At one instant, force $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathit{N}$acts on a 0.25kgobject that has position vector $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{r}}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)}\mathit{m}$and velocity vector$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{v}}\mathbf{=}\mathbf{(}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{)}\mathit{m}\mathbf{/}\mathit{s}$About the origin and in unit-vector notation, (a) What is the object’s angular momentum? (b) What is the torque acting on the object?

The mass of the object is m = 0.25 kg

The force acting on the object is$\stackrel{\rightharpoonup }{F}=4.0\hat{j}N$

The position vector is$\stackrel{\rightharpoonup }{r}=\left(2.0m\right)\hat{i}-\left(2.0m\right)\hat{k}$

The velocity vector is$\stackrel{\rightharpoonup }{v}=\left(-5.0m/s\right)\hat{i}+\left(5.0m/s\right)\hat{k}$

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. It also deals with the torque. It is a measure of the force that can cause an object to rotate about an axis

Formulae:

$$\begin{gathered} \stackrel{\rightharpoonup }{l}=m\left(\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{v}\right) \\ \stackrel{\rightharpoonup }{\tau }=\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{F} \end{gathered}$$

Let position vector is $\stackrel{\rightharpoonup }{r}=x\hat{i}+y\hat{j}+z\hat{k}$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, y = 0 and v_y = 0m/s. Then,

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

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

$\begin{array}{rcl}\stackrel{\rightharpoonup }{l}& =& m\left(\stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{v}\right)\\ & =& m\left(z{v}_x-x{v}_z\right)\hat{j}\\ & =& 0.25kg\left[\left(-2.0m\right)\left(-5.0m/s\right)-\left(2.0m\right)\left(5.0m/s\right)\right]\hat{j}\\ & =& 0kg\frac{m^2}{s}\end{array}$

To find torque acting on the object, force acting on it will be $\stackrel{\rightharpoonup }{F}={\stackrel{\rightharpoonup }{F}}_x\hat{i}+{\stackrel{\rightharpoonup }{F}}_y\hat{j}+{\stackrel{\rightharpoonup }{F}}_z\hat{k}$.

The expression of torque is,

$\begin{array}{rcl}\stackrel{\rightharpoonup }{\tau }& =& \stackrel{\rightharpoonup }{r}\times \stackrel{\rightharpoonup }{F}\\ & =& \left(y{F}_z-z{F}_y\right)\hat{i}+\left(z{F}_x-x{F}_z\right)\hat{j}+\left(x{F}_y-y{F}_x\right)\hat{k}\end{array}$

In the given position and force vector, y= 0m and

$\begin{array}{rcl}{F}_x& =& F_z\\ & =& 0N\\ \stackrel{\rightharpoonup }{\tau }& =& \left(-z{F}_y\right)\hat{i}+\left(x{F}_y\right)\hat{k}\\ & =& -\left(-2.0m\right)\left(4.0N\right)\hat{i}+\left(2.0m\right)\left(4.0N\right)\hat{k}\\ & =& \left(8.0N.m\right)\hat{i}+\left(8.0N.m\right)\hat{k}\end{array}$