91Ó°ÊÓ

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 concept which says that an object will be in equilibrium when the net force acting on it is zero. Also, use the relation which says that torque is equal to product of moment of inertia and angular acceleration.

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

Here, the center of the disk will not move up or down, therefore net force on the disk system will be zero.

Thus, the force that is pulling the string will be equal to weight of the disk.

Mathematically, tension in the string $T$will be equal to $mg.$

The expression which relates weight of the disk and tension in the string is,

$\begin{array}{rcl}\mathrm{Ï„}& =& mgR\\ & =& I\mathrm{Î\pm }\\ & & \end{array}$

Here,$m$is the mass of the disk,g is the acceleration due to gravity,$I$is the moment of inertia and$\mathrm{Î\pm }$is the angular acceleration of the disk,

Thus, the angular acceleration of the object is,

$\mathrm{Î\pm }=\frac{mgR}{I}$

The angular speed after time interval is $t$,

$\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }t$

Here,${\mathrm{Ó\neg }}_0$is the initial angular speed, and$\mathrm{Ó\neg }$is the final angular speed.

Substitute

$\begin{array}{rcl}m& =& 1.2\text{kg}\\ I& =& 0.0015\text{kg}·{\text{m}}^{\text{2}}\\ g& =& 9.8{\text{m/s}}^{\text{2}}\\ R& =& 0.06\text{m}\\ & & \end{array}$

$\begin{array}{rcl}\mathrm{Î\pm }& =& \frac{mgR}{I}\\ \mathrm{Î\pm }& =& \frac{\left(1.2\text{kg}\right)\left(9.8{\text{m/s}}^{\text{2}}\right)\left(0.06\text{m}\right)}{\left(0.0015\text{kg}·{\text{m}}^2\right)}\\ & =& 470.4{\text{rad/s}}^{\text{2}}\\ & & \end{array}$

The angular speed of the disk after time$t$is $\mathrm{Ó\neg }={\mathrm{Ó\neg }}_0+\mathrm{Î\pm }t$,

Substitute

$\begin{array}{rcl}{\mathrm{Ó\neg }}_0& =& 12\text{rad/s}\\ \mathrm{Î\pm }& =& 470.4{\text{rad/s}}^{\text{2}}\\ t& =& 0.1\text{s}\\ & & \end{array}$

$\begin{array}{rcl}\mathrm{Ó\neg }& =& {\mathrm{Ó\neg }}_0+\mathrm{Î\pm }t\\ \mathrm{Ó\neg }& =& \left(12\text{rad/s}\right)+\left(470.4{\text{rad/s}}^{\text{2}}\right)\left(0.1\text{s}\right)\\ & =& 59\text{rad/s}\\ & & \end{array}$

Therefore, angular speed of the object after a time interval is $59\text{rad/s}$.

Determine the angle of rotation of the yo-yo by using the following formula:

$\mathrm{Δ}\mathrm{θ}=\left(\frac{{\mathrm{Ó\neg }}_i+{\mathrm{Ó\neg }}_f}{2}\right)\mathrm{Δ}t$

Here, $\mathrm{Δ}\mathrm{θ}$represents the change in angle, $\mathrm{Δ}t$is elapsed time, and ${\mathrm{Ó\neg }}_i$are the initial and ${\mathrm{Ó\neg }}_f$final angular speeds of the yo-yo respectively.

Substitute

$\begin{array}{rcl}{\mathrm{Ó\neg }}_i& =& 12\text{rad/s}\\ {\mathrm{Ó\neg }}_f& =& 59\text{rad/s}\\ \mathrm{Δ}t& =& 0.1\text{s}\\ & & \end{array}$

$\begin{array}{rcl}\mathrm{Δ}\mathrm{θ}& =& \left(\frac{12\text{rad/s}+59\text{rad/s}}{2}\right)\left(0.1\text{s}\right)\\ & =& \left(\frac{71}{2}\right)\left(0.1\right)\text{rad}\\ & =& 3.55\text{rad}\\ & & \end{array}$

Therefore, the angle through which the yo-yo turns is$3.55\text{rad}$.

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

$\begin{array}{rcl}\mathrm{Δ}\mathrm{θ}& =& \left(3.55\text{rad}\right)\left(\frac{360°}{2\mathrm{π}\text{rad}}\right)\\ & =& \left(3.55\right)\left(\frac{7×360}{2×22}\right)\\ & =& 203.3°\\ & & \end{array}$

Therefore, the angle through which the yo-yo turns is $203.3°$.