Torque

rotation/rotation.1.eps

We have just seen how Newton's Law can be applied to collections of particles as if the entire collection were a ``particle'' with the mass of the collection located at the center of mass of the collection. This trick works very well for solid objects made up of many parts, but isn't as useful for collections like a box containing a few loose marbles. This is because rigid objects retain their shape at all times - we only need to keep track of the position of the center of mass and the orientation of the entire object around the CM.

For this reason, we can break the general motion of rigid objects up into two parts: translation of the CM (the object as a whole) through space and rotation of the object about the CM. In the last section we learned how to treat the translation part in detail. In this section we will learn how to treat the dynamics of rotation in detail.

The figure above is a ``generic'' figure that we will use to think of rotation. You may imagine it to be a big, flat disk. If we color a dot on the disk with a marker pen, we can see that the mass underneath the mark is very small and localized - we can think of it as being a ``particle'' of mass $dm$ and the whole disk consisting of the sum of all the little dots like this that we could draw that would eventually exactly cover the disk. Summing these little dots is done mathematically with integration. We say:

$$M_{\rm tot}= \int dm$$ (5.1)

Of course, written this way the integral is almost tautological (and hence useless). To make use of it, we need to find some way of expressing the integral in coordinates. Fortunately this integral is not too difficult to figure out for simple coordinate systems and figures.

In the case of the circular disk above, we note that the mass under our colored dot must be the total mass $M_{\rm tot}$ times the fraction of the total area of the disk $A$ that happens to lie under our dot, $dA$. The only remaining chore is to figure out what $dA$ (the area element of a disk) is in suitable coordinates. Let's look at it close up:

rotation/rotation.2.eps

We see that a little box with differential sides in polar coordinates has area $dA = (dr)\times(rd\theta) = r dr d\theta$. If we add all the little boxes like this that exist in a disk with radius $R$,

$$A = \int_0^R \int_0^{2\pi} r dr d\theta = (2\pi)(\frac{R^2}{2}) = \pi R^2$$ (5.2)

we indeed get the area of a circle! Similarly we could integrate over a circular annulus (with inner radius $R_1$ and outer radius $R_2$) or over a pie-shaped wedge and calculate their area. The area element $dA$ looks useful!

From this we see that the differential mass $dm = \frac{M}{\pi R^2} rdrd\theta$ where we have now put in the area of the disk and its mass explicitly. We are now prepared to work out the dynamics of this little point of mass, subject to the constraint that the disk is rigid, so that all that it can do is translate and rotate.

Let us apply an external force to this point and write Newton's second law for this tiny chunk of mass (calling it the $i$-th chunk even though we're going to eventually integrate over all of them instead of doing a discrete sum. The other similar chunks of mass that make up the whole object we'll index with $j$.):

rotation/rotation.3.eps

$$\vec{\bf F}_i = \vec{\bf F} + \sum_{j\ne i} \vec{\bf F}_{ij}$$ (5.3)

$$= dm \vec{\bf a}$$ (5.4)

As before (when we've seen internal forces), when we sum all of the force acting on the object due to the single external force $\vec{\bf F}$, the internal forces cancel due to Newton's Third law. We get:

$$\vec{\bf F} = \int dm \vec{\bf a}_{\rm cm} = M\vec{\bf a}_{\rm cm}$$ (5.5)

which describes the motion of the center of mass itself due to the application of the external force.

What about the motion of the little bit of mass with respect to the center of mass? Well, from the figure above we can see that the net (external) force on it has two components. One ($\vec{\bf a}_r$) acts along the line from the center of mass to the little chunk. If the object were not rigid, this force would pull the little chunk away from the center. However, the object is rigid, so this component just accelerates the chunk and the center of mass in together in such a way that their distance of separation remains constant.

The other, $\vec{\bf a}_t$, is of considerable interest and the subject of this chapter as the motion of the center of mass itself was the subject of the last chapter. First, we note that this component describes the rotation of this little chunk around the center of mass. Second, we note that the whole (rigid) mass rotates together, so that (once the axis of rotation is known) a single coordinate describes its orientation. This suggests that we find a clever way to write Newton's second law in this one-dimensional coordinate system:

$$\vec{\bf F}_t = dm \vec{\bf a}_t$$ (5.6)

$$\vert\vec{\bf r}\vert\vert\vec{\bf F}_t\vert = \vert\vec{\bf r}\vert dm \vert\vec{\bf a}_t\vert$$ (5.7)

$$= r^2 dm \frac{\vert\vec{\bf a}_t\vert}{r}$$ (5.8)

$$= (r^2 dm) \alpha$$ (5.9)

where we have used

$$s_t = r\theta$$ (5.10)

$$v_t = \frac{ds_t}{dt} = r\frac{d\theta}{dt} = r\omega$$ (5.11)

$$a_t = \frac{dv_t}{dt} = \frac{d^2s_t}{dt^2} = r\frac{d\omega}{dt} = r\frac{d^2\theta}{dt^2} = r\alpha$$ (5.12)

to relate coordinates on the particular arc that contains $dm$ to angular coordinates that apply to the whole object. We use Newton's second law rewritten in this way to define the torque:

$$\tau = (r^2 dm) \alpha$$ (5.13)