10.1 The outboard motor (pictured below) is used to propel boats through water and has a 4-stroke engine - NSC Technical Mathematics - Question 10 - 2022 - Paper 2

To find the angular velocity, we use the relationship between linear velocity and angular velocity:

$V = r \omega$

Where:

• $V$ is the linear velocity (8.33 m/s)
• $r$ is the radius (180 m)
• $ext{Rearranging gives: } \omega = \frac{V}{r} = \frac{8.33}{180} \approx 0.0463 \text{ rad/s}$

To convert degrees to radians, we use the formula:

$\text{radians} = \text{degrees} \times \frac{\pi}{180}$

Thus,

$210\degree = 210 \times \frac{\pi}{180} = \frac{7\pi}{6} \text{ rad}$

The length of an arc is given by:

$s = r \theta$

For arc BC:

• $r = 5$ cm
• $\theta = \frac{7\pi}{6}$ rad

Thus,

$s = 5 \times \frac{7\pi}{6} = \frac{35\pi}{6} \approx 18.33 \text{ cm}$

Given the area of the shaded sector is 54 cm²:

$A = \frac{1}{2} r^2 \theta$

Where $r = 9$ cm. Rearranging gives:

$\theta = \frac{2A}{r^2} = \frac{2 \times 54}{9^2} = \frac{108}{81} = \frac{4}{3} \text{ rad}$

We can use the formula for the length of the chord:

$l = 2r \sin\left(\frac{\theta}{2}\right)$

Using $r = 7$ cm and $\theta = \frac{4\pi}{3}$:

$l = 2 \times 7 \times \sin\left(\frac{2\pi}{3}\right) = 14 \times \frac{\sqrt{3}}{2} = 7\sqrt{3} \approx 12.12 ext{ cm}$

The total length of the rubber belt is 140 cm. To find the length that is not in contact with the pulleys, we will subtract the lengths of the arcs from the total length:

$\text{Length that is not in contact} = 140 - 4.19 - \text{(length of arc BC + length of arc JK)}$