2018 A/L Physics Past Paper · Question 15 (English Medium)

Answer: (4) $C$

What the question asks

We want: the centre of gravity of eleven equal spheres joined by light rods.

Which of O, A, B, C, D is it nearest to?

Given information

From the figure:

One sphere is at O.

Eight spheres sit on the two diagonal lines, two on each half-line.

One more sphere is a distance $L$ above O.

One more sphere is a distance $L$ to the right of O.

Relevant theory

The centre of gravity of a body is the point at which the resultant force of gravity or weight of the body acts.

Reasoning and calculation

Look at the X part first.

For every sphere, there is another sphere on the opposite side of O, at the same distance.

Each opposite pair balances at O.

There are 4 pairs, and one sphere is at O itself.

So these nine spheres act like one mass $9m$ at O.

So we do not need the lengths of the diagonal arms.

Only three masses are left:

$9m$ at O.

$m$ at $(0, L)$, above O.

$m$ at $(L, 0)$, to the right of O.

Required equations

$\bar{x} = \frac{9m(0) + m(0) + m(L)}{11m} = \frac{L}{11}$

$\bar{y} = \frac{9m(0) + m(L) + m(0)}{11m} = \frac{L}{11}$

$d = \frac{\sqrt{2}\,L}{11} \approx 0.13L$

Reading the options

The centre is on the 45° line, up and to the right of O.

It is only about $0.13L$ from O.

On that line, the marked point nearest to O is C.

B and D are much farther away. A is on the upright arm, not on the 45° line.

Checks

$sqrt(2) / 11$

A quick check.

The upper arm and the right arm are the same: same length, same mass.

So the centre moves up as much as it moves right: $\bar{x}$ equals $\bar{y}$.

So it must be on the 45° line, where C is.

Conclusion

The centre is at $(L/11,\ L/11)$.

That is about $0.13L$ from O on the 45° line.

This is point C. The answer is option (4).