The minute arm of a tower clock is 1.5 m long, which is the radius r.

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Here's the solution for question 2.1:

The minute arm of a tower clock is 1.5 m long, which is the radius $r$ .
A clock face has 12 hour marks. The total angle of a circle is $2\pi$ radians.
The angle between each hour mark is $\frac{2\pi}{12} = \frac{\pi}{6}$ radians.

2.1.1. The arc displacement in radians

Determine the number of intervals moved.
The minute arm moves from 1 to 8.
Number of intervals = $8 - 1 = 7$ intervals.
Calculate the total angular displacement.
The angular displacement $\theta$ is the number of intervals multiplied by the angle per interval.
$\theta = 7 \times \frac{\pi}{6} \text{ rad}$ $\theta = \frac{7\pi}{6} \text{ rad}$ $\theta \approx \frac{7 \times 3.14159}{6} \text{ rad}$ $\theta \approx 3.665 \text{ rad}$
The arc displacement is $\boxed{\frac{7\pi}{6} \text{ rad}}$ or approximately $\boxed{3.665 \text{ rad}}$ .

2.1.2. The circumferential length produced by the tip of the arm
Use the formula for arc length.
The circumferential length (arc length) $s$ is given by $s = r\theta$ , where $r$ is the radius and $\theta$ is the angular displacement in radians.
Given $r = 1.5$ m and $\theta = \frac{7\pi}{6}$ rad.
Calculate the circumferential length.
$s = 1.5 \text{ m} \times \frac{7\pi}{6} \text{ rad}$ $s = \frac{1.5 \times 7\pi}{6} \text{ m}$ $s = \frac{10.5\pi}{6} \text{ m}$ $s = 1.75\pi \text{ m}$ $s \approx 1.75 \times 3.14159 \text{ m}$ $s \approx 5.498 \text{ m}$
The circumferential length is $\boxed{1.75\pi \text{ m}}$ or approximately $\boxed{5.498 \text{ m}}$ .

The minute arm of a tower clock is 1.5 m long, which is the radius r.

Step-by-step solution

Determine the number of intervals moved.

Calculate the total angular displacement.

Use the formula for arc length.

Calculate the circumferential length.

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