Physics 9702/23 — October/November 2013

Background Concept

For an object of uniform density $\rho$, its mass $m$ is found from its volume $V$ using

$$m = \rho V$$

Weight $W$ is the gravitational force on the mass in a gravitational field of strength $g$:

$$W = mg$$

Here, $\rho$ is in $\text{kg m}^{-3}$, so $V$ must be in $\text{m}^3$ to keep units consistent.

A cylindrical disc is a cylinder, so its volume is

$$V = \pi r^2 h$$

where $r$ is the radius and $h$ is the thickness (height).

Understanding the Question

You are given:

• diameter $= 28\ \text{mm}$ (so you must halve it to get radius),
• thickness $= 12\ \text{mm}$,
• density $\rho = 6.8 \times 10^{3}\ \text{kg m}^{-3}$.

You must calculate the weight in newtons, to two significant figures. So the path is:

1. convert mm to m,
2. find $V$ of the cylinder,
3. find $m = \rho V$,
4. find $W = mg$.

Approach

• Convert all lengths to metres because the density is in SI units.
• Use the cylinder formula $V = \pi r^2 h$.
• Multiply by density to get mass.
• Multiply by $g$ (take $g = 9.81\ \text{m s}^{-2}$ unless otherwise stated) to get weight.
• Round the final value to 2 s.f.

Step-by-Step Reasoning

1. Convert dimensions:

• Radius is half the diameter:

$$r = \frac{28\ \text{mm}}{2} = 14\ \text{mm} = 14 \times 10^{-3}\ \text{m} = 0.014\ \text{m}$$

• Thickness:

$$h = 12\ \text{mm} = 12 \times 10^{-3}\ \text{m} = 0.012\ \text{m}$$

1. Volume of cylinder:

$$V = \pi r^2 h = \pi (0.014)^2(0.012)$$

Compute:

$$(0.014)^2 = 1.96 \times 10^{-4}$$ $$1.96 \times 10^{-4} \times 0.012 = 2.352 \times 10^{-6}$$ $$V = \pi (2.352 \times 10^{-6}) = 7.39 \times 10^{-6}\ \text{m}^3$$

1. Mass from density:

$$m = \rho V = (6.8 \times 10^3)(7.39 \times 10^{-6}) = 5.02 \times 10^{-2}\ \text{kg}$$

1. Weight:

$$W = mg = (5.02 \times 10^{-2})(9.81) = 4.92 \times 10^{-1}\ \text{N} = 0.492\ \text{N}$$

To two significant figures:

$$W = 0.49\ \text{N}$$

Key Takeaways

• Always convert to SI units before using formulas involving $\rho$ in $\text{kg m}^{-3}$.
• For a disc-shaped cylinder, use $V = \pi r^2 h$ with $r = \tfrac{1}{2}$diameter.
• Weight is a force: $W = mg$ and is measured in newtons.

Common Mistakes

• Using diameter instead of radius in $\pi r^2 h$ (this makes the volume 4 times too big).
• Forgetting to convert mm to m, leading to a volume error of $10^9$.
• Giving the final answer in kg (mass) instead of N (weight).
• Rounding too early (can shift the final 2 s.f. answer).

Things to Be Careful About

• Unit consistency: $\text{mm} \rightarrow \text{m}$ is essential because $\text{m}^3$ is required.
• Significant figures: only the final answer needs to be to 2 s.f.; keep extra digits during intermediate steps.
• Use the value of $g$ expected by the exam (typically $9.81\ \text{m s}^{-2}$ unless told otherwise).