A uniform rod AB of cross-sectional area S=4 cm², length

A uniform rod $AB$ of cross-sectional area $S=4\ \text{cm}^2$, length $l=20$ cm, and density $\gamma=7.8\ \text{g/cm}^3$ rotates with angular velocity $\omega=10\pi\ \text{s}^{-1}$ about an axis through one end, perpendicular to the rod. Find its kinetic energy of rotation.

Step 1: Mass of the rod

\[M=\gamma Sl=7.8\times4\times20=624\ \text{g}=0.624\ \text{kg}.\]

Step 2: Moment of inertia about one end

With linear density $\lambda=M/l$, a mass element at distance $r$ from the pivot end is $dm=\lambda\,dr$:

\[I=\int_0^l r^2\cdot\frac{M}{l}\,dr=\frac{M}{l}\left[\frac{r^3}{3}\right]_0^l=\frac{Ml^2}{3}.\]

\[\boxed{I=\frac13Ml^2}\]

Step 3: Numerical value of $I$

\[I=\frac13\times0.624\times(0.2)^2=\frac{0.624\times0.04}{3}\approx8.32\times10^{-3}\ \text{kg}\cdot\text{m}^2.\]

Step 4: Kinetic energy

\[E_k=\frac12I\omega^2=\frac{Ml^2\omega^2}{6},\qquad \omega^2=(10\pi)^2=100\pi^2.\]

\[E_k=\frac{0.624\times0.04\times100\pi^2}{6}=0.416\pi^2\approx4.11\ \text{J}.\]

\[\boxed{E_k=\frac{Ml^2\omega^2}{6}=0.416\pi^2\ \text{J}\approx4.11\ \text{J}}\]

Step 5: Comparison with rotation about the center

Rotating about the rod's center instead gives $I_{\text{center}}=\tfrac1{12}Ml^2$, exactly a quarter of the end-pivot value, so the kinetic energy at the same $\omega$ would be $E_k/4\approx1.03$ J — rotation about an end requires four times as much energy as rotation about the center, since the $r^2$-weighting sends mass elements farther from the axis in the end-pivot case.