Figure shows an overhead view of a ring that can rotate about its centre like a merry- go-round. Its outer radius ${\mathbf{R}}_{\mathbf{2}}$isrole="math" localid="1660990374636" $\mathbf{0.800}\mathbf{}\mathit{m}$, its inner radius${\mathbf{R}}_{\mathbf{1}}$ is${\mathbf{R}}_{\mathbf{2}}\mathbf{/}\mathbf{2}\mathbf{.00}$, its mass$\mathbf{M}$ is$\mathbf{8.00}\mathbf{}\mathbf{kg}$, and the mass of the crossbars at its centre is negligible. It initially rotates at an angular speed of 8.00 rad/s with a cat of mass$\mathbf{m}\mathbf{}\mathbf{=}\mathbf{}\mathbf{M}\mathbf{/}\mathbf{4.00}$on its outer edge, at radius${\mathbf{R}}_{\mathbf{2}}$. By how much does the cat increase the kinetic energy of the cat ring system if the cat crawls to the inner edge, at radius?

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

Where, R is radius, ${\mathrm{\omega }}_{\mathrm{i}}$ 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,

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

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

By simplifying above equation,

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

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

$\text{Increasein}\mathrm{KE}=\mathrm{\Delta KE}={\mathrm{KE}}_2-{\mathrm{KE}}_1$

$\mathrm{\Delta KE}=\left(\frac{1}{2}\right)\times {\mathrm{I}}_2\times {\mathrm{\omega }}_2^2-\left(\frac{1}{2}\right)\times {\mathrm{I}}_1\times {\mathrm{\omega }}_1^2$

$\mathrm{\Delta KE}=\left(\frac{1}{2}\right)\times 3.52\times 10{.184}^2-\left(\frac{1}{2}\right)\times 4.48\times 8^2$

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

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