91Ó°ÊÓ

m = 2.0 kg

v_x = 30 m/s

v_y = 60 m/s

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

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:

$\stackrel{â‡\P Ä}{l}=m\left(\stackrel{â‡\P Ä}{r}×\stackrel{â‡\P Ä}{v}\right)$

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

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

$\stackrel{â‡\P Ä}{r}×\stackrel{â‡\P Ä}{v}=\left(y{v}_z-z{v}_y\right)\hat{i}+\left(z{v}_x-x{v}_z\right)\hat{j}+\left(x{v}_y-y{v}_x\right)\hat{k}$

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

$\stackrel{â‡\P Ä}{r}×\stackrel{â‡\P Ä}{v}=\left(x{v}_y-y{v}_x\right)\hat{k}$

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

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

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

According to the vector subtraction law,

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

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

$\stackrel{â‡\P Ä}{r}\text{'}×\stackrel{â‡\P Ä}{v}=\left(y\text{'}{v}_z-z\text{'}{v}_y\right)\hat{i}+\left(z\text{'}{v}_x-x\text{'}{v}_z\right)\hat{j}+\left(x\text{'}{v}_y-y\text{'}{v}_x\right)\hat{k}$

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

$\stackrel{â‡\P Ä}{r}\text{'}×\stackrel{â‡\P Ä}{v}=\left(x\text{'}{v}_y-y\text{'}{v}_x\right)\hat{k}$

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

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