A disk of radius $\mathbf{\text{0.2 m}}$and moment of inertia $\mathbf{\text{1.5 kg}}\mathbf{·}{\mathbf{\text{m}}}^{\mathbf{\text{2}}}$ is mounted on a nearly frictionless axle (Figure 11.106). A string is wrapped tightly around the disk, and you pull on the string with a constant force of $\mathbf{\text{25 N}}$. After a while the disk has reached an angular speed of$\mathbf{2}\mathit{r}\mathit{a}\mathit{d}\mathbf{/}\mathit{s}\mathbf{.}$What is its angular speed $\mathbf{0}\mathbf{.}\mathbf{1}$seconds later? Explain briefly.

Angular speed measures how fast the central angle of a rotating body changes with respect to time.

The torque acting on a disk is defined as the product of the force acting on the torque and the perpendicular distance. It is expressed as,

$\tau =FR$

Here, $\tau$is the torque on the disc, $F$is the force acting on the disc, and $R$is the perpendicular distance of the acting point of force from the axis of rotation.

The relation between angular acceleration and torque is,

$\tau =I\alpha$

Here,localid="1668512897485" $I$is the moment of inertial of the disc andlocalid="1668512903156" $\alpha$is the angular acceleration.

Substitute $FR$for $\tau$in the above equation,

$FR=I\alpha$

Rearrange above equation forin the above equation.

$\alpha =\frac{FR}{I}$ …… (1)

As given,

The force, $F=25\text{N}$

The radius, $R=0.2\text{m}$

The angular momentum, $I=1.5\text{kg}·{\text{m}}^2$

So substitute these values into equation (1).

$\begin{array}{rcl}\alpha & =& \frac{\left(25\text{N}\right)\left(0.2\text{m}\right)}{1.5\text{kg}·{\text{m}}^2}\\ & =& 3.3{\text{3s}}^{-\text{2}}\\ & & \end{array}$

Relation between final an initial angular velocity is,

${\omega }_f={\omega }_i+\alpha t$ ….. (2)

Here,${\omega }_f$is final angular velocity,${\omega }_i$is initial angular velocity having a value of 2 rad/s, $\alpha$is angular acceleration, and$t$ is the time having a values of $0.1s$.

Substitute known values into equation (2).

role="math" localid="1668514392912" $\begin{array}{rcl}{\omega }_f& =& \left(2\right)+\left(3.33\right)\left(0.1\right)\\ & =& 2.33\\ & & \end{array}$

Hence, angular speed of the disc after $\text{0.1 s}$is 2rad/s.