Parallel and Perpendicular Axis Theorems: notes and previous year questions

1Why we need shortcuts

A rod has $I = ML^2/12$ about its middle. Slide the axis toward one end and $I$ grows, up to $ML^2/3$ at the end. Two theorems give $I$ about a new axis without a new integral:

• Parallel axis theorem: moves an axis sideways, keeping its direction. Works for any body.
• Perpendicular axis theorem: links the axis out of a flat body to two axes lying in it. Works for flat bodies only.

2The parallel axis theorem

Why (energy): spin a body about an axis P a distance $d$ from its centre of mass. The centre of mass goes round P in a circle at speed $\omega d$, and the body also turns about its own centre at $\omega$. So $\frac12 I\omega^2 = \frac12 M(\omega d)^2 + \frac12 I_{cm}\omega^2$; cancel $\frac12\omega^2$.

Why (sums): for each bit, $\vec r\,' = \vec r + \vec d$, so $r'^2 = r^2 + d^2 + 2\vec r\cdot\vec d$. Summing, $I = I_{cm} + Md^2 + 2\vec d\cdot\sum m\vec r$, and $\sum m\vec r = 0$ measured from the centre of mass.

• Of all parallel axes, $I$ is smallest through the centre of mass.
• One of the two axes must pass through the centre of mass. To move between two axes that both miss it, go through it in two steps.
• $d$ is the perpendicular gap between the axes.

3The perpendicular axis theorem

Lay a flat body in the $xy$ plane with $z$ out of it. A bit $dm$ at $(x, y)$ is $y$ from the $x$-axis, $x$ from the $y$-axis and $\sqrt{x^2 + y^2}$ from the $z$-axis. Adding all the bits:

4Diameters and plates

For a ring or disc, every diameter is alike, so $I_x = I_y$ and $I_z = 2I_{diameter}$:

A rectangular plate $a \times b$ acts like a rod of length $b$ about the $x$-axis ($Mb^2/12$) and like a rod of length $a$ about the $y$-axis ($Ma^2/12$), so

5Both theorems together

Find $I$ about a centre-of-mass axis with the perpendicular theorem, then shift it with the parallel theorem.

Because $I = I_{cm} + Md^2$, the axis through a square plate that gives the largest $I$ is at a corner, the point farthest from the centre. For a ring, the tangent in the plane and the tangent through the plane give $\frac{3MR^2/2}{2MR^2} = \frac34$.

6Frames and holes

For a body made of parts, find each part's $I$ about the same axis and add. For a hole, subtract the missing part.

In the triangle, each rod's centre is $L/(2\sqrt3)$ from G, so each gives $\frac{ML^2}{12} + \frac{ML^2}{12}$. About a vertex, directly: two rods about their ends ($2ML^2/3$) plus the far rod, whose centre is $\frac{\sqrt3}{2}L$ away ($\frac{ML^2}{12} + \frac{3ML^2}{4}$), which again gives $\frac{3ML^2}{2}$.