2024 A/L Physics Past Paper · Question 12 (English Medium)

Answer: (4) $D$

Given information

This question is about the balance point of a circular plate after a piece is cut out.

The large plate has radius $2r$. A smaller circle of radius $r$ has been removed from its left side. The material and thickness are uniform.

Imagine holding the remaining plate so it can hang freely. Its centre of gravity is the point through which its resultant weight acts.

What we must find: which of the marked points, $A$ to $E$, is closest to that point.

How we find it: compare the masses of the two circles using their areas. Then put the missing part back in a moment calculation to recover the original disc.

Take the centre $C$ of the original disc as $x=0$, with right positive. The removed circle is tangent at the left edge and reaches $C$, so its centre $A$ is at $x=-r$.

What the question asks

$r$ ($\text{m}$): radius of the removed circle.

$M$ ($\text{kg}$): mass of the original full disc.

$m$ ($\text{kg}$): mass of the removed part.

$x$ ($\text{m}$): signed distance of the remaining centre from $C$.

$g$ ($\text{m s}^{-2}$): gravitational acceleration.

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.

3. Uniform circular disc | G - Centre of disc

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

3. The algebraic sum of the moments of all the forces about any point in their plane should be zero.

What is happening: taking material away on the left changes the distribution of weight. The remaining centre lies on the horizontal symmetry line, to the right of the original centre.

We can locate it by imagining the cut-out part put back. The completed disc would once again have its centre of gravity at $C$. Its weight therefore has zero moment about $C$.

For example, cut this shape from a sheet of uniform card. Hang it from a small pin near its upper edge and let it settle.

A string carrying a small weight hangs vertically from the same pin. At rest, the weight acts along that line, so it passes through the centre of gravity.

In words: the part left behind hangs with its centre of gravity below the pin. The missing circle has changed where that point is.

Required equations

$\dfrac{M}{m}=\dfrac{\pi(2r)^2}{\pi r^2}$

$(M-m)gx+mg(-r)=Mg(0)$

$x=\dfrac{mr}{M-m}$

Both masses are positive. The minus sign belongs to the position on the left, not to the mass.

Applying it to this question

Because the material and thickness are uniform, the mass ratio is the area ratio. The larger radius is twice the smaller radius, so the full disc has four times the removed mass.

$\dfrac{M}{m}=\dfrac{\pi(2r)^2}{\pi r^2}$

$\dfrac{M}{m}=4$

$4m$

$M$

$M_{\rm remaining}=M-m$

$M_{\rm remaining}=4m-m$

$3m$

$M_{\rm remaining}$

$(M-m)gx+mg(-r)=Mg(0)$

$3mgx-mgr=0$

$3mgx=mgr$

$x=\dfrac{mgr}{3mg}$

$\dfrac{r}{3}$

$x$

Reasoning and calculation

The positive result lies a short distance to the right of $C$. The marked point nearest $r/3$ on that side is $D$. This is option (4).

$A$ is the centre of the removed circle, at $x=-r$. This confuses the missing part with the remaining plate. Adding the missing part back would then leave the combined centre on the left, not at $C$.

$B$ is left of $C$. This moves the centre towards the part taken away. Removing mass on the left makes the remaining centre lie to the right: the equation gives $x>0$, not a negative value.

$C$ was the centre of the uncut disc. Keeping it ignores the removal. If the remaining mass $3m$ still acted at $C$, adding mass $m$ at $-r$ would give a combined centre at $-r/4$, instead of the original $0$.

$E$ is farther right than the position $r/3$ near $D$. Merely choosing a point on the right misses the size of the shift. For a proposed position $x_E>r/3$, the restored plate would have $3mgx_E-mgr>0$, so its moment about $C$ would not be zero.

Checks

$(3*x-1)/4$

$x_{\rm restored}=\dfrac{3m(r/3)+m(-r)}{4m}$

$x_{\rm restored}=\dfrac{mr-mr}{4m}$

$x_{\rm restored}=0$

The restored centre is exactly $C$, as required for the full uniform disc. This reverses the construction and checks the proposed position.

The ratio of masses cancels, leaving the length $r/3$. Thus $x$ has the same unit as $r$.

The remaining shape is symmetric above and below the horizontal axis, so its centre lies on that axis. Removal on the left shifts it to the right. The computed $x=r/3$ is inside the remaining material for this cut.

Conclusion

$x=\dfrac{r}{3}$

The centre of gravity lies at $x=r/3$ to the right of $C$, nearest the marked point $D$. The answer is (4).