2020 A/L Physics Past Paper · Question 19 (English Medium)

Answer: (2) $\dfrac{5r}{4}$

Given information

We know:

Four rings of the same uniform wire — $A$ ($r$), $B$ ($2r$), $D$ ($3r$) on the left of $X$; $C$ ($2r$) on the right.

All touch at $X$, with their centres on one straight line.

What the question asks

We want: the distance between the centre of gravity $G$ of the composite body and $X$.

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.

2. Uniform ring | G - Centre of ring

A ring's weight is spread evenly around its circumference, but since the circumference is symmetric about its own centre, the resultant weight of a uniform ring acts at its centre — the same place as its own centre of gravity.

Required equations

$W_i = (2\pi r_i)\,\lambda g \propto r_i$

$\sum W_i x_i = \left(\sum W_i\right) x$

$x = \frac{\sum W_i x_i}{\sum W_i}$

Applying it to this question

Body: the composite made of all four rings.

The weight of each ring acts at its own centre.

Since the wire is uniform, mass per unit length $\lambda g$ is constant; so weight is proportional to radius: $A$ – $w$, $B$ – $2w$, $C$ – $2w$, $D$ – $3w$.

Reasoning and calculation

Condition: the moment of the resultant weight $8w$ about $X$ equals the sum of the moments of each individual weight.

$B$ and $C$ are equal weights at equal distance $2r$ on either side of $X$ — their moments cancel each other out.

$A: r,\; B: 2r,\; C: -2r,\; D: 3r$

$10wr$

$w(r) + 2w(2r) + 2w(-2r) + 3w(3r)$

$8w$

$w + 2w + 2w + 3w$

$x = \frac{10wr}{8w}$

$1.25r$

$x = \frac{5r}{4}$

Checks

Physical sense: $G$ lies on the side of $A$, $B$, $D$, closer to $X$ than the biggest ring $D$'s centre ($3r$) — because $A$'s centre is only at $r$.

$\dfrac{5r}{4}$ lies between the two ✓.

Second route: removing $B$, $C$, the combined $G_1$ of $A$ ($w$, $r$) and $D$ ($3w$, $3r$) is $\dfrac{w r + 3w \cdot 3r}{4w} = 2.5r$.

The combined $G_2$ of $B$, $C$ ($4w$) is at $X$. Between these two equal weights ($4w$ each): $\dfrac{2.5r}{2} = 1.25r$ ✓.

$(a * a + b * b + d * d - c * c) / (a + b + c + d)$

Conclusion

$x = \frac{5r}{4}$

Weights $w$, $2w$, $2w$, $3w$; moment about $X$: $10wr = 8w \cdot x$ ⇒ $x = \dfrac{5r}{4}$ — the answer is (2).

Giving all four rings equal weight.

The weight of a ring made of the same wire is proportional to its circumference, that is, its radius.