Rotational Inertia

Qualitatively, what is rotational inertia?

                                            This video was produced by Doc Schuster.

Put simply, the rotational inertia (represented by \(I\)) of an object is a measure of how much a spinning object will "resist" deviating from a uniform and constant angular velocity \(\vec{ω}\). Rotational inertia plays the same role in rotational motion as mass plays in linear motion. Both are a measure of how much an object resists changing from uniform motion. In linear motion, the inertia (how much the object will resist changing from uniform motion in a straight line at a constant speed) is just the mass \(m\). But in rotational motion, how much a spinning object will resist changing from a state of uniform angular velocity \(\vec{ω}\) (which is what the quantity \(I\) measures) depends on not just the mass \(m\) of the object but also on how that mass is distributerd. We already have some intuition of this notion. For example, suppose that I place two spheres of equal mass on the table which have a constant angular velocity of \(\vec{ω}=0\) (in other words, they are not spinning). But suppose that one of the spheres is solid and that the other is completely hollow. Even though both spheres are equally massive, if you tried to spin the hollow sphere you would find that it would be much more difficult to get it spinning than the solid sphere. This is one example of how the rational inertia of an object (the object's resistance to change from a constant velocity \(\vec{ω}\)) depends on how the mass of the object is distributed in space.

The rotational inertia of a system of \(n\) particles is defined as

$$I≡\sum_{i=1}^nm_ir_i^2,\tag{1}$$

where \(m_i\) is the mass of each particle and \(r_i\) is the perpendicular distance (or lever arm) between the \(i^{th}\) particle and the axis of the system's rotation. For a continuous mass distribution (i.e. a bowling ball or a spinning top), the rotational inertia is defined as

$$I≡\int{r^2dm},\tag{2}$$

where \(dm\) is the mass of each infinitesimal mass element and \(r\) is the perpendicular distance between that mass element and the object's axis of rotation. Equations (1) and (2) capture how the quantity \(I\) depends on the mass of the system or object and how that mass is distributed in space by the distances \(r_i\) (for a system of \(n\) particles) or \(r\) (for an object).

 
Figure 1: The rotational inertia of a hollow sphere is given by \(I=\frac{2}{3}MR^2\) and the rotational inertia of a solid ball is given by \(\frac{2}{5}MR^2\). Since the lever arms \(r\) (perpendicular distance from the axis of rotation) of each mass element in the solid sphere is smaller (because each mass is closer to the axis of rotation) than the lever arms in the hollow sphere (since each mass is farther away from the axis of rotation), the solid sphere must have a smaller rotational inertia. According to Equation (2), if \(r\) is smaller, then the function \(r^2\) that we are taking the area under is smaller.

Figure 1: The rotational inertia of a hollow sphere is given by \(I=\frac{2}{3}MR^2\) and the rotational inertia of a solid ball is given by \(\frac{2}{5}MR^2\). Since the lever arms \(r\) (perpendicular distance from the axis of rotation) of each mass element in the solid sphere is smaller (because each mass is closer to the axis of rotation) than the lever arms in the hollow sphere (since each mass is farther away from the axis of rotation), the solid sphere must have a smaller rotational inertia. According to Equation (2), if \(r\) is smaller, then the function \(r^2\) that we are taking the area under is smaller.

 

Just by looking at these equations qualitatively, we can understand why a hollow ball has more rotational inertia than a solid ball of the same mass. The perpendicular distance \(r\) of each mass element comprising the solid sphere is bigger than the lever arm \(r\) between each mass element and the axis of rotation. As you can see visually in Figure 1, for the hollow sphere all of the mass elements are located on the outer surface of the sphere; this is why most of them are pretty far away from the axis of rotation. But for a solid sphere, most of the mass elements are located inside of the sphere. The mass elements inside of the sphere can be quite close to the axis of rotation as illustrated in Figure 1. Thus, most of the mass elements in the solid sphere will have a smaller leaver arm \(r\) than the mass elements in the hollow sphere which are along the distant outer edge of the sphere.

What does a bigger or smaller value of \(r\) mean? If the function \(r(m)\) is small (meaning that \(r^2\) is also small), then the area \(\int{r^2dm}\) must also be small and, according to Equation (2), the value of \(I\) must be small. If \(r(m)\) is big (the case for the hollow sphere), then the area underneath that function, \(\int{r^2dm}\), must also be big and, according to Equation (2), that object must have a large amount of rotational inertia \(I\).


Calculating the rotational inertia of a hollow sphere

We know qualitatively that the hollow sphere has more rotational inertia than the solid ball; it is therefore more difficult to speed up or to slow down its spin. But let's prove this quantitively by solving the integral in Equation (2) to find the rotational inertia for a hollow sphere; then, after that, we'll do the same thing for the solid sphere. In Equation (2), \(r\) represents the perpendicular distance of each mass element from the axis of rotation. We know that the radius of the sphere, \(R\), represents the total distance of each mass element from the center of the sphere. If we let \(θ\) denote the angle between the axis of rotation and the distance \(R\) of each particle, then the lever arm of each particle is simply just \(r=Rsinθ\). Substituting this into Equation (2), we have

$$I=R^2\int{sin^2θdm},\tag{3}$$

To solve the integral in Equation (3), let's express both the integrand and the limits of integration with respect to \(θ\). If the hollow sphere has a very thin shell, then we can ignore the shell's thickness and approximate all of the sphere's mass as residing on the sphere's surface. This means that each mess element \(dm\) resides on the infinitesimal surface element \(Rdθ\). If we let \(σ\) denote the mass per unit area on the hollow sphere, then the mass \(dm\) of each mass element can be expressed as

$$dm=σ(2πRsinπ·Rdθ)=\biggl(\frac{M}{4πR^2}\biggr)(2πRsinπ·Rdθ)=\frac{M}{2}sinθdθ.$$

Substituting this for \(dm\) into Equation (3), we have

$$I=\frac{MR^2}{2}\int_{0}^{π}sin^3θdθ,\tag{4}$$

where the angle associated with \(R\) and the axis of rotation for each mass element \(dm\) runs from values of \(0°\) to \(90°\). Solving the definite integral in Equation (4), we have

$$I=\frac{MR^2}{2}\int_{0}^{π}sin^3θdθ=\frac{MR^2}{2}\biggl[\frac{cos^3θ}{3}-cosθ\biggr]_0^π$$

$$=\frac{MR^2}{2}\biggl[\biggl(\frac{cos^3π}{3}-cosπ\biggr)-\biggl(\frac{cos^30}{3}-cos0\biggr)\biggr]$$

Simplifying the cosine functions above, we have

$$I=\frac{MR^2}{2}\biggl[\biggl(\frac{-1}{3}-(-1)\biggr)-\biggl(\frac{1}{3}-1\biggr)\biggr]$$

$$=\frac{MR^2}{2}\biggl[\biggl(\frac{2}{3}\biggr)-\biggl(\frac{-2}{3}\biggr)\biggr]=\frac{MR^2}{2}\biggl(\frac{4}{3}\biggr)$$

Simplifying the above expression, we find that the rotational inertia for a hollow sphere is given by

$$I_{\text{Hollow Sphere}}=\frac{2}{3}MR^2.\tag{5}$$

Given that we know the mass \(M\) and radius \(R\) of a hollow sphere, one can use Equation (5) to calculate the rotational inertia of that sphere.


This article is licensed under a CC BY-NC-SA 4.0 license.