Questions: At the instant of the figure, a 1.20 kg particle P has a position vector r of magnitude 7.30 m and angle θ₁=47.0° and a velocity vector v of magnitude 4.10 m / s and angle θ₂=28.0°. Force F, of magnitude 4.70 N and angle θ₃=28.0° acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle?

At the instant of the figure, a 1.20 kg particle P has a position vector r of magnitude 7.30 m and angle θ₁=47.0° and a velocity vector v of magnitude 4.10 m / s and angle θ₂=28.0°. Force F, of magnitude 4.70 N and angle θ₃=28.0° acts on P. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle?

Transcript text: At the instant of the figure, a 1.20 kg particle $P$ has a position vector $\vec{r}$ of magnitude 7.30 m and angle $\theta_{1}=47.0^{\circ}$ and a velocity vector $\vec{v}$ of magnitude $4.10 \mathrm{~m} / \mathrm{s}$ and angle $\theta_{2}=28.0^{\circ}$. Force $\vec{F}$, of magnitude 4.70 N and angle $\theta_{3}=28.0^{\circ}$ acts on $P$. All three vectors lie in the xy plane. About the origin, what are the magnitude of (a) the angular momentum of the particle and (b) the torque acting on the particle?

Solution

Solution Steps

Step 1: Determine the position vector components

Given:

Position vector $ \vec{r} $ with magnitude 7.30 m and angle $ \theta_1 = 47.0^\circ $

Calculate the components of $ \vec{r} $: \[ r_x = 7.30 \cos(47.0^\circ) \] \[ r_y = 7.30 \sin(47.0^\circ) \]

Step 2: Determine the velocity vector components

Given:

Velocity vector $ \vec{v} $ with magnitude 4.10 m/s and angle $ \theta_2 = 28.0^\circ $

Calculate the components of $ \vec{v} $: \[ v_x = 4.10 \cos(28.0^\circ) \] \[ v_y = 4.10 \sin(28.0^\circ) \]

Step 3: Calculate the angular momentum

Given:

Mass $ m = 1.20 $ kg

The angular momentum $ \vec{L} $ is given by: \[ \vec{L} = m (\vec{r} \times \vec{v}) \]

Calculate the cross product $ \vec{r} \times \vec{v} $: \[ \vec{r} \times \vec{v} = (r_x v_y - r_y v_x) \hat{z} \]

Substitute the values: \[ L_z = 1.20 \left[ (7.30 \cos(47.0^\circ) \cdot 4.10 \sin(28.0^\circ)) - (7.30 \sin(47.0^\circ) \cdot 4.10 \cos(28.0^\circ)) \right] \]

Final Answer

The magnitude of the angular momentum $ L $ is: \[ L = 1.20 \left[ (7.30 \cos(47.0^\circ) \cdot 4.10 \sin(28.0^\circ)) - (7.30 \sin(47.0^\circ) \cdot 4.10 \cos(28.0^\circ)) \right] \]

Calculate the numerical value: \[ L \approx 1.20 \left[ (7.30 \cdot 0.682 \cdot 4.10 \cdot 0.469) - (7.30 \cdot 0.731 \cdot 4.10 \cdot 0.883) \right] \] \[ L \approx 1.20 \left[ 9.48 - 23.52 \right] \] \[ L \approx 1.20 \left[ -14.04 \right] \] \[ L \approx -16.85 \, \text{kg} \cdot \text{m}^2/\text{s} \]

The magnitude of the angular momentum is approximately $ 16.85 \, \text{kg} \cdot \text{m}^2/\text{s} $.

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