Question: At time t, the vector $\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}{\mathbf{t}}^{\mathbf{2}}\hat{\mathbf{i}}\mathbf{-}\left(2.0t+6.0t^2\right)\hat{\mathbf{j}}$ gives the position of a3 .0 kgparticle relative to the origin of ancoordinate system (is in meters and tis in seconds). (a) Find an expression for the torque acting on the particle relative to the origin. (b) Is the magnitude of the particle’s angular momentum relative to the origin increasing, decreasing, or unchanging?

The mass of the particle is$m=3.0kg$

The position vector is$\overrightarrow{r}=4.0t^2\hat{i}-\left(2.0t+6.0t^2\right)\hat{j}$

Find the torque acting on the particle by using concept of Newton’s second law in angular form.

Formula:

$\frac{d\overrightarrow{l}}{dt}={\overrightarrow{\tau }}_{net}$

(a)

The expression of the velocity is,

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

Putting the position vector in above equation

role="math" localid="1661255607475" $$\begin{gathered} \overrightarrow{v}=\frac{d\left[4.0t^2\hat{i}-\left(2.0t+6.0t^2\right)\hat{j}\right]}{dt} \\ \Rightarrow \overrightarrow{v}=8.0t\hat{i}-\left(2.0+12t\right)\hat{j} \end{gathered}$$

Let position vector be$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$and velocity vector be$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}+v_z\hat{k}$.

The cross product of the position vector and velocity vector is

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

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

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

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

$$\begin{gathered} \overrightarrow{l}=m\left(\overrightarrow{r}\times \overrightarrow{v}\right) \\ \Rightarrow \overrightarrow{l}=m\left(x{v}_y-y{v}_x\right)\hat{k} \\ \Rightarrow \overrightarrow{l}=3.0kg\left[\left(4.0t^2\right)\left(-2.0-12t\right)+\left(2.0t+6.0t^2\right)\left(8.0t\right)\right]\hat{k} \\ \Rightarrow \overrightarrow{l}=3.0kg\left(-8.0t^2-48t^3+16t^2+48t^3\right) \end{gathered}$$

According to Newton’s second law in angular form, the sum of all torques acting on a particle is equal to the time rate of the change of the angular momentum of that particle.

$$\begin{gathered} \Rightarrow \overrightarrow{\tau }=\frac{d\overrightarrow{l}}{dt} \\ \Rightarrow \overrightarrow{\tau }=\left(24\hat{k}\right)\frac{d{t}^2}{dt} \\ \Rightarrow \overrightarrow{\tau }=48t\hat{k} \end{gathered}$$

(b)

The magnitude of the particle’s angular momentum relative to origin increases in proportion to the t².