Vintage Watchstraps

Straps for Vintage Fixed Wire Lug Trench Watches or Officer's Wristwatches

# Moment of Inertia

The ‘moment of inertia’ of a rotating body is the body's resistance to angular acceleration. It is the rotational equivalent of inertia or mass in systems involving linear acceleration.

The moment of inertia of a point mass is often stated, without any justification, as being the mass of the particle $$m$$ multiplied by the square of its distance $$k$$ from the axis of rotation, or $$mk^2$$. The moment of inertia of more complex body is then defined as the sum of the moments of inertia of all the individual elements, $$I=\sum_0^k{mk^2}$$. But the basic definition $$I=mk^2$$ begs the question; why is the square of the distance from the centre of rotation used?

There are many explanations of this on the web, some involving work done or angular momentum, which are rather circular, whereas others say that it just is in order to make sums come out right. But in fact the squared term comes simply from the equations of motion, as shown in the section below.

## Why does the Moment of Inertia Depend on Distance Squared?

Referring to the diagram; a point mass $$m$$ constrained by a light stiff connector to move in a circle of radius $$r$$ is accelerated by torque $$\tau.$$

The tangential force acting on the mass is the torque divided by the distance $$r$$ of the mass from the centre of rotation:

$F_t=\tau/r$

(In case you are tempted to multiply the torque by the radius, note that a given torque, e.g. on a machine nut, can be produced using a longer arm and less force. To produce the same torque, the force decreases as the arm gets longer. This is why bigger spanners are longer, not just because it looks right.)

The tangential force causes tangential acceleration $$a_t$$, which is related to angular acceleration $$\alpha$$ radians per second by:

$a_t = \alpha r$

(Note that the mass also experiences centripetal acceleration as a result of being constrained to follow a circular path. This does not involve the torque, it is caused by a centripetal force shown as $$F_c$$ which is caused by strain in the connector between the mass and the centre of rotation.)

Newton's second law:

$F_t=ma_t$

Substituting $$\tau/r$$ for $$F_t$$ and $$\alpha r$$ for $$a_t$$ gives:

$\tau/r=m\alpha r$

Multiply both sides by $$r$$ and rearrange:

$\tau=m r^2 \alpha$

This shows that rotational inertia which resists angular acceleration depends on the square of the distance of the centre of mass from the axis of rotation.

For any object, no matter what its shape, there exists a distance $$k$$ called the radius of gyration which can be used with the mass of the object to calculate its rotational inertia. For some objects, e.g. a watch balance where the effective mass can be assumed to be concentrated in the thin ring of the rim, the radius of gyration can be estimated by inspection. For other objects of more complicated shapes it must be calculated, or even found by measurement for very complicated shapes.

In engineering a ‘moment of force’, often shortened to ‘moment’ is a measure of the tendency of the force to cause a body to rotate about an axis. It is used interchangeably with ‘torque’. The term $$mr^2$$ is called the ‘moment of inertia’ $$I$$, and Newton's second law expressed for rotational motion becomes:

$\tau=I\alpha$

This is the same form as Newton's second law of linear motion, $$F=ma$$, with the moment of inertia $$I$$ in the rotational form taking place of the mass $$m$$ in the linear equation.

### Verify the Result

Checking that units on either side of the = sign are consistent is useful to verify that the result makes sense. Torque is measured in units of newton metres. A newton is defined as the force that gives a mass of 1kg an acceleration of 1 metre per second squared, so its units are $$kg. metre / second^2$$. Torque therefore has units:

$Torque\ \tau: Newton.metre = \frac{kg.metre}{second^2}.metre = \frac{kg.metre^2}{second^2}$

A radian is a ratio, a fraction of a circle, and therefore dimensionless. The units of angular acceleration are simply $$1/second^2$$ and therefore the units of rotational mass acceleration are:

$Moment\ of\ Inertia\ I \times angular\ acceleration\ = m r^2 \alpha = \frac{kg.metre^2}{second^2}$

Since the unnits of torque and rotational mass acceleration are both $$kg.metre^2/second^2$$ the expressions on both side of the = sign are consistent.

## Moment of Inertia of a Thin Rod

A thin rod, that is a rod with length much greater than its thickness, of length $$l$$ and mass $$m$$ has a mass per unit length of $$m/l$$. The moment of inertia of a piece of the rod at distance $$x$$ from one end about that end is $$x^2m/l$$. The moment of inertia of the whole rod about the end is then the sum of the individual moments of inertia from $$x=0$$ to $$x=l$$.

$I = \int_0^l \frac{x^2m}{l} \, dx = \frac{m}{l} \left [ \frac{x^3}{3} \right ]_{0}^{l} = \frac{1}{3} ml^2$

The radius of gyration $$k$$ with respect to the point of rotation is given by:

$I = mk^2 = \frac{1}{3} ml^2$ $\therefore k = \frac{l}{\sqrt{3}} = \approx \frac{l}{1.7} = \approx 0.58l$

The radius of gyration, the distance from the point of rotation which would give the same moment of inertia if all the mass was concentrated there, is at nearly 60% of the length of the rod.

If the length of the rod is ten times greater than its thickness, using the formula for rotation of a cylinder about its end rather than a thin rod makes less than 1% difference to the answer, so for most rods the simpler formula gives reasonable accuracy.

## Moment of Inertia of a Disc About its Centre

The moment of inertia of an object is the sum of all the elemental particles which make it up. A disc can be thought of as being composed of many rings of radius $$r$$ and thickness $$\delta r$$. The moment of inertia of each ring is $$mr^2$$. The mass of a ring is its volume $$2\pi r\delta rt$$ multiplied by its density $$\rho$$. The moment of inertia of the ring $$mr^2$$ is then $$2\pi r^3 \delta r t \rho$$. The moment of inertia of the disc is then found by integrating this expression between the limits $$r=0$$ and $$r=r$$.

$I = \int_0^r 2\pi t\rho r^3\delta r = 2\pi t\rho \int_0^r r^3\delta r = \frac{\pi t\rho r^4}{2}$

The total mass of the disc is its volume multiplied by its density, $$\pi r^2 t \rho$$. Substituting $$m$$ for this expression in the equation above gives the expression:

$I = \frac{1}{2} m r^2$

Here $$r$$ is the radius of the disc. The radius of gyration of the disc is therefore $$\frac{r}{\sqrt{2}}$$.

## Moment of Inertia of a Thick Ring

A ring of height $$h$$, internal radius $$r_1$$ and external radius $$r_2$$ is imagined to be composed of many infinitesimally thin rings of radius $$r$$ and width $$\delta r$$.

The plan area of each ring is $$\pi(r+ \delta r)^2 - \pi r^2 = \pi(r^2 + 2r \delta r + \delta r^2) - \pi r^2 = \pi(2r \delta r + \delta r^2)$$

Since $$\delta r$$ is infinitesimally small $$\delta r^2 = 0$$ and the plan area of the ring is $$2\pi r \delta r$$. Its volume is $$2\pi r \delta r h$$ and its mass is $$2\pi r \delta r h \rho$$, where $$\rho$$ is the density of the material.

The moment of inertia of each ring $$mr^2$$ is therefore $$2\pi r^3 \delta r h \rho$$. The moment of inertia of the ring is found by integrating the moment of inertia of all the rings of thickness $$\delta r$$ between $$r=r_1$$ and $$r=r_2$$.

$I = 2\pi h\rho \int_{r_1}^{r_2} r^3\delta r$ $I = 2\pi h\rho \left [ \frac{r^4}{4} \right ]_{r_1}^{r_2} = \frac{\pi h\rho (r_2^4 - r_1^4) }{2}$

Density $$\rho$$ is mass divided by volume:

$\rho = \frac{m}{\pi (r_2^2 - r_1^2) h}$

Substituting this into the integrated solution:

$I = \frac{\pi h (r_2^4 - r_1^4) }{2} \frac{m}{\pi (r_2^2 - r_1^2) h}$

Using the simplification $$(a^4-b^4) = (a^2+b^2)(a^2-b^2)$$:

$I = \frac{1}{2} m(r_2^2 + r_1^2)$

This is a useful equation, because for a disc $$r_1=0$$ and $$r_2 = r$$, which gives the result found above for a disc $$I=\frac{1}{2}mr^2$$, and for a thin ring both $$r_1$$ and $$r_2 = r$$ so $$r_2^2 + r_1^2 = 2r^2$$, giving $$I=mr^2$$.

If you have any comments or questions, please don't hesitate to to get in touch via my Contact Me page.