Question

A 2.0kg particle-like object moves in a plane with velocity components v_x = 30m/sand v_y = 60m/sas it passes through the point with (x,y)coordinates of (3.0, -4.0)m.Just then, in unit-vector notation, (a) what is its angular momentum relative to the origin and (b) what is its angular momentum relative to the point located at (-2.0, -2.0)m?

Short answer

1. The angular momentum of the object relative to the origin is$\stackrel{\rightharpoonup }{l}=\left(60\times {10}^{2}kg.\frac{m^2}{s}\right)\hat{k}$
2. The angular momentum of the object relative to the point located at (-2.0,-2.0)m is$\stackrel{\rightharpoonup }{l}=\left(7.2\times {10}^{2}kg.\frac{m^2}{s}\right)\hat{k}$

Step-by-step solution

Identification of given data

m = 2.0 kg

v_x = 30 m/s

v_y = 60 m/s

(x,y) = (3.0, -4.0)m

To understand the concept linear momentum

The problem deals with the calculation of angular momentum. The angular momentum of a rigid object is product of the moment of inertia and the angular velocity. It is analogous to linear momentum.

Formula:

$\stackrel{\rightharpoonup }{l}=m\left(\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{v}\right)$

(a) Determining the angular momentum of the z = 0m object relative to origin

Let position vector is and velocity vector is $\stackrel{\rightharpoonup }{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$.

The cross product of the position vector and velocity vector is,

$\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{v}=\left(y{v}_z-z{v}_y\right)\hat{i}+\left(z{v}_x-x{v}_z\right)\hat{j}+\left(x{v}_y-y{v}_x\right)\hat{k}$

In the given position and velocity vector are z = 0 m and v = 0m/s. Then,

$\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{v}=\left(x{v}_y-y{v}_x\right)\hat{k}$

The angular momentum of the object with position vector and velocity vector is

$\begin{array}{rcl}\stackrel{\rightharpoonup }{l}& =& m\left(\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{v}\right)\\ & =& m\left(x{v}_y-y{v}_x\right)\hat{k}\\ & =& 2.0kg\left[\left(3.0m\right)\left(60m/s\right)-\left(-4.0m\right)\left(30m/s\right)\right]\hat{k}\\ & =& \left(6.0\times {10}^{2}kg.\frac{m^2}{s}\right)\hat{k}\end{array}$

(b) Determining the angular momentum of the object relative to the point located at (-2.0,-2.0)m

The position vectors are $\stackrel{\rightharpoonup }{r}=\left(3.0m\right)\hat{i}-\left(4.0m\right)\hat{j}$and ${\stackrel{\rightharpoonup }{r}}_0=\left(-2.0m\right)\hat{i}+\left(-2.0m\right)\hat{j}$

According to the vector subtraction law,

$\begin{array}{rcl}\stackrel{\rightharpoonup }{r}\text{'}& =& \hat{r}-{\hat{r}}_0\\ & =& \left[\left(3.0m\right)\hat{i}-\left(4.0m\right)\hat{j}\right]-\left[\left(-2.0m\right)\hat{i}+\left(-2.0m\right)\hat{j}\right]\\ & =& \left(5.0m\right)\hat{i}-\left(2.0m\right)\hat{j}\end{array}$

Let position vector be $\stackrel{\rightharpoonup }{r}\text{'}=x\text{'}\hat{i}+y\text{'}\hat{j}+z\text{'}\hat{k}$and velocity vector be $\stackrel{\rightharpoonup }{v}\text{'}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$. The cross product of the position vector and velocity vector is

$\stackrel{\rightharpoonup }{r}\text{'}\times \stackrel{\rightharpoonup }{v}=\left(y\text{'}{v}_z-z\text{'}{v}_y\right)\hat{i}+\left(z\text{'}{v}_x-x\text{'}{v}_z\right)\hat{j}+\left(x\text{'}{v}_y-y\text{'}{v}_x\right)\hat{k}$

In the given position and velocity vector, z = 0 m and v = 0 m/s. Then

$\stackrel{\rightharpoonup }{r}\text{'}\times \stackrel{\rightharpoonup }{v}=\left(x\text{'}{v}_y-y\text{'}{v}_x\right)\hat{k}$

The angular momentum of the object with position vector and velocity vector is

$\begin{array}{rcl}\stackrel{\rightharpoonup }{l}& =& m\left(\stackrel{\rightharpoonup }{r}\text{'}\times \stackrel{\rightharpoonup }{v}\right)\\ & =& m\left(x\text{'}{v}_y-y\text{'}{v}_x\right)\hat{k}\\ & =& 2.0kg\left[\left(5.0m\right)\left(60m/s\right)-\left(-2.0m\right)\left(30m/s\right)\right]\hat{k}\\ & =& \left(7.2\times {10}^{2}kg.\frac{m^2}{s}\right)\hat{k}\end{array}$