, you can treat the rolling wheel as a translation of the centre of mass +the rotation about the centre of mass. There is another way to treat a moving body. It can be treated as a pure rotational motion about an axis called Instantaneous Axis Of Rotation(IAOR). In your case of the rolling wheel, the IAOR passes through the point of contact of the wheel with the incline. Moreover, the $\omega$ for rotation about IAOR is the same as the rotation about COM in your case.

Let us use the first method to analyse your case. The forces acting are:-
1)$F_g$ which acts through the centre of mass and hence can provide no torque for the rotation.
2)$N$ or the normal reaction which again passes through COM and hence produces no torque.
3)The only force that can provide any torque is the friction $f$.
Now, all of these can accelerate the centre of mass. But in the direction perpendicular to the incline, all forces are balanced and there is no acceleration. Along the incline, the translational acceleration is $a=\frac{F_g\sin\theta-f}{m}$. To find the friction, remember, friction tries to prevent relative motion. It can do so by making the rolling a pure rolling. Here the point of contact is (momentarily) stationary and hence no relative motion. for pure rolling, $v=\omega R$ and $a=\alpha R$ must hold true. From this $\alpha=\frac{fR}{I}$, where $I$ is the moment of inertia about COM axis, $R$ the radius and $\alpha$ the angular acceleration. This determines all the variables with $I$.

Using IAOR in your analysis, the torques change but the physical quantities $a$ and $speed$ remain the same. You can try that out for yourself.

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Thanks Mr. Pasani. Just one more thing. You said that friction prevents relative motion, right. Does this mean that friction acts in the same direction of the wheel's motion or opposite like in my drawing? I think, Since there's no sliding, it must be same direction as motion then. – user990692 Oct 7 '13 at 20:08
@user990692 The direction depends on the conditions. But in all cases, friction always tries to induce pure rolling. For example, you provide a force to the wheel in such a way, that the torque of that force is insufficient to cause enough angular acceleration for pure rolling, the friction will act in such a way so as to increase the angular acceleration and decrease translational acceleration, to bring about pure rolling. If the torque of the applied force produces a $\alpha$ greater than what is required by pure rolling, the friction tries to reduce $\alpha$ or increase $a$. – Satwik Pasani Oct 8 '13 at 1:29
@user990692 Therefore, the direction is not only dependent on how much and where you apply the force but on the object itself and its moment of inertia about the COM. – Satwik Pasani Oct 8 '13 at 1:32

See, the body rolling down the hill is due to the friction acting in it. In a simple way the necessary torque is required to rolling the body is provided by friction force. If there would no friction the body will have only translation motion.

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To answer your question as simply as possible: the torque comes from the force of gravity. For rolling to occur, the force of gravity must not be able to overcome the static friction at the point of contact with the ground, otherwise sliding occurs. Thus, we can say the contact point is a fixed pivot. Then the force due to gravity (including the normal force) acting through the CoM of the object becomes a force perpendicular to the normal axis and offset from the pivot. Because the pivot is free to rotate, it does. This causes a new point to be the contact point, wash, rinse and repeat.

Let me stress in contradiction to the previous answer that the force of gravity through the CoM can and does provide torque for the rotation because the rotation is centred at the contact point with the ground, not at the CoM.

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Thank for this very short and easy explanation. I vote on the other question just because it provides a bit more information. – user990692 Oct 7 '13 at 20:03
@JimtheEnchanter I forgot to mention, but my arguments about the torques are true only in the COM frame of reference. – Satwik Pasani May 20 '15 at 13:46