Physics 5054/22 — October/November 2020

Walkthrough

This part asks for the definition of the centre of mass. The centre of mass is a specific point in an object where all its mass can be considered to be concentrated for the purpose of calculating moments and gravitational forces. Practically, it is the point through which the object's weight acts, and if supported at this point, the object will balance perfectly in any orientation.

Key Takeaways

• The centre of mass is the average position of all the parts of the system, weighted according to their masses.
• For a uniform object like a metre rule, the centre of mass is at its geometric centre (the $50 \text{ cm}$ mark).

Common Mistakes

• Saying 'the point where gravity acts' is not precise enough; it is where the weight appears to act or where the mass is concentrated.
• Confusing centre of mass with centre of gravity (though they are effectively the same in a uniform gravitational field, the O Level mark scheme specifically looks for 'mass' or 'balances').

Things to Be Careful About

• Use the exact wording expected by the mark scheme: 'where the mass appears to be concentrated' or 'where the object balances'.

Walkthrough

To find the mass of the metre rule, we apply the principle of moments about the pivot. The rule is balanced, meaning the total clockwise moment equals the total anticlockwise moment.

First, we identify the forces and their positions. The weight of the beaker and its contents ($410 \text{ g}$) acts downwards at the centre of the beaker. Since the beaker extends from $84 \text{ cm}$ to $92 \text{ cm}$, its centre is at $\frac{84 + 92}{2} = 88 \text{ cm}$. The pivot is at $80 \text{ cm}$, so the perpendicular distance (moment arm) for the beaker is $88 - 80 = 8 \text{ cm}$. This creates a clockwise moment.

Second, the weight of the metre rule acts downwards at its centre of mass, which is given as the $50 \text{ cm}$ mark. The distance from the pivot ($80 \text{ cm}$) to the rule's centre of mass is $80 - 50 = 30 \text{ cm}$. This creates an anticlockwise moment.

Setting the moments equal: $m_{\text{rule}} \times 30 = 410 \times 8$. Solving for $m_{\text{rule}}$ gives $109.33 \text{ g}$, which rounds to $110 \text{ g}$ to 2 significant figures.

Key Takeaways

• When an object is placed on a rule, its effective weight acts at its own centre of mass (midpoint if uniform).
• The principle of moments requires using perpendicular distances from the pivot to the line of action of each force.
• Gravitational field strength $g$ cancels out when balancing masses on the same side of the equation.

Common Mistakes

• Using the edge of the beaker ($84 \text{ cm}$ or $92 \text{ cm}$) instead of its centre ($88 \text{ cm}$) for the moment calculation.
• Forgetting that the rule's own weight acts at its centre of mass ($50 \text{ cm}$ mark), not at the end.
• Calculating the distance incorrectly (e.g., using $92 - 80 = 12 \text{ cm}$ instead of $88 - 80 = 8 \text{ cm}$).

Things to Be Careful About

• Always find the centre of mass of the beaker itself before calculating its moment arm.
• Ensure distances are measured from the pivot, not from the $0 \text{ cm}$ mark.
• Round the final answer to an appropriate number of significant figures (2 s.f. is standard here, giving $110 \text{ g}$).