Why a Tightrope Walker Carries a Long Pole

A spinning top, a swinging door and a see-saw all rotate as well as move. Physics handles them by treating a body's mass as if concentrated at one special point, the centre of mass, and by describing turning with two new quantities, torque and angular momentum.

This lesson covers locating the centre of mass, torque and angular momentum as cross products, and the conditions for a rigid body to stay in equilibrium.

**The centre of mass is the point where a system's whole mass can be treated as concentrated; for two particles it lies on the line joining them at $x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$, and for a uniform symmetric body it lies at the geometric centre.

Particle systems:**

$$x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}{\vec{r}}_{cm}=\frac{\sum m_i{\vec{r}}_i}{M}$$

- The centre of mass lies closer to the heavier particle, with distances in the inverse ratio of the masses

Uniform symmetric bodies:

- Rod: at its midpoint; ring, disc or sphere: at the geometric centre
- Rectangular plate: where the diagonals cross; triangular plate: at the centroid

Worked example 1. Masses of 2.0 kg and 3.0 kg sit at $x=0$ and $x=5.0$ m:

$$x_{cm}=\frac{2.0\times 0+3.0\times 5.0}{5.0}=3.0\text{ m}$$

It is 3.0 m from the lighter mass and 2.0 m from the heavier one — the inverse of the mass ratio.

Worked example 2 — three particles. Masses of 1.0 kg at (0, 0), 1.0 kg at (4.0, 0) and 2.0 kg at (0, 4.0) m:

$$x_{cm}=\frac{1.0\times 4.0}{4.0}=1.0\text{ m}{y}_{cm}=\frac{2.0\times 4.0}{4.0}=2.0\text{ m}$$

An everyday example. A tightrope walker at a village fair holding a long bamboo pole shifts the pole to keep the combined centre of mass directly above the rope.

The substance. The centre of mass need not lie inside the body — for a ring it sits in the empty space at the middle.

**Torque is the turning effect of a force, $\vec{\tau }=\vec{r}\times \vec{F}$, angular momentum is the turning analogue of momentum, $\vec{L}=\vec{r}\times \vec{p}$, and the net torque equals the rate of change of angular momentum.

Torque:**

$$\vec{\tau }=\vec{r}\times \vec{F}\tau =rF\sin \theta$$

- Here $\vec{r}$ runs from the axis to the point where the force acts, and $r\sin \theta$ is the perpendicular distance to the force's line of action
- The direction follows the right-hand rule; the unit is N m

Angular momentum: $\vec{L}=\vec{r}\times \vec{p}$, which for circular motion gives $L=mvr$, in kg m${}^2$ s${}^{-1}$.

Relating the two. Differentiating $\vec{L}=\vec{r}\times \vec{p}$:

$$\frac{d\vec{L}}{dt}=\vec{v}\times m\vec{v}+\vec{r}\times \vec{F}=\vec{\tau }$$

Worked example 1 — opening a door. A 20 N force is applied at the handle, 0.80 m from the hinges:

- Perpendicular to the door: $\tau =0.80\times 20\times \sin 90^{\circ }=16$ N m
- At 30° to the door: $\tau =0.80\times 20\times \sin 30^{\circ }=8.0$ N m
- Perpendicular but only 0.20 m from the hinges: $\tau =0.20\times 20=4.0$ N m

Worked example 2 — a cross product. For $\vec{r}=2\hat{i}$ m and $\vec{F}=3\hat{j}$ N, $\vec{\tau }=2\hat{i}\times 3\hat{j}=6\hat{k}$ N m.

An everyday example. A mechanic loosening a stuck wheel nut reaches for a long spanner, because doubling the lever arm doubles the torque for the same effort.

The substance. A force through the axis produces no torque, however large — pushing a door at its hinges does not turn it at all.

A rigid body is in mechanical equilibrium when both the net external force and the net external torque on it are zero, so it has neither linear nor angular acceleration.

The two conditions:

- Translational equilibrium: $\sum \vec{F}=0$
- Rotational equilibrium: $\sum \vec{\tau }=0$ about any axis

A couple. Two equal and opposite forces along different lines give zero net force but a non-zero torque, so the body is in translational but not rotational equilibrium — as when both hands turn a steering wheel.

Worked example 1 — a see-saw. A 30 kg child sits 2.0 m from the pivot. For a 40 kg child to balance it:

$$30\times 9.8\times 2.0=40\times 9.8\times d\Rightarrow d=1.5\text{ m}$$

The pivot must also push up with $70\times 9.8=686$ N to satisfy the force condition.

Worked example 2 — a loaded plank. A 4.0 m uniform plank weighing 200 N rests on supports at ends A and B, with a 600 N load 1.0 m from A. Taking moments about A:

$$N_B\times 4.0=200\times 2.0+600\times 1.0\Rightarrow N_B=250\text{ N}{N}_A=800-250=550\text{ N}$$

An everyday example. The beam balance at a kirana shop settles level when the moments of the goods and the weights on either side of the pivot are equal.

The substance. Zero net force does not guarantee equilibrium — a couple can set a body spinning even though the forces on it add to zero.