Calculate the kinetic energy of a uniform rectangular

Calculate the kinetic energy of a uniform rectangular plate with sides $a$ and $b$ ($a>b$), rotating with constant angular velocity $\omega$ about the axis through its center parallel to the longer side $a$.

Step 1: Set up the moment-of-inertia integral

With $-a/2\le x\le a/2$, $-b/2\le y\le b/2$, uniform density $\sigma=M/(ab)$, and distance-from-axis $r=|y|$:

\[I=\int_{-a/2}^{a/2}\!\!\int_{-b/2}^{b/2}y^2\cdot\frac{M}{ab}\,dy\,dx=\frac{M}{ab}\cdot a\int_{-b/2}^{b/2}y^2\,dy=\frac{M}{b}\int_{-b/2}^{b/2}y^2\,dy.\]

Step 2: Evaluate the $y$-integral

\[\int_{-b/2}^{b/2}y^2\,dy=\left[\frac{y^3}{3}\right]_{-b/2}^{b/2}=\frac{1}{3}\cdot\frac{b^3}{4}=\frac{b^3}{12}.\]

Step 3: Moment of inertia

\[I=\frac{M}{b}\cdot\frac{b^3}{12}=\frac{Mb^2}{12}.\]

\[\boxed{I=\frac1{12}Mb^2}\]

Step 4: Kinetic energy

\[E_k=\frac12I\omega^2=\frac{Mb^2\omega^2}{24}.\]

\[\boxed{E_k=\frac{Mb^2\omega^2}{24}}\]

Step 5: Remark on which dimension matters

The result depends only on $b$ — the dimension perpendicular to the rotation axis — not on $a$, since mass located anywhere along the axis direction contributes no distance from that axis. Rotating instead about the axis parallel to the shorter side $b$ would give $I=\tfrac1{12}Ma^2$ and, since $a>b$, a larger kinetic energy for the same $\omega$.