# Distinction between torque and force

Distinction between Torque and Force.

Consider a case when a cylinder is rolling down a rough inclined plane. While analysing its motion, we consider the force applied by the friction and the component of $mg$ down the slope, for finding the translation of the center of mass. We again consider the torque of friction for calculating the rotation.

Isn't it like double counting the force of friction? Why? After all they are same force, only different aspects of it.

• Torque is the force times the lever arm ($\tau=F\times l$), and it is measured in Newton-meters, not Newtons. – HDE 226868 Sep 28 '14 at 14:34
• Yeah I understand, but aren't we considering the same force twice, by counting it once for the motion of CM, and once again for the rotation of the body about the CM? – Mriganka Basu Roy Chowdhury Sep 28 '14 at 14:36
• On an inclined plane (unrelated example), gravity can be used to calculate how Earth attracts an object, but we also take it into consideration when calculating the friction opposing that object (because gravity is related in this case to normal force, and friction is defined as $f_k=\mu_kF_N$). As you put it, we "double-count" gravity there, but it still works. – HDE 226868 Sep 28 '14 at 14:39
• I think of it like, the friction force depends on the normal force. So the more you increase the normal force, the friction increases, but the normal force doesn't disappear. It is defined as $\mu N$. But in this case, we make friction responsible for two different motions altogether. – Mriganka Basu Roy Chowdhury Sep 28 '14 at 14:41

## 3 Answers

I suppose you could try to think of the problem in another way.

What would happen to a cylinder (or sphere) if you put it on a frictionless inclined plane? Would it still roll or just slide?

The imbalance in forces acting on the cylinder at different points, with respect to its center of mass, are what lead to the rotation. Gravity acts on the center of mass while friction is acting at some distance, R, from the center of mass.

Here is another fun thought experiment. Suppose you have a cylinder of radius R and length L. Now at its center of mass (i.e., at L/2) imagine it is incredibly thin (i.e., r $\ll$ R) for a very small length (i.e., $\Delta$l $\ll$ L). Put this cylinder on a thin rod of the same diameter as $\Delta$l.

Now, would the cylinder roll or slide?

Let's cheat and imagine r to be infinitesimally thin so as to be effectively at the objects center of mass. Then ask the above question to yourself again.

I would try to explain very briefly.

When you apply a force onto a body, it causes the body to move in the line of application of force. This introduces translational motion into the body.

When you apply a torque onto a body, it causes the body to rotate about a point. This introduces rotational motion into the body.

Torque in rotation is analogous to what force in translation.

Force represents an energy transfered (and applied in general in a straight line). On the other hand Torque represents a force acting on a point on a straight line but the effects are applied elsewhere (in rotating the body). As such there is (at least) this subtle difference.

The force is applied to a point of the body which then (by internal constraints fo the body amnd system) this force is transformed into rotational energy producing angular momentum etc, as such the radius of the body (or lever arm wrt the center of application) enters the picture.