Gears and belt drives

Read each dimension arrow before you read its number. The 200 mm on the driver spans the whole pulley between two tangent lines, so it is a diameter. The 250 mm on the driven pulley runs from the centre dot to the rim, so it is a radius. Convert both to diameters:

$$\begin{aligned} \text{VR} &= \frac{\text{driven diameter} }{\text{driver diameter} } \\[10pt] &= \frac{2 \times 250}{200} \\[10pt] &= 2.50 \end{aligned}$$

(B) and (C) are what you get by treating both numbers as diameters. (A) is the right ratio upside down.

The wheel has 18 teeth. The worm is single-start, so it has one start. The three loops you can see are one thread wrapped three times round the worm, not three starts.

$$\begin{aligned} \text{VR} &= \frac{\text{teeth on the wheel} }{\text{starts on the worm} } \\[10pt] &= \frac{18}{1} \end{aligned}$$

(D) is what you get by counting the loops as starts. (A) and (B) come from dividing the 18 teeth by the three loops.

(a) A double-threaded worm has two starts:

$$\begin{aligned} \text{VR} &= \frac{\text{worm wheel teeth} }{\text{worm starts} } \\[10pt] &= \frac{40}{2} \\[10pt] &= 20 \end{aligned}$$

1 mark Provides correct working to give the correct gear ratio (VR).

$$\begin{aligned} \eta &= \frac{\text{MA} }{\text{VR} } \\[10pt] 0.45 &= \frac{\text{MA} }{20} \\[10pt] \text{MA} &= 9 \end{aligned}$$

1 mark Provides the correct formula and substituted values for mechanical advantage.

$$\begin{aligned} F_L &= \text{MA} \times F_E \\[10pt] &= 9 \times 50 \\[10pt] &= 450\ \text{N} \end{aligned}$$

1 mark Provides correct working to give the correct load.

$$\begin{aligned} F &= mg \\[10pt] m &= \frac{450}{9.8} \\[10pt] &= 45.92 \end{aligned}$$

1 mark Provides the correct formula and substituted values for mass.

1 mark Gives the correct answer to the nearest whole unit with the correct unit: 46 kg.

(b) This is QCAA’s published working from the 2020 mark guide, and it earns the marks. Efficiencies in series multiply, so find the efficiency of the added effort wheel first:

$$\begin{aligned} \eta_{\text{total} } &= \eta_1 \times \eta_2 \\[10pt] 0.36 &= 0.45 \times \eta_2 \\[10pt] \eta_2 &= 0.80 \end{aligned}$$

1 mark Provides the correct formula and substituted values for the effort wheel efficiency.

1 mark Provides correct working to give the correct effort wheel efficiency.

$$\text{VR} = \frac{\text{MA} }{\eta} = \frac{9}{0.80} = 11.25$$

1 mark Provides the correct formula and substituted values for velocity ratio.

1 mark Provides correct working to give the correct answer to two decimal places.

Treat (b) as the way to earn these marks, not as a general method. It reuses the MA of 9 from the worm with the efficiency of the effort wheel, and the question gives you no MA for the added component. The step you can reuse elsewhere is the first one, that efficiencies in series multiply.

The pedal gear is the driver. Each pedal turn moves 150 teeth past the 50-tooth wheel gear, so the wheel turns 3 times. Four pedal turns give 12 wheel turns:

$$\begin{aligned} \text{distance} &= 12 \times \pi \times 0.725 \\[10pt] &= 27.33\ \text{m} \end{aligned}$$

(A) uses the ratio upside down. (B) is one pedal turn instead of four. (C) leaves out the $\pi$.

A VR of 1:3 means the wheel turns 3 times for each pedal turn, so three pedal turns give 9 wheel turns:

$$\begin{aligned} \text{distance} &= 9 \times \pi \times 0.74 \\[10pt] &= 20.92\ \text{m} \end{aligned}$$

That is 21 m. (A) is double the answer. (C) leaves out the $\pi$. (D) is about one turn of the wheel, 2.32 m.

First decide which gear is the driver. The motor drives the large 32-tooth gear, so that is the driver. The small 8-tooth gear at the wheel hub is the driven. On this bicycle the driver is the bigger gear, which is the opposite of a winch.

(a)

$$\begin{aligned} \text{GR} &= \frac{\text{teeth on driven} }{\text{teeth on driver} } \\[10pt] &= \frac{8}{32} \\[10pt] &= 0.25 \end{aligned}$$

1 mark Determines the VR or GR of the driven to driver gears.

$$\begin{aligned} \pi d &= \pi \times 0.7 \\[10pt] &= 2.199\ \text{m} \\[10pt] \text{driven rotations} &= \frac{15}{2.199} \\[10pt] &= 6.82 \end{aligned}$$

$$\begin{aligned} \text{driver turns} &= 6.82 \times 0.25 \\[10pt] &= 1.7 \end{aligned}$$

1 mark Determines the answer to one decimal place.

Reading the ratio the other way gives 27.3 driver rotations instead of 1.7.

(b) The drive shaft and the driver gear are on the same shaft, so they turn together:

$$\begin{aligned} \text{shaft turns} &= \text{driver turns} \\[10pt] &= 1.7 \end{aligned}$$

1 mark Correct interpretation of the mechanical system.

1 mark Determines the answer to the nearest whole unit: 2 rotations.

(c) The 20 mm is the shaft’s diameter, so each rotation moves the effort $\pi \times 0.02$ m. QCAA used the 2 rotations from (b):

$$\begin{aligned} d &= 2 \times \pi \times 0.02 \\[10pt] &= 0.126\ \text{m} \end{aligned}$$

1 mark Determines the distance moved by the motor drive shaft.

$$\begin{aligned} W &= Fs \\[10pt] &= 600 \times 0.126 \\[10pt] &= 75.6\ \text{J} \end{aligned}$$

1 mark Determines the work.

The stem gives the time, $t = 2$ s:

$$\begin{aligned} P &= \frac{W}{t} \\[10pt] &= \frac{75.6}{2} \\[10pt] &= 37.8 \approx 38\ \text{W} \end{aligned}$$

1 mark Determines the answer in W to the nearest whole unit.

Using the unrounded 1.7 rotations gives 0.107 m, 64.3 J and 32 W. The guide accepts that too. It allows distances from 0.1068 to 0.13 m, work from 64 to 75.6 J and power from 32 to 38 W. The 2025 subject report advises leaving rounding until the end.

Here is a response that earns all five marks:

To climb an incline the cyclist needs a high mechanical advantage. The gear attached to the pedals is made smaller, with fewer teeth, than the gear attached to the rear wheel. The smaller the pedal gear is compared with the rear wheel gear, the higher the mechanical advantage, so less force is needed to turn the pedals.

Because the pedal gear is smaller, the pedals have to make more rotations to move the bicycle the same distance. The work done on the pedals stays the same, because the smaller force on the pedals moves through a greater distance.

Sketch a small pedal gear linked by a chain to a larger rear wheel gear, and label which is which.

1 mark Explains MA with wording that the pedal gear is smaller than the rear wheel gear. The guide also accepts tooth counts, radii or diameters, or the reverse wording.

1 mark States that the work done remains the same.

1 mark States that less force is needed to move the pedals.

1 mark States that more rotations of the pedals are needed.

1 mark Provides a sketch with the larger rear wheel gear linked to the smaller pedal gear.