A Steel Bangle Is Harder to Spin Than a Coin of the Same Mass

A see-saw balances only when the turning effects on both sides match. A wheel resists being spun up — and how much it resists depends on where its mass sits, not just how much mass it has.

The first idea is rotational equilibrium; the second is moment of inertia, the rotational version of mass.

This part covers equilibrium of rigid bodies, moment of inertia of particles, radius of gyration, and moments of inertia of standard bodies.

A rigid body is in equilibrium only when the net force is zero (translational equilibrium) and the net torque about any point is zero (rotational equilibrium).

$$\sum \vec{F}=0\sum \vec{\tau }=0$$

Take $g=10$ m/s${}^2$.

Worked example 1 — a see-saw. A $30$ kg child sits $2$ m from the pivot. Where must a $40$ kg child sit?

$$30\times 2=40\times x\Rightarrow x=1.5\text{ m}$$

Worked example 2 — a metre rule. A uniform $0.2$ kg metre rule is pivoted at the $40$ cm mark. Its weight acts at $50$ cm, $10$ cm from the pivot. A $0.1$ kg mass hung $d$ cm on the other side balances it when

$$0.1\times d=0.2\times 10\Rightarrow d=20\text{ cm (at the }20\text{ cm mark)}$$

Worked example 3 — a plank on two supports. A $4$ m, $20$ kg plank rests on supports at its ends, and a $60$ kg person stands $1$ m from the left end. Taking torques about the left end:

$$4R_R=200(2)+600(1)\Rightarrow R_R=250\text{ N}{R}_L=800-250=550\text{ N}$$

An everyday example. A shopkeeper's beam balance settles level only when the torques from the two pans are equal.

The substance. A couple has zero net force but a non-zero torque, so zero net force alone does not guarantee equilibrium.

**The moment of inertia about an axis is $I=\sum m_i{r}_i^2$, where each $r_i$ is a particle's perpendicular distance from that axis; it measures how strongly the body resists changes in rotation.**

The unit is kg m${}^2$.

Worked example 1 — two masses on a rod. Two $2$ kg masses sit at the ends of a light $1$ m rod.

$$I_{\text{centre}}=2(0.5{)}^2+2(0.5{)}^2=1{\text{ kg m}}^2{I}_{\text{one end}}=2(0{)}^2+2(1{)}^2=2{\text{ kg m}}^2$$

The same body has twice the moment of inertia about an end.

Worked example 2 — a triangle. Three $1$ kg masses sit at the corners of an equilateral triangle of side $0.5$ m. About an axis through one corner, perpendicular to the plane:

$$I=0+1(0.5{)}^2+1(0.5{)}^2=0.5{\text{ kg m}}^2$$

An everyday example. A tightrope performer at a street show carries a long bamboo pole — its mass far from the centre gives a large moment of inertia, so the performer tips over slowly enough to correct.

The substance. Moment of inertia depends on the axis, unlike mass, which is fixed.

**The radius of gyration $k$ is the distance from the axis at which the whole mass $M$ could be placed to give the same moment of inertia, so $I=M{k}^2$ and $k=\sqrt{\frac{I}{M}}$.

Worked example 1 — the two-mass rod.** Total mass $M=4$ kg.

$$k_{\text{centre}}=\sqrt{\frac{1}{4}}=0.5\text{ m}{k}_{\text{end}}=\sqrt{\frac{2}{4}}\approx 0.71\text{ m}$$

Worked example 2 — a ring and a disc. A ring about its axis has $I=M{R}^2$, so $k=R$. A disc has $I=\frac{1}{2}M{R}^2$, so

$$k=\frac{R}{\sqrt{2}}\approx 0.707R$$

For a disc of radius $0.2$ m, $k\approx 0.14$ m.

An everyday example. The heavy iron wheel of a hand-operated sewing machine has its mass concentrated near the rim, giving a large radius of gyration and a smooth, steady spin.

The substance. Radius of gyration is not the distance to the centre of mass — for a disc spinning about its own axis, the centre of mass is on the axis, yet $k\approx 0.707R$.

**For uniform bodies of mass $M$, the standard results are: rod about its centre $\frac{M{L}^2}{12}$ and about an end $\frac{M{L}^2}{3}$; ring about its axis $M{R}^2$; disc about its axis $\frac{1}{2}M{R}^2$; solid sphere $\frac{2}{5}M{R}^2$ and hollow sphere $\frac{2}{3}M{R}^2$ about a diameter.

More results:

- Ring about a diameter**: $\frac{1}{2}M{R}^2$
- Disc about a diameter: $\frac{1}{4}M{R}^2$
- Solid cylinder about its axis: $\frac{1}{2}M{R}^2$

Worked example 1 — a rod. $M=0.5$ kg, $L=1.2$ m:

$$I_{\text{centre}}=\frac{0.5\times 1.44}{12}=0.06{\text{ kg m}}^2{I}_{\text{end}}=\frac{0.5\times 1.44}{3}=0.24{\text{ kg m}}^2$$

Worked example 2 — spheres. A $5$ kg sphere of radius $0.2$ m:

$$I_{\text{solid}}=\frac{2}{5}(5)(0.04)=0.08{\text{ kg m}}^2{I}_{\text{hollow}}=\frac{2}{3}(5)(0.04)\approx 0.133{\text{ kg m}}^2$$

An everyday example. A steel bangle is harder to set spinning than a solid coin-shaped disc of the same mass and radius, because all the bangle's mass sits at the rim.

The substance. The further the mass lies from the axis, the larger the numerical factor — from $\frac{2}{5}$ for a solid sphere up to $1$ for a ring.