2023 A/L Physics Past Paper · Question 22 (English Medium)

Answer: (3) $C$

What the question asks

We want: the point where the centre of gravity of the sheet is.

What each letter means:

$R$ ($\text{cm}$): the radius of the sheet.

$G$: the centre of gravity of the sheet.

$w$ ($\text{N}$): the weight of one thin rod of the sheet.

$l$ ($\text{cm}$): the length of a rod.

$y$ ($\text{cm}$): the height of a rod above the straight edge.

$W$ ($\text{N}$): the weight of the whole sheet.

$\bar{y}$ ($\text{cm}$): the height of $G$ above the straight edge.

Given information

What we know:

The sheet is uniform and thin, and it is half of a circle.

The points $A$, $B$, $C$, $D$ and $E$ are on the middle line, from the straight edge up to the top of the curve.

We take a sample radius $R = 10\ \text{cm}$ to put numbers on the heights.

Relevant theory

A body is made by several particles, each attracted towards the centre of the Earth by a force of gravity (given by Newton's universal law of gravitation) and thus the body experiences a resultant force of gravity (weight). That force is called the weight of the body.

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

Moment of force F about the point 'O' $= F \times d$ where 'd' is the perpendicular distance from 'O' to the force.

The first quote says a body is many small parts, and each part has its own pull of gravity. The second quote says the centre of gravity is the point where the total weight acts.

The third quote says the moment of a force is the force times its distance. So the moment of the total weight acting at $G$ must equal the moments of all the parts added up.

For example, think of a see-saw with many children sitting on it. One heavy grown-up can make the same turn as all of them, if the grown-up sits at just the right place.

The point where the whole weight can sit and make the same turn is the centre of gravity.

Reasoning and calculation

Cut the sheet into thin rods, all parallel to the straight edge.

Each rod is a thin uniform rod. Its weight acts at its middle point. The sheet is the same on the left and on the right, so every middle point is on the line from $A$ to $E$.

So $G$ is on the line from $A$ to $E$. All five points are on that line, so we still need the height of $G$.

The rods near the straight edge are long, so they are heavy. The rods near the top are short, so they are light.

There is more weight low down than high up. So $G$ is below half the height, $0.5R$. This rules out $D$ and $E$.

Now we use moments about the straight edge to find the height.

Required equations

$w \propto l$

$W = \sum w$

$W \bar{y} = \sum w y$

Every rod has the same small height, so we can use the lengths in $\text{cm}$ in place of the weights. The same unit cancels in $\dfrac{\sum w y}{\sum w}$, and $\bar{y}$ comes out in $\text{cm}$.

$79.30$

$\sum w = 19.90 + 19.08 + 17.32 + 14.28 + 8.72$

$342.18$

$\sum w y = 19.90 + 57.24 + 86.60 + 99.96 + 78.48$

$4.3$

$\bar{y} = \dfrac{342.18}{79.30}$

$m / w$

Reading the options

Now match it with the five options.

On the figure, with $R = 10\ \text{cm}$, $B$ is about $2\ \text{cm}$ up, $C$ about $4\ \text{cm}$, $D$ about $7\ \text{cm}$ and $E$ about $9\ \text{cm}$.

(1) $A$ is on the straight edge, at $0\ \text{cm}$. It does not match.

(2) $B$ is too low. It does not match.

(3) $C$ is at about $4\ \text{cm}$. This matches.

(4) $D$ is above half the height. It does not match.

(5) $E$ is near the top. It does not match.

Checks

Check with a rectangle. A rectangle has the same weight at every height, so its $G$ is at half the height, $0.5R$. The semi-circle loses weight near the top, so its $G$ must be lower than $0.5R$. $4.3\ \text{cm}$ is lower than $5\ \text{cm}$.

Check with more rods. With many very thin rods the height comes out as $\dfrac{4R}{3\pi} = 4.24\ \text{cm}$, very close to our $4.3\ \text{cm}$. Five rods are enough to pick $C$.

Conclusion

$\bar{y} \approx 0.43R$

Conclusion

The quotes give the centre of gravity as the point where the total weight acts, and the moment as force times distance. Cutting the sheet into thin rods and adding their moments about the straight edge is our own working, and so is the check value $\dfrac{4R}{3\pi}$. The book table has no semi-circular sheet.

Teacher's note

Cut the shape into thin rods. The middle points of the rods show the line that $G$ is on.

Then ask where most of the weight is. A shape that gets narrow at the top has its $G$ below half the height.

Exam tip