In Fig. $\mathbf{10}\mathbf{-}\mathbf{35}$, three $\mathbf{0.0100}\mathbf{\text{\hspace{0.17em}kg}}\mathbf{}$particles have been glued to a rod of length $\mathbf{L}\mathbf{=}\mathbf{}\mathbf{6}\mathbf{.00}\mathbf{\text{\hspace{0.17em}cm}}\mathbf{}$and negligible mass and can rotate around a perpendicular axis through point $\mathit{O}$at one end. How much work is required to change the rotational rate

(a) from $\mathbf{0}$torole="math" localid="1660925307834" $\mathbf{20.0}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}\mathbf{\text{s}}$,

(b) from$\mathbf{20.0}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}\mathbf{\text{s}}$to$\mathbf{40.0}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}\mathbf{\text{s}}$, and

(c) from$\mathbf{40.0}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}\mathbf{\text{s}}$to$\mathbf{60.0}\mathbf{\text{\hspace{0.17em}rad}}\mathbf{/}\mathbf{\text{s}}$?

(d) What is the slope of a plot of the assembly’s kinetic energy (in joules) versus the square of its rotation rate (in radians squared per second squared)?

The length of rod is,$\mathrm{L}=6.00\text{\hspace{0.17em}cm}=0.06\text{\hspace{0.17em}m}$ .

The mass of particles is, $\mathrm{m}=0.0100\text{\hspace{0.17em}kg}$.

Find the rotational inertia using the formula for it. Then, using the work-energy theorem, we can find the work done from rotational K.E at different rotation rates. Then, from formula for rotational K.E, we can find the slopeof a plot of the assembly’s K.E vs the square of its rotation rate.

Formula:

$\begin{array}{c}\mathbf{I}\mathbf{=}{\mathbf{mR}}^{\mathbf{2}}\\ \mathbf{W}\mathbf{=}\mathbf{\Delta K}\mathbf{.}\mathbf{E}\\ \mathbf{P}\mathbf{=}\left|\frac{W}{t}\right|\end{array}$

The distance between two successive particles is,

$\mathrm{d}=\frac{\mathrm{L}}{3}=\frac{0.06\text{\hspace{0.17em}m}}{3}=0.02\text{\hspace{0.17em}m}$

Rotational inertia of the system is

$\mathrm{I}={\sum }^{\text{}}{\mathrm{mR}}^2$

$\begin{array}{l}\mathrm{I}={\mathrm{md}}^2+\mathrm{m}{\left(2\mathrm{d}\right)}^2+\mathrm{m}{\left(3\mathrm{d}\right)}^2\\ \mathrm{I}=14{\mathrm{md}}^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{l}\mathrm{I}=14\left(0.01\text{\hspace{0.17em}kg}\right){\left(0.02\text{\hspace{0.17em}m}\right)}^2\\ \mathrm{I}=0.000056{\text{\hspace{0.17em}kg.m}}^2\end{array}$

${\mathrm{\omega }}_{\mathrm{i}}=0\text{\hspace{0.17em}rad}/\text{s}$and${\mathrm{\omega }}_{\mathrm{f}}=20.0\text{\hspace{0.17em}rad}/\text{s}$

According to the work-energy theorem,

role="math" localid="1660926663039" $\begin{array}{l}\mathrm{W}=\mathrm{\Delta K}.\mathrm{E}\\ W=\frac{1}{2}I{\omega }_f^2-\frac{1}{2}I{\omega }_i^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\text{W}=\frac{1}{2}(0.000056{\text{\hspace{0.17em}kg.m}}^{\text{2}}){(20\text{\hspace{0.17em}rad}/\text{s})}^2-0\\ \text{W}=0.0112\text{\hspace{0.17em}J}\\ =11.2\text{\hspace{0.17em}mJ}\end{array}$

Therefore, work required to change the rotational rate from$0$ to $20.0\text{\hspace{0.17em}rad}/\text{s}$is,$11.2\text{\hspace{0.17em}mJ}$ .

${\mathrm{\omega }}_{\mathrm{i}}=20\text{\hspace{0.17em}rad}/\text{s}$and${\mathrm{\omega }}_{\mathrm{f}}=40.0\text{\hspace{0.17em}rad}/\text{s}$

According to the work-energy theorem,

role="math" localid="1660926632385" $\begin{array}{l}\mathrm{W}=\mathrm{\Delta K}.\mathrm{E}\\ W=\frac{1}{2}I{\omega }_f^2-\frac{1}{2}I{\omega }_i^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\text{W}=\frac{1}{2}(0.000056{\text{\hspace{0.17em}kg.m}}^{\text{2}}){(40\text{\hspace{0.17em}rad}/\text{s})}^2-\frac{1}{2}(0.000056{\text{\hspace{0.17em}kg.m}}^{\text{2}}){(20\text{\hspace{0.17em}rad}/\text{s})}^2\\ \text{W}=33.6\text{\hspace{0.17em}mJ}\end{array}$

Therefore, work required to change the rotational rate from$20.0\text{\hspace{0.17em}rad}/\text{s}$ to $40.0\text{\hspace{0.17em}rad}/\text{s}$is, $33.6\text{\hspace{0.17em}mJ}$.

$\begin{array}{c}{\mathrm{\omega }}_{\mathrm{i}}=40\text{\hspace{0.17em}rad}/\text{s}\\ {\mathrm{\omega }}_{\mathrm{f}}=60.0\text{\hspace{0.17em}rad}/\text{s}\end{array}$

According to the work-energy theorem,

$\begin{array}{l}\mathrm{W}=\mathrm{\Delta K}.\mathrm{E}\\ \mathrm{W}=\frac{1}{2}{\mathrm{I\omega }}_{\mathrm{f}}^2-\frac{1}{2}{\mathrm{I\omega }}_{\mathrm{i}}^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\text{W}=\frac{1}{2}(0.000056{\text{\hspace{0.17em}kg.m}}^{\text{2}}){(60\text{\hspace{0.17em}rad}/\text{s})}^2-\frac{1}{2}(0.000056{\text{\hspace{0.17em}kg.m}}^{\text{2}}){(40\text{\hspace{0.17em}rad}/\text{s})}^2\\ \text{W}=56.0\text{\hspace{0.17em}mJ}\end{array}$

Therefore, work required to change the rotational rate from$40.0\text{\hspace{0.17em}rad}/\text{s}$ to$60.0\text{\hspace{0.17em}rad}/\text{s}$ is, $56.0\text{\hspace{0.17em}mJ}$.

Assembly’s K.E is,

$K.E=\frac{1}{2}I{\omega }^2$

So, the slope of the graph$K.E$vs${\mathrm{\omega }}^2$is$\frac{1}{2}I$.

i.e,

$\begin{array}{c}\frac{1}{2}I=\left(0.5\right)\left(0.000056{\text{\hspace{0.17em}kg.m}}^{\text{2}}\right)\\ =2.8\times {10}^{-5}{\text{\hspace{0.17em}Js}}^{\text{2}}/{\text{rad}}^{\text{2}}\end{array}$

Therefore, the slope of the plot of the assembly’s K.E vs the square of its rotation rate is $2.8\times {10}^{-5}{\text{\hspace{0.17em}Js}}^{\text{2}}/{\text{rad}}^{\text{2}}$.