91Ó°ÊÓ

The given data is listed below as,

• The moment of inertia of the string is,$I=0.0015\mathrm{kg}-{\mathrm{m}}^2$
• The mass of the object is,m=1.5kg
• Initially, the height of the hand above the floor is,$y_0=0.25\mathrm{m}$
• The initial angular speed of the object is, ${\mathrm{Ó\neg }}_0=12\mathrm{rad}/\mathrm{s}$
• Finally, the height of the hand above the floor is, y=0.35 m

Angular speed is described as the ratio of the change in the angular rotation to the time. However, it is also known as rotational or angular velocity. This is also helps in understanding the rotation of an object.

The equation of the change in energy of the system is:

$W=K_{trans}+K_{rot}$ …(¾±)

Here, $K_{tras}$is transitional kinetic energy is zero because mass is not moving and $K_{rot}$is rotational kinetic energy.

Equation (i) can be expressed as follows:

$k_{rot}=W$ …(¾±¾±)

The equation of the force can be calculated as:

$F=m.g$

Here, m is the mass of the object, and g is the acceleration due to gravity.

The equation of the work done is expressed as:

$W=F\left(y-y_0\right)$

Here, $F$is the force exerted, y is the final height and $y^0$is the initial height.

Substitute all the values in above equation.

$W=m.g.\left(y-y_0\right)$

As the moment of inertia is given, the equation of the rotational kinetic energy becomes:

$K_{rot}=\frac{1}{2}I{\mathrm{Ó\neg }}^2-\frac{1}{2}I{{\mathrm{Ó\neg }}^2}_0$

Here, I is the moment of inertia, $\mathrm{Ó\neg }$is the final angular speed, and ${\mathrm{Ó\neg }}^2$is the initial angular speed.

Substitute all the values in equation (ii).

$$\begin{gathered} \frac{1}{2}I{\mathrm{Ó\neg }}^2-\frac{1}{2}I{{\mathrm{Ó\neg }}^2}_0=m.g.\left(y-y_0\right) \\ I\left({\mathrm{Ó\neg }}^2-{{\mathrm{Ó\neg }}^2}_0\right)=2.m.g.\left(y-y_0\right) \\ \left({\mathrm{Ó\neg }}^2-{{\mathrm{Ó\neg }}^2}_0\right)=\frac{2.m.g.\left(y-y_0\right)}{I} \\ \mathrm{Ó\neg }=\sqrt{{{\mathrm{Ó\neg }}_0}^2+\frac{2.m.g.\left(y-y_0\right)}{I}} \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{Ó\neg }=\sqrt{{\left(12\mathrm{rad}/\mathrm{s}\right)}^2+\frac{2×\left(1.2\mathrm{kg}\right)×\left(0.35\mathrm{m}-0.25\mathrm{m}\right)}{0.0015\mathrm{kg}.{\mathrm{m}}^2}} \\ =\sqrt{144{\mathrm{rad}}^2/{\mathrm{s}}^2+\frac{2×\left(1.2\mathrm{kg}\right)×\left(0.98{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{0.0015\mathrm{kg}.{\mathrm{m}}^2}} \\ =\sqrt{144{\mathrm{rad}}^2/{\mathrm{s}}^2+\frac{\left(23.52\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\right)}{0.0015\mathrm{kg}.{\mathrm{m}}^2}} \\ =144\mathrm{rad}/s \end{gathered}$$

Thus, the angular speed of the object is $144\mathrm{rad}/s$.