A $1.3$kg ball on the end of a lightweight rod is located at $(x,y)=(0.3m,2.0m)$where the $y$axis is vertical. The other end of the rod is attached to a pivot at $(x,y)=(0m,3.0m)$. What is the torque about the pivot? Write your answer using unit vectors.

The rotating equivalent of force is torque. It is provided by.

$\overrightarrow{r}=\overrightarrow{r}\times \overrightarrow{F}$

The force is $\overrightarrow{F}$, and the distance between the pivot and the site of application of force is $\overrightarrow{r}$.

The right-hand thumb rule determines the direction of torque, which is a vector quantity.

For the application of force, see the diagram below:

The position vector of the mass is:

$\overrightarrow{r}=\left(x_2-x_1\right)\widehat{i}+\left(y_2-y_1\right)\widehat{j}$

Substitute $(3.0\mathrm{m},2.0\mathrm{m})$for $\left(x_2,y_2\right)$and $(0.0\mathrm{m},3.0\mathrm{m})$for $\left(x_1,y_1\right)$

$\overrightarrow{r}=(3.0\mathrm{m}-0.0\mathrm{m})\widehat{i}+(2.0\mathrm{m}-3.0\mathrm{m})\widehat{j}$

$=3.0\mathrm{m}\widehat{i}-1.0\mathrm{m}\widehat{j}$

On the mass, the force vector is:

$\overrightarrow{F}=m\overrightarrow{a}$

Here, $m$and $\overrightarrow{a}$are the mass and acceleration of the ball, respectively.

Substitute for $1.3\mathrm{kg}\text{for}m\text{and}-9.8\mathrm{m}/{\mathrm{s}}^2\widehat{j}\text{for}\overrightarrow{a}\text{.}$

$\overrightarrow{F}=-\left(1.3\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\widehat{j}$

The torque acting on mass is

$\overrightarrow{r}=\overrightarrow{r}\times \overrightarrow{F}$

Substitute $-\left(1.3\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\widehat{j}\text{for}\overrightarrow{F}\text{and}3.0\mathrm{m}\widehat{i}-1.0\mathrm{m}\widehat{j}\text{for}\overrightarrow{r}$

$\overrightarrow{t}=\left(\right(3.0\widehat{i}-1.0\widehat{j}\left)\mathrm{m}\right)\times \left(-\left(1.3\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\widehat{j}\right)$

$=-(38\mathrm{N}\cdot \mathrm{m})\widehat{k}$