A disk of radius$\mathbf{\text{8 cm}}$is pulled along a frictionless surface with a force of$\mathbf{\text{10 N}}$ by a string wrapped around the edge (Figure 11.102). $\mathbf{\text{24 cm}}$of string has unwound off the disk. What are the magnitude and direction of the torque exerted about the center of the disk at this instant?

Torque is the measure of the force that can cause an object to rotate about an axis. Force is what causes an object to accelerate in linear kinematics. Similarly, torque is what causes an angular acceleration. Hence, torque can be defined as the rotational equivalent of linear force.

Angular momentum for the given system with the momentum of inertia$\left(I\right)$and angular velocity $\left(\omega \right)$is given as

$L=I\omega$

Torque is defined as rate of change of angular momentum of the system.

$\tau =\frac{dL}{dt}$

The other expression for the torque is,

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$ ….. (1)

Here, $L$is the angular momentum, $\tau$is the torque, $t$is time,$\overrightarrow{r}$ is the distance of disk having a value $\left(\text{8 cm}\right)\hat{j}$, and$\overrightarrow{F}$ is the force having a value of $\left(10\text{N}\right)\hat{i}$.

Assume right direction as positive $x-$direction, and upward direction as positive$y-$direction.

Consider the unit vector towards right, and upwards directions as $\hat{i}$and $\hat{j}$respectively.

Substitute $\left(10\text{N}\right)\hat{i}$for$\overrightarrow{F},$and$\left(\text{8 cm}\right)\hat{j}$ for$\overrightarrow{r}$into equation (1).

$\begin{array}{rcl}\overrightarrow{\tau }& =& \left(\text{8 cm}\right)\hat{j}\times \left(10\text{N}\right)\hat{i}\\ & =& -\left[\left(\text{8 cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(10\text{N}\right)\right]\hat{k}\\ & =& -\text{0.8 N}·\text{m}\hat{k}\\ & & \end{array}$

Hence, the vector form of torque is that is torque direction is into the page.