Question

For a thin rod of length and uniform densityfind

(a) M,

(c)${\mathbf{l}}_{\mathbf{m}}$about an axis perpendicular to the rod,

(d)labout an axis perpendicular to the rod and passing through one end (see Problem 1).

Short answer

(a) Mass of the thin rod is M = lp.

(b) about an axis perpendicular to the rod is $\frac{1}{12}{\mathrm{ML}}^2$.

(c) about an axis perpendicular to the rod and passing through one end is $\frac{1}{3}{\mathrm{ML}}^2$.

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.

Mass of the rod

The thin rod has a length L and density is p.

So, the mass of the rod is calculated using the formula,

$$\begin{gathered} \mathrm{Mass}=\mathrm{Length}\times \mathrm{density} \\ \mathrm{M}=\mathrm{L}\times \mathrm{p} \end{gathered}$$

Therefore, the mass of the thin rod is obtained as $\mathrm{M}=\mathrm{L}\times \mathrm{p}$.

Step 3: lmabout an axis perpendicular to the rod

The center of mass of the rod is at the middle so the moment of inertia is,

$$\begin{gathered} {\mathrm{l}}_{\mathrm{m}}={\int }_{\frac{\mathrm{L}}{2}}^{\frac{\mathrm{L}}{2}}{\mathrm{px}}^2\mathrm{dx} \\ ={\overline{)\frac{\mathrm{p}}{3}{\times }^3}}_{\frac{\mathrm{L}}{2}}^{\frac{\mathrm{L}}{2}} \\ =\frac{1}{12}{\mathrm{pL}}^3 \\ =\frac{1}{12}{\mathrm{ML}}^3 \end{gathered}$$

Therefore, ${\mathrm{l}}_{\mathrm{m}}$about an axis perpendicular to the rod is calculated to be$\frac{1}{12}{\mathrm{ML}}^3$ .

Step 4:  about an axis perpendicular to the rod and passing through one end

Applying the parallel axis theorem, we get,

$$\begin{gathered} \mathrm{l}={\mathrm{Md}}^2+{\mathrm{l}}_{\mathrm{m}} \\ =\frac{{\mathrm{ML}}^2}{4}+\frac{1}{12}{\mathrm{ML}}^2 \\ =\frac{1}{3}{\mathrm{ML}}^2 \end{gathered}$$

Therefore, about an axis perpendicular to the rod and passing through one end is calculated using the parallel axis theorem and is obtained as $\frac{1}{3}{\mathrm{ML}}^2$.