String is wrapped around an object of mass M and moment of inertia I (the density of the object is not uniform). With your hand you pull the string straight up with some constant force F such that the center of the object does not move up or down, but the object spins faster and faster (Figure 9,62). This is like a${\mathit{y}}_{\mathbf{0}}\mathbf{-}{\mathit{y}}_{\mathbf{0}}$; nothing but the vertical string touches the object.

When your hand is a height${\mathit{y}}_{\mathbf{0}}$above the floor, the object has an angular speed${\mathit{\omega }}_{\mathbf{0}}$. When your hand has risen to a height y above the floor, what is the angular speed$\mathit{\omega }$of the object? Your result should not contain$\mathit{F}$or the (unknown) radius of the object. Explain the physics principles you are using.

The given data is listed below as follows,

• The moment of inertia of the string is, I
• The mass of the object is, M
• The force that pulls the string is,F
• Initially, the height of the hand above the floor is,$y_0$
• Initially, the height of the hand above the floor is,y
• The initial angular speed of the object is,${\omega }_0$

Angular speed is the ratio of change in angular rotation to time. In physics, it is also known as angular velocity and rotational velocity. The magnitude of this is based on how an object rotates or revolves.

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

$W=K_{trans}+K_{rot}$ …(i)

Here,$W$is the amount of the work done, $K_{tras}$is transitional kinetic energy is zero because mass is not moving and$K_{rot}$is rotational kinetic energy.

Substitute all the values in equation (i).

$K_{rot}=W$ …(ii)

The equation of the work done is expressed as:

$W=F.d$ …(iii)

Here, F is the force exerted and d is the distance through which the force is exerted.

The equation of the force can be calculated as:

$F=m.g$

Here, g is the acceleration due to gravity.

The equation of the work done is expressed as:

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

Here, Fis the force exerted,y is the final height and $y_0$is the initial height.

Substitute all the values in the equation.

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

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

$K_{rot}=\frac{1}{2}I{\omega }^2-\frac{1}{2}I{{\omega }^2}_0$

Here, I is the moment of inertia, $\omega$is the final angular speed, and ${\omega }_0$is the initial angular speed

Substitute all the values in equation (ii).

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

Thus, the angular speed of an object is $\sqrt{\frac{2mg\left(y-y_0\right)}{I}}+{\omega }_0^2$.