Learn how the distribution of mass can affect the difficulty of causing angular acceleration.

What is rotational inertia?

Rotational inertia is a property of any object which can be rotated. It is a scalar value which tells us how difficult it is to change the rotational velocity of the object around a given rotational axis.
Rotational inertia plays a similar role in rotational mechanics to mass in linear mechanics. Indeed, the rotational inertia of an object depends on its mass. It also depends on the distribution of that mass relative to the axis of rotation.
When a mass moves further from the axis of rotation it becomes increasingly more difficult to change the rotational velocity of the system. Intuitively, this is because the mass is now carrying more momentum with it around the circle (due to the higher speed) and because the momentum vector is changing more quickly. Both of these effects depend on the distance from the axis.
Rotational inertia is given the symbol II. For a single body such as the tennis ball of mass mm (shown in Figure 1), rotating at radius rr from the axis of rotation the rotational inertia is
I=mr2I = mr^2
and consequently rotational inertia has SI units of kgm2\mathrm{kg\cdot m^2}.
Rotational inertia is also commonly known as moment of inertia. It is also sometimes called the second moment of mass; the 'second' here refers to the fact that it depends on the length of the moment arm squared.
Figure 1: A tethered tennis ball rotating about a central point.

How does rotational inertia relate to Newton's 2ⁿᵈ law?

Rotational inertia takes the place of mass in the rotational version of Newton's 2ⁿᵈ law.
Consider a mass mm attached to one end of a massless rod. The other end of the rod is hinged so that the system can rotate about the central hinge point as shown in Figure 2.
Figure 2: A mass rotating due to a tangential force.
We now start rotating the system by applying a tangential force FTF_T to the mass. From Newton’s 2ⁿᵈ law,
FT=maTF_T = m a_T.
this can also be written as
FT=m(rα)F_T = m (r \alpha).
Newton's 2ⁿᵈ law relates force to acceleration. In rotational mechanics torque τ\tau takes the place of force. Multiplying both sides by the radius gives the expression we want.
FTr=m(rα)rτ=mr2ατ=Iα\begin{aligned} F_T r &= m (r \alpha) r\\ \tau &= m r^2 \alpha \\ \tau &= I \alpha\end{aligned}
This expression can now be used to find the behavior of a mass in response to a known torque.
Exercise 1a:
A motor capable of producing a constant torque of 100 Nm100~\mathrm{Nm} and a maximum rotation speed of 150 rad/s150~\mathrm{rad/s} is connected to a flywheel with rotational inertia 0.1 kgm20.1~\mathrm{kg m^2}. What angular acceleration will the flywheel experience as the motor is switched on?
Solution:
Rearranging the rotational version of Newton's 2ⁿᵈ law and substituting in the numbers we find:
α=τI=100 Nm0.1 kgm2=1000 rad/s2\begin{aligned} \alpha &= \frac{\tau}{I} \\ &= \frac{100~\mathrm{Nm}}{0.1~\mathrm{kg m^2}} \\ &= 1000 ~\mathrm{rad/s^2}\end{aligned}
Exercise 1b:
How long will the flywheel take to reach a steady speed if starting from rest?
ω=ω0+αt\omega = \omega_0 + \alpha t
Since we know the maximum rotational velocity of the motor, we can solve to find the time taken to accelerate up to that rotational velocity.
t=ωmaxα=150 rad/s1000 rad/s2=0.15 s\begin{aligned} t &= \frac{\omega_\mathrm{max}}{\alpha} \\ &= \frac{150~\mathrm{rad/s}}{1000~\mathrm{rad/s^2}} \\ &= 0.15~\mathrm{s}\end{aligned}

How can we calculate rotational inertia in general?

Often mechanical systems are made of many masses connected together, or complex shapes.
It is possible to calculate the total rotational inertia for any shape about any axis by summing the rotational inertia of each mass.
I=m1r12+m2r22+=Σmiri2\begin{aligned} I &= m_1 r_1^2 + m_2 r_2^2 + \ldots \\ &= \Sigma m_i r_i^2 \end{aligned}
Figure 3: A rigid system of masses shown with two different rotation axes.
Exercise 2a:
Consider the object shown in figure 3(a). What is its rotational inertia?
Summing all the mr2mr^2 terms for each mass,
I=(1 kg12 m2)+(1 kg1.52 m2)+(1 kg0.752 m2)+(2 kg0.752 m2)=4.9375 kgm2\begin{aligned} I &= (1~\mathrm{kg}\cdot 1^2~\mathrm{m^2}) + (1~\mathrm{kg}\cdot 1.5^2~\mathrm{m^2}) + (1~\mathrm{kg}\cdot 0.75^2~\mathrm{m^2}) + (2~\mathrm{kg}\cdot 0.75^2~\mathrm{m^2}) \\ &= 4.9375~\mathrm{kg\cdot m^2}\end{aligned}.
Exercise 2b:
Consider the alternate case of Figure 3(b) of the same system rotating about a different axis. What would you expect the rotational inertia to be in this case?
Although the system of masses is the same as before, they are now rotating about a much closer axis. Because of the dependence on the square of the distance to the rotational axis, we would expect the rotational inertia to be significantly lower.
I=(2 kg0.52 m2)+(1 kg0.52 m2)+(1 kg0.52 m2)+(1 kg0.52 m2)=1.25 kgm2\begin{aligned} I &= (2~\mathrm{kg}\cdot 0.5^2~\mathrm{m^2}) + (1~\mathrm{kg}\cdot 0.5^2~\mathrm{m^2}) + (1~\mathrm{kg}\cdot 0.5^2~\mathrm{m^2}) + (1~\mathrm{kg}\cdot 0.5^2~\mathrm{m^2}) \\ &= 1.25~\mathrm{kg\cdot m^2}\end{aligned}

How can we find the rotational inertia of complex shapes?

For more complicated shapes, it is generally necessary to use calculus to find the rotational inertia. However, for many common geometric shapes it is possible to find tables of equations for the rotational inertia in textbooks or other sources. These typically give the moment of inertia for a shape rotated about its centroid (which often corresponds with the shapes center of mass).
For example, the rotational inertia of a solid cylinder with radius rr rotated about a central axis is
I=12mr2I = \frac{1}{2}m r^2
and for a hollow cylinder with inner and outer radii rir_i and ror_o respectively,
I=m(ri2+ro2)2I = \frac{m(r_i^2 + r_o^2)}{2}
Expressions for other simple shapes are shown in Figure 4.
Figure 4: Equations for the rotational inertia of some simple shapes under rotation.
Complex shapes can often be represented as combinations of simple shapes for which there exists a known equation for rotational inertia. We can then combine these rotational inertia to find that of the composite object.
It is also sometimes possible to calculate rotational inertia for uniform density objects with voids by considering the void to be a shape with negative rotational inertia.
The problem that we will likely run into when combining simple shapes is that the equations tell us the rotational inertia as found about the centroid of the shape and this does not necessarily correspond to the axis of rotation of our composite shape. We can account for this using the parallel axis theorem.
The parallel axis theorem allows us to find the moment of inertia of an object about a point oo as long as we known the moment of inertia of the shape around its centroid cc, mass mm and distance dd between points oo and cc.
Io=Ic+md2\boxed{I_o = I_c + md^2}
Exercise 3:
If the shape shown in Figure 5 is made by welding three 10 mm10~\mathrm{mm} thick metal discs (each with mass 50 kg50~\mathrm{kg}) to a metal ring with mass 100 kg100~\mathrm{kg}. If rotated about a central axis (out of the page), what is the rotational inertia of the object?
Solution:
We begin by finding the rotational inertia of the four components separately.
Using the previously given equation for a hollow cylinder, we can find the rotational inertia IbI_b of the big disc. This centroid of this component already coincides with the rotational axis of the final part so no correction is necessary.
Ib=12m(ri2+ro2)=12(100 kg)(0.752+12) m278.125 kgm2\begin{aligned}I_b &= \frac{1}{2} m (r_i^2 +r_o^2) \\ &= \frac{1}{2} (100~\mathrm{kg})\cdot(0.75^2 + 1^2)~\mathrm{m^2} \\ &\simeq 78.125~\mathrm{kg\cdot m^2} \end{aligned}
Using the same procedure for the smaller discs, for now centered on their own rotational axes:
Is=12mrc2=12(50kg)(0.30 m)2=2.25 kgm2\begin{aligned} I_s &= \frac{1}{2} m r_c^2 \\ &= \frac{1}{2}\cdot (50\cdot \mathrm{kg}) (0.30~\mathrm{m})^2 \\ &= 2.25~\mathrm{kg\cdot m^2}\end{aligned}
Because each of these three discs are rotating about a point distance dd away from their respective centroids we can use the parallel axis theorem to find the effective rotational inertia IsI_s'.
d=12(1+0.75)=0.875 md = \frac{1}{2} (1 + 0.75) = 0.875~\mathrm{m}
Is=Is+md2=(2.25 kgm2)+(50 kg)(0.875 m)240.5 kgm2\begin{aligned} I_s' &= I_s + md^2 \\ &= (2.25 ~\mathrm{kg\cdot m^2}) + (50~\mathrm{kg})\cdot (0.875~\mathrm{m})^2 \\ &\simeq 40.5~\mathrm{kg\cdot m^2}\end{aligned}
As each of the three small discs are at the same radius from the rotational axis of the part, we can treat them as one part with three times the rotational inertia. Finally, the rotational inertial of the object is:
I78+340.5200 kgm2\begin{aligned} I &\simeq 78 + 3\cdot 40.5 \\ &\simeq 200~\mathrm{kg\cdot m^2}\end{aligned}
Figure 5: A system of one large hollow disc and three smaller filled discs.

Where else does rotational inertia come up in physics?

Rotational inertia is important in almost all physics problems that involve mass in rotational motion. It is used to calculate angular momentum and allows us to explain (via conservation of angular momentum) how rotational motion changes when the distribution of mass changes. It also is needed to find the energy which is stored as rotational kinetic energy in a spinning flywheel.
Loading