10.1 The pictures below show a wheelbarrow and an enlargement of the wheel of the wheelbarrow - NSC Technical Mathematics - Question 10 - 2019 - Paper 2

For the chord KL, which is the tangent to the rim, we can again use the circle properties. With chord KL being 32 cm, we derive:

Using Pythagorean theorem in triangle OAB, where OA = radius of the rim (OC = 20 cm), we have:

Using Pythagorean theorem:

$AB^2 = OA^2 - (KL / 2)^2$ Substituting in:

$AB^2 = 20^2 - (32/2)^2 = 400 - 256 = 144$ Thus:

$AB = \sqrt{144} = 12 cm - 1.5 cm (radius of hole) = 10.5 cm$

To find the rotational frequency, we use the angular velocity in revolutions per minute:

Given the angular velocity of 647 revolutions per minute, we convert this into radians per second:

$\omega = 647 \times \frac{2\pi}{60} \approx 67.7 rad/s$ Thus, the rotational frequency ν = 647 rev/min.

The circumferential velocity (v) can be calculated using the formula:

$v = R \cdot \omega$ Where R is the radius (20 cm) and (\omega \approx 67.7 rad/s) Substituting these values:

$v = 0.2 \cdot 67.7 = 13.54 m/s \approx 252.66 cm/min$

To find angle AOB, we compute the angle in radians:

Given the length of the spokes (OA and OB) is 5.2 cm:

Using the relationship of arc length:

$\theta = \frac{s}{r} = \frac{4}{20} \cdot \pi = \frac{\pi}{5} \approx 1.4 rad$

Using the previously calculated angle and radius, we find the length of arc AB:

$s = r \cdot \theta = 5.2 \cdot 1.4 \approx 7.3 cm$

The area of minor sector AOB can be found using the formula:

$Area = \frac{1}{2} r^2 \theta = \frac{1}{2} \cdot (5.2^2) \cdot (1.4) \approx 19 cm²$