# 5.5: Torque

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#### Rotational Newton's Second Law

As we saw for linear motion, we can only go so far with energy conservation. If we want to analyze aspects of motion such as elapsed time and direction of motion, we need more than mechanical energy conservation to work with. In the linear case, we found that this meant that we had to use Newton's Second Law. We now seek the rotational equivalent of that law.

The rotational equivalent of the Newton's Second Law must relate the reaction of the system (rotational acceleration) to an external influence (rotational force), with the degree of this effect being determined by an internal property of the system (rotational mass). That is, we need a rotational substitute for all of the participants of this formula:

\[ \overrightarrow a_{cm} = \dfrac{\overrightarrow F_{net}}{m} \]

We already found a rotational version of acceleration in our discussion of rotational kinematics – it is the angular acceleration. We even defined a direction for this vector using the right-hand rule. The center of mass qualification in the case above is unneeded for the rotational case, because the angular acceleration is the same about every point on a rigid object.

We have also determined an appropriate candidate for the "rotational mass" – the rotational inertia. This is certainly a reasonable choice, for a couple of reasons. First, from our direct experience we know that it is easier to swing an object (e.g. a baseball bat) when holding the heavier end than when holding the lighter end, so the degree to which an extended object "resists" angular acceleration is determined by the distribution of mass. Second, if the physics is to remain consistent, why would the quantity that plays the role of mass in kinetic energy be different from the quantity that plays the role of mass for the second law?

With those two quantities established, we can now get a glimpse into what the "rotational force" is by examining the units:

\[ \left[\alpha\right] = \dfrac{\left[rotational\;force\right]}{\left[I\right]} \;\;\; \Rightarrow \;\;\; \left[rotational\;force\right] = \left[\dfrac{rad}{s^2}\right]\left[kg\cdot m^2\right] = \dfrac{kg \cdot m^2}{s^2} \]

This is weird... These are units of energy! We'll need to chalk this up to coincidence, since clearly the vector quantity of rotational force cannot be a measure of energy. One way to see the difference is to remember the presence of radians in the numerator, even though they are not physical units. We will soon see the source of this coincidence, and it shouldn't take long before the apparent ambiguity between this quantity and energy fades away.

Alert

*While the physical units are the same as energy, we never refer to the SI units of this quantity as "joules." Using this term implies that we are talking about energy, which we are not. Generally we stick to "Newton-meters."*

We can't continue calling this vector "rotational force" forever, so we will henceforth refer to it by its proper name: *torque*. In keeping with our tradition of using greek variables for rotational quantities, we will represent torque with \(\overrightarrow \tau\), giving as our rotational Newton's second law:

\[\overrightarrow \alpha = \dfrac{\overrightarrow \tau_{net}}{I} \]

#### Torque

In the cases of acceleration and inertia, we found a direct relationship between the linear and rotational quantities, so we would expect there to be a similar relationship between force and torque. Furthermore, since the linear/rotational bridge for acceleration and inertia both require a point of reference (the pivot), we would expect the same to be true for the bridge between force and torque.

The first thing we notice is that an object can experience no net force and yet still experience a nonzero rotational acceleration: