A box and its contents have a total mass$\mathbf{M}$. A string passes through a hole in the box (Figure$\mathbf{9}\mathbf{.}\mathbf{57}$), and you pull on the string with a constant force$\mathbf{F}$(this is in outer space—there are no other forces acting).

(a) Initially the speed of the box was${\mathbf{v}}_{\mathbf{i}}$. After the box had moved a long distance$\mathbf{w}$, your hand had moved an additional distance$\mathbf{d}$(a total distance of$\mathbf{w}\mathbf{+}\mathbf{d}$), because additional string of length$\mathbf{d}$came out of the box. What is now the speed${\mathbf{v}}_{\mathbf{i}}$of the box? (b) If we could have looked inside the box, we would have seen that the string was wound around a hub that turns on an axle with negligible friction, as shown in Figure$\mathbf{9}\mathbf{.}\mathbf{58}$. Three masses, each of mass, are attached to the hub at a distance$\mathbf{r}$from the axle. Initially the angular speed relative to the axle was${\mathbf{\omega }}_{\mathbf{1}}$. In terms of the given quantities, what is the final angular speed relative to the axis,${\mathbf{\omega }}_{\mathbf{f}}$?

The given data is listed below as,

• The total mass of the box and its contents is, $M$
• The force required to pull the string is,$F$
• The initial speed of the box is,$V_i$
• Initially, the distance moved by the box is, $W$
• The distance moved by the hand is,$d$
• The total distance moved by the hand and the box is$w+d$
• The mass of the three masses is,$m$
• The distance at which the masses are attached is,$r$
• The angular speed relative to the axle is,${\omega }_i$

According to the work-energy theorem, the work done on an object is the same as that of the change in kinetic energy of the object.

The law of angular motion states that a body will continue to rotate with a constant angular speed unless an external torque acts on it.

The work-energy theorem gives the final speed and the law of angular motion gives the final angular speed of the box.

The equation of work done in the box is expressed as,

$$\begin{gathered} \mathbf{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Mv}}^{\mathbf{2}}\mathbf{}{\mathbf{}}_{\mathbf{f}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Mv}}_{\mathbf{i}}^{\mathbf{2}} \\ \mathbf{F}\mathbf{.}\mathbf{d}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{M}\left(v^2{}_f-v_i^2\right) \end{gathered}$$

Here,$M$ is the total mass of the box,$v_f$ is the final velocity of the box, $v_I$and is the initial velocity of the box.

Substitute w for $d$in the above expression.

$$\begin{gathered} F.w=\frac{1}{2}M\left(v_f^2-v_i^2\right) \\ F.w+\frac{1}{2}M{v}_i^2=\frac{1}{2}M{v}_i^2 \\ {v}_f^2=\frac{F.w+\frac{1}{2}M{v}_i^2}{\frac{1}{2}M} \\ {v}_f=\sqrt{\frac{F.w+\frac{1}{2}M{v}_i^2}{\frac{1}{2}M}} \\ =\sqrt{\frac{2Fw}{M}}+v_i^2 \end{gathered}$$

Thus, the speed of the box is $\sqrt{\frac{2Fw}{M}}+v_i^2$.

The expression for the final angular speed of the box can be expressed as,

$\omega =\frac{v}{r}$

Here,$v$ is the final velocity and $r$is the distance of the masses from the axle.

Substitute all the values in the above expression

$\omega =\frac{\sqrt{{\displaystyle \frac{2Fw}{M}+v_i^2}}}{r}$

Substitute for and for in the above expression.

$$\begin{gathered} \omega =\frac{\sqrt{{\displaystyle \frac{2Fw}{M}+v_i^2}}}{r} \\ =\sqrt{\frac{2Fd}{3m{r}^2}+\frac{v_i^2}{r_i^2}} \\ =\sqrt{\frac{2Fd}{3m{r}^2}+}{\omega }_i^2 \end{gathered}$$

Thus, the final angular speed relative to the axis is$\sqrt{\frac{2Fd}{3m{r}^2}+}{\omega }_i^2$.