Question

jogger 1 in the figure has a mass of 61.3 kg and runs in a straight line with a speed of 3.35 m/s.

what is the magnitude of the joggers angular momentum with respect to the origin, \\(o\\)?

\\(l =\\) \\(\text{kg} \cdot \text{m}^2/\text{s}\\)

Explanation:

Identify the given parameters

We are given the following physical quantities for Jogger 1:

• Mass of Jogger 1, \(m = 61.3\text{ kg}\).
• Speed of Jogger 1, \(v = 3.35\text{ m/s}\).
• From the figure, Jogger 1 (labeled as ①) is running parallel to the x-axis in the positive x-direction.
• The position of Jogger 1's line of motion is at a constant y-coordinate, \(y = 5.00\text{ m}\).

Formulate the angular momentum equation

The angular momentum \(\vec{L}\) of a point particle relative to the origin \(O\) is defined by the cross product:

$$\vec{L} = \vec{r} \times \vec{p}$$

where \(\vec{r}\) is the position vector of the particle and \(\vec{p} = m\vec{v}\) is its linear momentum.

Calculate the magnitude of angular momentum

For a particle moving parallel to the x-axis, the perpendicular distance (lever arm) from the origin to the line of motion is the absolute value of its y-coordinate:

$$r_{\perp} = |y| = 5.00\text{ m}$$

The magnitude of the angular momentum \(L\) is given by:

$$L = m v r_{\perp}$$

Compute the numerical value

Substitute the given values into the magnitude formula:

$$L = 61.3\text{ kg} \times 3.35\text{ m/s} \times 5.00\text{ m}$$

$$L = 1026.775\text{ kg}\cdot\text{m}^2/\text{s}$$

Rounding to three significant figures (matching the given values \(61.3\), \(3.35\), and \(5.00\)):

$$L \approx 1030\text{ kg}\cdot\text{m}^2/\text{s}$$

Answer:

What is the magnitude of the jogger's angular momentum with respect to the origin, \(O\)?

\(L =\) <blank>\(1030\)</blank> \(\text{kg}\cdot\text{m}^2/\text{s}\)