If you did not already do problem P63 do it now. Also calculate numerically the angle through which the apparatus turns, in radians and degrees.

The multiplication of mass and the square of a distance of the particle from the rotation axis are known as the moment of inertia.

Use the relation which says that torque is equal to product of moment of inertia and angular acceleration. Also, use the expression which relates the final angular speed, initial angular speed, angular acceleration, and time elapsed.

The diagram represents a disk like object that can rotate about a vertical axis passing through its center. It is wrapped by a string with its one end in the hand and pulling it in horizontal direction by a force $\text{'}F\text{'}$.

The expression which relates torque acting on the object, tension in the string, moment of inertia of the object, and the angular acceleration is,

$\begin{array}{rcl}\tau & =& I\alpha \\ & =& FR\\ & & \end{array}$

Here, $F$is the force acting on the string, $R$is the radius of the disk, $I$is the moment of inertia of the disk, and $\alpha$is the angular acceleration of the disk.

Thus, the angular acceleration of the object is,

$\alpha =\frac{FR}{I}$

The expression for the angular speed after time interval $t$is,

${\omega }_i={\omega }_f+\alpha t$

Here,${\omega }_i$is the initial angular speed, and ${\omega }_f$is the final angular speed.

Substitute, $\frac{FR}{I}$for $\alpha$ in

$\begin{array}{rcl}{\omega }_f& =& {\omega }_i+\alpha t\\ {\omega }_f& =& {\omega }_i+\left(\frac{FR}{I}\right)t\\ & & \end{array}$

The total moment of inertia of the disk-four masses system is,

$I=I_{disk}+4I_{mass}$

The moment of inertia of the disk is$I_{disk}=\frac{1}{2}M{R}^2$

Here,$M$is the mass of the disk.

The moment of inertia of a mass about the center of the disk is,

$I_{mass}=m{b}^2$

Here,$m$is the mass of the small ball, and$b$is the distance of the mass from the center of the disk.

Thus, total moment of inertia of the system is,

$I=\frac{1}{2}M{R}^2+4m{b}^2$

Substitute $I$in $\omega ={\omega }_0+\left(\frac{FR}{I}\right)t$

$\begin{array}{rcl}{\omega }_f& =& {\omega }_i+\left(\frac{FR}{\frac{1}{2}M{R}^2+4m{b}^2}\right)t\\ \text{where,}{\omega }_0& =& 0\\ F& =& 21\text{N}\\ R& =& 0.11\text{m}\\ b& =& 0.14\text{m}\\ M& =& 2\text{kg}\\ m& =& 0.4\text{kg}\\ t& =& 0.2\text{s}\end{array}$

$\begin{array}{rcl}{\omega }_i& =& \left(0\right)+\left(\frac{\left(21\text{N}\right)\left(0.11\text{m}\right)}{\frac{1}{2}\left(1.2\text{kg}\right){\left(0.11\text{m}\right)}^2+4\left(0.4\text{kg}\right){\left(0.14\text{m}\right)}^2}\right)\left(0.2\text{s}\right)\\ & =& 12\text{rad/s}\\ & & \end{array}$

Therefore, the final angular speed is $12\text{rad/s}$.

Determine the angle of rotation of the apparatus by using the following formula:

$\Delta \theta =\left(\frac{{\omega }_i+{\omega }_f}{2}\right)\Delta t$

Here,represents the change in angle,is elapsed time.

$\begin{array}{rcl}{\omega }_i& =& 0\text{rad/s}\\ {\omega }_f& =& 12\text{rad/s}\\ \Delta t& =& 0.2\text{s}\\ & & \end{array}$

Substitutethese values,

$\begin{array}{rcl}\Delta \theta & =& \left(\frac{0\text{rad/s}+12\text{rad/s}}{2}\right)\left(0.2\text{s}\right)\\ & =& \left(6\right)\left(0.2\right)\text{rad}\\ & =& 1.2\text{rad}\\ & & \end{array}$

Therefore, the angle through which the apparatus turns is$1.2\text{rad}$.

Convert the angle of rotation into degrees by using the following conversion:

$\begin{array}{rcl}\Delta \theta & =& \left(1.2\text{rad}\right)\left(\frac{360°}{2\pi \text{rad}}\right)\\ & =& \left(1.2\right)\left(\frac{7\times 360}{2\times 22}\right)\\ & =& 68.7°\\ & & \end{array}$

Hence, the angle through which the apparatus turns is $68.7°$.