QuestionProblem 6: The figure shows a projectile with the velocities at three different location. a) Find the angular momentum when the projectile is at the origin, $b$ ) find the angular momentum when the projectile is at its maximum height, and c ) when the projectile lands. Note the angular momentum is measured relative to the origin. - Also, $\mathrm{v}_{0}=30.0 \mathrm{~m} / \mathrm{s}, \theta=30.0^{\circ}$ , and $\mathrm{m}=2.50 \mathrm{~kg}$ , and $\mathrm{R}=20.0 \mathrm{~m}$ .

Studdy Solution

STEP 1

1. The projectile is subject to uniform gravitational acceleration.
2. Air resistance is negligible.
3. Angular momentum $\mathbf{L}$ is calculated with respect to the origin.
4. The angular momentum $\mathbf{L}$ is given by $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ , where $\mathbf{r}$ is the position vector and $\mathbf{p}$ is the linear momentum.

STEP 2

1. Calculate the initial conditions and components.
2. Find the angular momentum at the origin.
3. Find the angular momentum at maximum height.
4. Find the angular momentum at the landing point.

STEP 3

Calculate the initial velocity components:
Given: $v_0 = 30.0 \, \text{m/s}, \quad \theta = 30.0^\circ$
Calculate: $v_{0x} = v_0 \cos(\theta)$ $v_{0y} = v_0 \sin(\theta)$
$v_{0x} = 30.0 \cos(30.0^\circ) = 30.0 \times \frac{\sqrt{3}}{2} = 25.98 \, \text{m/s}$ $v_{0y} = 30.0 \sin(30.0^\circ) = 30.0 \times \frac{1}{2} = 15.0 \, \text{m/s}$

STEP 4

Find the angular momentum at the origin:
At the origin, $\mathbf{r} = \mathbf{0}$ , so: $\mathbf{L} = \mathbf{r} \times \mathbf{p} = \mathbf{0} \times m\mathbf{v}_0 = \mathbf{0}$

STEP 5

Find the angular momentum at maximum height:
At maximum height, $v_{1y} = 0$ and $v_{1x} = v_{0x}$ .
Position at maximum height: $r_x = \frac{v_{0x} \cdot t}{2}$ $r_y = \frac{v_{0y}^2}{2g}$
Calculate time to reach maximum height: $t = \frac{v_{0y}}{g} = \frac{15.0}{9.81} \approx 1.53 \, \text{s}$
$r_x = v_{0x} \cdot t = 25.98 \times 1.53 \approx 19.89 \, \text{m}$ $r_y = \frac{15.0^2}{2 \times 9.81} \approx 11.47 \, \text{m}$
Angular momentum: $\mathbf{L} = \mathbf{r} \times m\mathbf{v} = (r_x \hat{i} + r_y \hat{j}) \times m(v_{0x} \hat{i})$ $\mathbf{L} = m \cdot r_y \cdot v_{0x} \hat{k} = 2.50 \cdot 11.47 \cdot 25.98 \hat{k} \approx 744.3 \, \text{kg} \cdot \text{m}^2/\text{s}$

STEP 6

Find the angular momentum at the landing point:
At landing, $\mathbf{r} = R \hat{i}$ and $\mathbf{v} = v_{0x} \hat{i} - v_{0y} \hat{j}$ .
Angular momentum: $\mathbf{L} = \mathbf{r} \times m\mathbf{v} = R \hat{i} \times m(v_{0x} \hat{i} - v_{0y} \hat{j})$ $\mathbf{L} = m \cdot R \cdot v_{0y} \hat{k} = 2.50 \cdot 20.0 \cdot 15.0 \hat{k} = 750.0 \, \text{kg} \cdot \text{m}^2/\text{s}$
The angular momentum values are: a) At the origin: $\mathbf{L} = 0$ b) At maximum height: $\mathbf{L} \approx 744.3 \, \text{kg} \cdot \text{m}^2/\text{s}$ c) At landing: $\mathbf{L} = 750.0 \, \text{kg} \cdot \text{m}^2/\text{s}$

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