Wheels Aand Bin Fig. 11-61 are connected by a belt that does not slip. The radius of Bis 3.00 times the radius of A. What would be the ratio of the rotational inertias${\mathbf{I}}_{\mathbf{A}}\mathbf{/}{\mathbf{I}}_{\mathbf{B}}$if the two wheels had (a) the same angular momentum about their central axes? (b) the same rotational kinetic energy?

Radius of B is 3.00 times the radius of A.

Using the formula $\mathrm{L}=\mathrm{I\omega }$and $\mathrm{K}=\frac{1}{2}{\mathrm{I\omega }}^2$we can find the ratio of the rotational inertias ${\mathrm{I}}_{\mathrm{A}}/{\mathrm{I}}_{\mathrm{B}}$if the two wheels had the same angular momentum about their central axes and the same rotational kinetic energy.

Formulae are as follow:

$$\begin{gathered} \mathrm{L}=\mathrm{I\omega } \\ \mathrm{K}=\frac{1}{2}{\mathrm{I\omega }}^2 \end{gathered}$$

where, $\mathrm{\omega }$is angular frequency, L is angular momentum, Iis moment of inertia and K is kinetic energy.

For${\mathrm{L}}_{\mathrm{A}}={\mathrm{L}}_{\mathrm{B}}$, the ratio of rotational inertias is,

$$\begin{gathered} \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}=\frac{\mathrm{L}/{\mathrm{\omega }}_{\mathrm{A}}}{\mathrm{L}/{\mathrm{\omega }}_{\mathrm{B}}} \\ \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}=\frac{{\mathrm{\omega }}_{\mathrm{A}}}{{\mathrm{\omega }}_{\mathrm{B}}}=\frac{\mathrm{v}/{\mathrm{R}}_{\mathrm{B}}}{\mathrm{v}/{\mathrm{R}}_{\mathrm{A}}} \\ \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}=\frac{{\mathrm{R}}_{\mathrm{A}}}{{\mathrm{R}}_{\mathrm{B}}} \\ \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}=\frac{1}{3}=0.333 \end{gathered}$$

Hence, the ratio of the rotational inertias ${\mathrm{I}}_{\mathrm{A}}/{\mathrm{I}}_{\mathrm{B}}$if the two wheels had the same angular momentum about their central axes is 0.333.

For${\mathrm{K}}_{\mathrm{A}}={\mathrm{K}}_{\mathrm{B}}$. , the ratio of rotational inertias is,

$$\begin{gathered} \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}=\frac{2\mathrm{K}/{\mathrm{\omega }}_{\mathrm{A}}^2}{2\mathrm{K}/{\mathrm{\omega }}_{\mathrm{B}}^2} \\ \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}=\frac{{\mathrm{\omega }}_{\mathrm{B}}^2}{{\mathrm{\omega }}_{\mathrm{A}}^2}=\frac{{\left(\mathrm{v}/{\mathrm{R}}_{\mathrm{B}}\right)}^2}{{\left(\mathrm{v}/{\mathrm{R}}_{\mathrm{A}}\right)}^2} \\ \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}=\frac{{\mathrm{R}}_{\mathrm{A}}^2}{{\mathrm{R}}_{\mathrm{B}}^2} \\ \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}}=\frac{1}{9}=0.111 \end{gathered}$$

Hence, the ratio of the rotational inertias ${\mathrm{I}}_{\mathrm{A}}/{\mathrm{I}}_{\mathrm{B}}$if the two wheels had the same rotational kinetic energy is 0.111.

Therefore, using the formula $\mathrm{L}=\mathrm{I\omega }$and $\mathrm{K}=\frac{1}{2}{\mathrm{I\omega }}^2$, it is possible to find the ratio of the rotational inertias if the two wheels had the same angular momentum about their central axes and the same rotational kinetic energy.