91Ó°ÊÓ

The length of rod is,$\mathrm{L}=6.00\text{ c³¾}=0.06\text{″¾}$ .

The mass of particles is, $\mathrm{m}=0.0100\text{ k²µ}$.

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:

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

The distance between two successive particles is,

$\mathrm{d}=\frac{\mathrm{L}}{3}=\frac{0.06\text{″¾}}{3}=0.02\text{″¾}$

Rotational inertia of the system is

$\mathrm{I}={∑}^{\text{​}}{\mathrm{mR}}^2$

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

Substitute all the value in the above equation.

$\begin{array}{l}\mathrm{I}=14\left(0.01\text{ k²µ}\right){\left(0.02\text{″¾}\right)}^2\\ \mathrm{I}=0.000056{\text{ k²µ.m}}^2\end{array}$

${\mathrm{Ó\neg }}_{\mathrm{i}}=0\text{ r²¹»å}/\text{s}$and${\mathrm{Ó\neg }}_{\mathrm{f}}=20.0\text{ r²¹»å}/\text{s}$

According to the work-energy theorem,

role="math" localid="1660926663039" $\begin{array}{l}\mathrm{W}=\mathrm{Δ°­}.\mathrm{E}\\ W=\frac{1}{2}I{\mathrm{Ó\neg }}_f^2−\frac{1}{2}I{\mathrm{Ó\neg }}_i^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\text{W}=\frac{1}{2}(0.000056{\text{ k²µ.m}}^{\text{2}}){(20\text{ r²¹»å}/\text{s})}^2−0\\ \text{W}=0.0112\text{ J}\\ =11.2\text{″¾´³}\end{array}$

Therefore, work required to change the rotational rate from$0$ to $20.0\text{ r²¹»å}/\text{s}$is,$11.2\text{″¾´³}$ .

${\mathrm{Ó\neg }}_{\mathrm{i}}=20\text{ r²¹»å}/\text{s}$and${\mathrm{Ó\neg }}_{\mathrm{f}}=40.0\text{ r²¹»å}/\text{s}$

According to the work-energy theorem,

role="math" localid="1660926632385" $\begin{array}{l}\mathrm{W}=\mathrm{Δ°­}.\mathrm{E}\\ W=\frac{1}{2}I{\mathrm{Ó\neg }}_f^2−\frac{1}{2}I{\mathrm{Ó\neg }}_i^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\text{W}=\frac{1}{2}(0.000056{\text{ k²µ.m}}^{\text{2}}){(40\text{ r²¹»å}/\text{s})}^2−\frac{1}{2}(0.000056{\text{ k²µ.m}}^{\text{2}}){(20\text{ r²¹»å}/\text{s})}^2\\ \text{W}=33.6\text{″¾´³}\end{array}$

Therefore, work required to change the rotational rate from$20.0\text{ r²¹»å}/\text{s}$ to $40.0\text{ r²¹»å}/\text{s}$is, $33.6\text{″¾´³}$.

$\begin{array}{c}{\mathrm{Ó\neg }}_{\mathrm{i}}=40\text{ r²¹»å}/\text{s}\\ {\mathrm{Ó\neg }}_{\mathrm{f}}=60.0\text{ r²¹»å}/\text{s}\end{array}$

According to the work-energy theorem,

$\begin{array}{l}\mathrm{W}=\mathrm{Δ°­}.\mathrm{E}\\ \mathrm{W}=\frac{1}{2}{\mathrm{IÓ\neg }}_{\mathrm{f}}^2−\frac{1}{2}{\mathrm{IÓ\neg }}_{\mathrm{i}}^2\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\text{W}=\frac{1}{2}(0.000056{\text{ k²µ.m}}^{\text{2}}){(60\text{ r²¹»å}/\text{s})}^2−\frac{1}{2}(0.000056{\text{ k²µ.m}}^{\text{2}}){(40\text{ r²¹»å}/\text{s})}^2\\ \text{W}=56.0\text{″¾´³}\end{array}$

Therefore, work required to change the rotational rate from$40.0\text{ r²¹»å}/\text{s}$ to$60.0\text{ r²¹»å}/\text{s}$ is, $56.0\text{″¾´³}$.

Assembly’s K.E is,

$K.E=\frac{1}{2}I{\mathrm{Ó\neg }}^2$

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

i.e,

$\begin{array}{c}\frac{1}{2}I=\left(0.5\right)\left(0.000056{\text{ k²µ.m}}^{\text{2}}\right)\\ =2.8×{10}^{−5}{\text{ J²õ}}^{\text{2}}/{\text{rad}}^{\text{2}}\end{array}$

Therefore, the slope of the plot of the assembly’s K.E vs the square of its rotation rate is $2.8×{10}^{−5}{\text{ J²õ}}^{\text{2}}/{\text{rad}}^{\text{2}}$.