91Ó°ÊÓ

$\begin{array}{l}{\mathrm{R}}_2=0.8\text{m}\\ {\mathrm{Ó\neg }}_{\mathrm{i}}=8\text{rad}/\text{s}\\ {\mathrm{R}}_1={\mathrm{R}}_2/2\\ {\mathrm{m}}_1={\mathrm{m}}_2/4\end{array}$

Where, R is radius, ${\mathrm{Ó\neg }}_{\mathrm{i}}$ is angular frequency and m is mass.

Calculate the angular speed by applying law of conservation of angular momentum. Using the formula for kinetic energy, find the kinetic energy and the difference in kinetic energy. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula is as follow:

$\mathbf{Initial}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}\mathbf{=}\mathbf{Final}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}$

According to the law of conservation of angular momentum,

$\text{Initialangularmomentum}=\text{Finalangularmomentum}$

$\begin{array}{l}{\mathrm{L}}_1={\mathrm{L}}_2\\ {\mathrm{I}}_1{\mathrm{Ó\neg }}_1={\mathrm{I}}_2{\mathrm{Ó\neg }}_2\\ \left({\mathrm{m}}_1{\mathrm{R}}_2^2+\left(\frac{1}{2}\right){\mathrm{m}}_2({\mathrm{R}}_1^2+{\mathrm{R}}_2^2)\right){\mathrm{Ó\neg }}_1=\left({\mathrm{m}}_1{\mathrm{R}}_1^2+\left(\frac{1}{2}\right){\mathrm{m}}_2({\mathrm{R}}_1^2+{\mathrm{R}}_2^2)\right){\mathrm{Ó\neg }}_1\end{array}$

By simplifying above equation,

$\begin{array}{c}\frac{{\mathrm{Ó\neg }}_2}{{\mathrm{Ó\neg }}_1}={\left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}\right)}^2\left(\frac{1+0.5\left(\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}\right)×\left({\left(\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\right)}^2+1\right)}{1+0.5\left(\frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}\right)×\left({\left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}\right)}^2+1\right)}\right)\\ \frac{{\mathrm{Ó\neg }}_2}{{\mathrm{Ó\neg }}_1}={\left(2\right)}^2\left(\frac{1+2(0.25+1)}{1+2(4+1)}\right)\\ \frac{{\mathrm{Ó\neg }}_2}{{\mathrm{Ó\neg }}_1}=1.273\\ {\mathrm{Ó\neg }}_2=8×1.273=10.184\text{rad/s}\end{array}$

$\begin{array}{l}{\mathrm{I}}_1=\left({\mathrm{m}}_1{\mathrm{R}}_2^2+\left(\frac{1}{2}\right)×{\mathrm{m}}_2({\mathrm{R}}_1^2+{\mathrm{R}}_2^2)\right)\\ {\mathrm{I}}_1=(2×0{.8}^2+\left(\frac{1}{2}\right)×8×(0{.8}^2+0{.4}^2)=4.48{\text{kg.m}}^{\text{2}}\\ {\mathrm{I}}_2=\left({\mathrm{m}}_1{\mathrm{R}}_1^2+\left(\frac{1}{2}\right)×{\mathrm{m}}_2({\mathrm{R}}_1^2+{\mathrm{R}}_2^2)\right)\\ {\mathrm{I}}_2=\left(2×0{.4}^2+\left(\frac{1}{2}\right)×8×(0{.8}^2+0{.4}^2)\right)=3.52{\text{kg.m}}^{\text{2}}\end{array}$

$\text{Increasein}\mathrm{KE}=\mathrm{Δ°­·¡}={\mathrm{KE}}_2−{\mathrm{KE}}_1$

$\mathrm{Δ°­·¡}=\left(\frac{1}{2}\right)×{\mathrm{I}}_2×{\mathrm{Ó\neg }}_2^2−\left(\frac{1}{2}\right)×{\mathrm{I}}_1×{\mathrm{Ó\neg }}_1^2$

$\mathrm{Δ°­·¡}=\left(\frac{1}{2}\right)×3.52×10{.184}^2−\left(\frac{1}{2}\right)×4.48×8^2$

Hence, the increase in the kinetic energy is $39.176\text{J}≈39.2\text{J}$.

Therefore, using the law of conservation of angular momentum and formula for kinetic energy, the difference in the kinetic energy can be found.