Review question

# If a disc is missing two smaller discs, where’s its centre of gravity? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6670

## Solution

$AB$ and $CD$ are two perpendicular diameters of a uniform circular metal disc of radius $\quantity{12}{in.}$ and centre $O$. Two circular holes of radii $\quantity{4}{in.}$ and $\quantity{2}{in.}$ are cut out of the disc, their centres being $\quantity{6}{in.}$ from $O$ along $OA$ and $\quantity{10}{in.}$ from $O$ along $OC$ respectively. Find

1. the distances of the centre of gravity, $G$, of the remaining portion of the disc from the diameters $CD$ and $AB$;

We begin by sketching this on coordinate axes. We have chosen to have $O$ at the origin, with $AB$ lying along the $y$-axis with $A$ at $(0,-12)$ and $C$ at $(-12,0)$.

We could equally well have drawn it with $A$ at $(0,12)$ and/or $C$ at $(12,0)$, or we could have swapped $A$ and $C$ around and have $AB$ on the $x$-axis, $CD$ on the $y$-axis. It makes no difference, as the axes are for our convenience only. We could also have placed the origin somewhere else, or not have had $AB$ lying along or parallel to an axis, but that would have made things harder, while our aim is to make things easier.

We therefore have a disc centred at $(0,0)$ of radius $12$ missing a disc centred at $(0,-6)$ of radius $4$, and also missing another disc centred at $(-10,0)$ of radius $2$.

We know how to calculate the centre of gravity (centre of mass) of a collection of objects by taking moments. We also know that the centre of gravity of a disc is at its centre, by symmetry.

If we complete the reduced disc to a complete disc by replacing the missing small discs, we can calculate the centre of gravity of the complete disc by thinking complete disc = reduced disc + small disc 1 + small disc 2.

So let’s suppose that the centre of gravity of the reduced disc is at $(X,Y)$, and that the lamina has density equal to $1$ per unit area. Then the mass of the complete disc is $12^2\times\pi=144\pi$, the smaller discs have masses $4^2\times\pi=16\pi$ and $2^2\times\pi=4\pi$, so the reduced disc has mass $144\pi-16\pi-4\pi=124\pi$.

Taking moments about $OA$ now gives

$144\pi\times 0= 124\pi\times X+16\pi\times0+4\pi\times(-10) \implies X = \dfrac{10}{31}.$

Similarly, taking moments about $OC$ gives

$144\pi\times 0= 124\pi\times Y+16\pi\times(-6)+4\pi\times0 \implies Y = \dfrac{24}{31}.$

Thus $G$ is the point $\left(\dfrac{10}{31},\dfrac{24}{31}\right)$, and these are the required distances from $AB$ and $CD$ respectively (in inches).

1. the angle that $OA$ will make with the vertical if this remaining portion is hung on a smooth pivot at $O$.

When a uniform lamina is hung from a smooth pivot, the centre of mass will always lie directly below the pivot.

The centre of gravity $G$ here is labelled, and the straight line from $O$ through $G$ is the downwards vertical. The angle that $OA$ makes with the (upwards) vertical is therefore $\theta$. Using trigonometry on the coordinates of $G$, we can find $\theta$ as $\theta= \tan^{-1} \frac{X}{Y} = \tan^{-1} \frac{10}{24} = 22.6^\circ \quad \text{to 1 d.p.}$