Question

In Fig.$10-49$, a small disk of radius $\mathbf{\text{r}}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{}\mathbf{\text{cm}}\mathbf{}$has been glued to the edge of a larger disk of radius $\mathbf{\text{R}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{cm}}\mathbf{}$so that the disks lie in the same plane. The disks can be rotated around a perpendicular axis through point$\text{O}$at the center of the larger disk. The disks both have a uniform density (mass per unit volume) of$\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{\times }\mathbf{}\mathbf{103}\mathbf{}\mathbf{}\mathbf{\text{kg}}\mathbf{/}{\mathbf{\text{m}}}^{\mathbf{2}}\mathbf{}$and a uniform thickness of $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{}\mathbf{\text{mm}}$. What is the rotational inertia of the two-disk assembly about the rotation axis through O?

Short answer

The rotational inertia of the two-disc assembly about the rotation axis through O is

$6.16\times {10}^{-5}kg{m}^2$

Step-by-step solution

Given

1. Thickness of the disc$\text{d}=5.0\times {10}^{-3}\text{m}$
2. Radius of big disc$\text{R}=0.04\text{m}$
3. Radius of small disc$r=0.02m$
4. Density of material$\rho =1400\text{kg}/{\text{m}}^3$
5. Rotational inertia of wheel$\text{I}=0.050{\text{kgm}}^2$
6. Magnitude of force$\text{P}=3.0\text{N}$

Understanding the concept

Use the parallel axis theorem to obtain the rotational inertia of the two-disc assembly.

Formula:

$$\begin{gathered} \text{I}=\frac{1}{2}{\text{mR}}^2+{\text{md}}^2 \\ \text{m}=\rho \text{v}=\rho \pi {\text{r}}^2\text{d} \end{gathered}$$

Calculate the rotational inertia of the two-disk assembly about the rotation axis through O 

For the rotational inertia of the two-disc assembly about the rotation axis through ,

$$\begin{gathered} \text{I}=\frac{1}{2}{\text{MR}}^2+\frac{1}{2}{\text{mr}}^2+\text{m}{\left(r+\text{R}\right)}^2 \\ \text{I}=\frac{1}{2}\rho \pi {\text{R}}^2{\text{dR}}^2+\frac{1}{2}\rho \pi {\text{r}}^2{\text{dr}}^2+\rho \pi {\text{r}}^2\text{d}{\left(\text{r}+\text{R}\right)}^2 \\ \text{I}=\left(\rho \pi \text{d}\right)\left[\frac{1}{2}{\text{R}}^4+\frac{1}{2}{\text{r}}^4+{\text{r}}^2{\left(\text{r}+\text{R}\right)}^2\right] \\ I=\left(1400kg/m^3\left)\pi \right(5.0\times {10}^{-3}m\right)\left[\frac{1}{2}{\left(0.04m\right)}^4+\frac{1}{2}{\left(0.02\text{m}\right)}^4+{\left(0.02\text{m}\right)}^2{\left(0.02\text{m}+0.04\text{m}\right)}^2\right] \\ \text{I}=6.16\times {10}^{-5}{\text{kg.m}}^2 \end{gathered}$$Therefore, the rotational inertia of the two-disc assembly about the rotation axis through O is$6.16\times {10}^{-5}kg{\text{m}}^2$