Question

Repeat Problem 3for a rod of length / with density varying uniformly from 2to 1.

Short answer

$$\begin{gathered} \left(\mathrm{a}\right)\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{thin}\mathrm{rod}\mathrm{is}\mathrm{M}=\frac{3}{2}\mathrm{I}. \\ \left(\mathrm{b}\right)\mathrm{The}\overline{\mathrm{x}}\mathrm{coordinate}\mathrm{for}\mathrm{the}\mathrm{centre}\mathrm{of}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{rod}\mathrm{is}\frac{4}{9}\mathrm{I}. \\ \left(\mathrm{c}\right){\mathrm{I}}_{\mathrm{m}}\mathrm{about}\mathrm{an}\mathrm{axis}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{rod}\mathrm{is}. \\ \left(\mathrm{d}\right)\mathrm{I}\mathrm{about}\mathrm{an}\mathrm{axis}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{rod}\mathrm{and}\mathrm{passing}\mathrm{through}\mathrm{one}\mathrm{end}\mathrm{is}0.2775{\mathrm{MI}}^2. \end{gathered}$$

Step-by-step solution

Density definition

The density of a substance is a measurement of how densely it is packed together. The mass per unit volume is how it's defined.

Determining mass of the rod

Consider the figure, as given below –

The thin rod has a length / and its density varies from 2-1.

So, the mass of the rod is calculated using the density function, which is –

$p=2-\frac{x}{I}$

Then, the total mass is -

$$\begin{gathered} M={\int }_0^{I}pdx \\ ={\int }_0^I\left(2-\frac{x}{I}\right)DX \\ =2I-\frac{1}{2}{I}^2 \\ =\frac{3}{2}I \end{gathered}$$

Therefore, the mass of the thin rod is obtained as $M=\frac{3}{2}I$.

Determining X coordinate of centre of mass

Assume the rod lies along the x-axis, then the centre of mass is at -

$$\begin{gathered} \overline{X}=\frac{1}{M}{\int }_0^{I}xpdx \\ =\frac{2}{3I}{\int }_0^I\left(2X-\frac{x^2}{I}\right)DX \\ =\frac{2}{3I}{\overline{)\left[X^2-\frac{X^3}{3I}\right]}}_0^I \\ =\frac{4}{9}I \end{gathered}$$

Therefore, the coordinate for the centre of mass of the rod is obtained as$\frac{4}{9}I$ .

Determining Imabout an axis perpendicular to the rod

The moment of inertia about the centre of mass –

$$\begin{gathered} p\left(\overline{x}\right)=2-\frac{1}{I}\left(\frac{4}{9}I+\overline{X}\right) \\ =\frac{14}{9}-\frac{\overline{X}}{I}{I}_m \\ ={\int }_{-\left(\frac{4}{9}\right)I}^{\left(\frac{5}{9}\right)I}{X}^2\left(\frac{14}{9}-\frac{X}{I}\right)dx \\ ={\overline{)\left[\frac{14}{27}{x}^3-\frac{x^4}{4I}\right]}}_{-\left(\frac{4}{9}\right)I}^{\left(\frac{5}{9}\right)I} \\ \approx 0.0651I^3+0.0553I^3 \\ =0.1204I^3\approx 0.08M{I}^2 \end{gathered}$$

Therefore, $I_m$about an axis perpendicular to the rod is calculated to be$0.08M{I}^2$ .

Determining  about an axis perpendicular to the rod and passing through the left (heavy) end

Applying the parallel axis theorem –

$$\begin{gathered} I=I_m+M{d}^2 \\ =0.08M{I}^2+M{\left(\frac{4}{9}\right)}^2 \\ \approx 0.2775M{I}^2 \end{gathered}$$

Therefore, I about an axis perpendicular to the rod and passing through one end is calculated using parallel axis theorem and is obtained as$0.2775M{I}^2$ .