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 III. For a single body such as the tennis ball of mass mmm (shown in Figure 1), rotating at radius rrr from the axis of rotation the rotational inertia is I = mr^2I=mr 2 I, equals, m, r, start superscript, 2, end superscript and consequently rotational inertia has SI units of \mathrm{kg\cdot m^2}kg⋅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 mmm 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 F_TF T F, start subscript, T, end subscript to the mass. From Newton’s 2ⁿᵈ law, F_T = m a_TF T =ma T F, start subscript, T, end subscript, equals, m, a, start subscript, T, end subscript. this can also be written as F_T = m (r \alpha)F T =m(rα)F, start subscript, T, end subscript, equals, m, left parenthesis, r, alpha, right parenthesis. Newton's 2ⁿᵈ law relates force to acceleration. In rotational mechanics torque \tauτtau takes the place of force. Multiplying both sides by the radius gives the expression we want. \begin{aligned} F_T r &= m (r \alpha) r\\ \tau &= m r^2 \alpha \\ \tau &= I \alpha\end{aligned} F T r τ τ =m(rα)r =mr 2 α =Iα 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~\mathrm{Nm}100 Nm and a maximum rotation speed of 150~\mathrm{rad/s}150 rad/s is connected to a flywheel with rotational inertia 0.1~\mathrm{kg m^2}0.1 kgm 2 . What angular acceleration will the flywheel experience as the motor is switched on? [Solution] \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? [Solution]
\omega = \omega_0 + \alpha t
omega, equals, omega, start subscript, 0, end subscript, plus, alpha, t
\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. \begin{aligned} I &= m_1 r_1^2 + m_2 r_2^2 + \ldots \\ &= \Sigma m_i r_i^2 \end{aligned} I =m 1 r 1 2 +m 2 r 2 2 +… =Σm i r i 2
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? [Solution] mr^2
m, r, start superscript, 2, end superscript
\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? [Solution] \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 rrr rotated about a central axis is I = \frac{1}{2}m r^2I= 2 1 mr 2 I, equals, start fraction, 1, divided by, 2, end fraction, m, r, start superscript, 2, end superscript and for a hollow cylinder with inner and outer radii r_ir i r, start subscript, i, end subscript and r_or o r, start subscript, o, end subscript respectively, I = \frac{m(r_i^2 + r_o^2)}{2}I= 2 m(r i 2 +r o 2 ) I, equals, start fraction, m, left parenthesis, r, start subscript, i, end subscript, start superscript, 2, end superscript, plus, r, start subscript, o, end subscript, start superscript, 2, end superscript, right parenthesis, divided by, 2, end fraction 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. [Hint] 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 ooo as long as we known the moment of inertia of the shape around its centroid ccc, mass mmm and distance ddd between points ooo and ccc. Io=Ic+md2 Exercise 3: If the shape shown in Figure 5 is made by welding three 10~\mathrm{mm}10 mm thick steel discs (each with mass 50~\mathrm{kg}50 kg) to a steel ring with mass 100~\mathrm{kg}100 kg. If rotated about a central axis (out of the page), what is the rotational inertia of the object? [Solution] I_b
I, start subscript, b, end subscript
\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}
\begin{aligned} I_s &= \frac{1}{2} m r_c^2 \\ &= \frac{1}{2}\cdot (50\cdot \mathrm{kg}) (0.8~\mathrm{m})^2 \\ &= 16~\mathrm{kg\cdot m^2}\end{aligned}
ddI_s'
I, start subscript, s, end subscript, prime
d = \frac{1}{2} (1 + 0.75) = 0.875~\mathrm{m}
\begin{aligned} I_s' &= I_s + md^2 \\ &= (16 ~\mathrm{kg\cdot m^2}) + (50~\mathrm{kg})\cdot (0.875~\mathrm{m})^2 \\ &\simeq 54~\mathrm{kg\cdot m^2}\end{aligned}
\begin{aligned} I &\simeq 78 + 3\cdot 54 \\ &\simeq 240~\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.