Moment Of Inertia (Angular & Rotational Inertia): Definition, Equation, Units

Whether it's an ice skater pulling in her arms and spinning faster as she does or a cat controlling how quickly it spins during a fall to ensure it lands on its feet, the concept of a moment of inertia is crucial to the physics of rotational motion.

Otherwise known as rotational inertia, the moment of inertia is the rotational analogue of mass in the second of Newton's laws of motion, describing the tendency of an object to resist angular acceleration.

The concept might not seem too interesting at first, but in combination with the law of the conservation of angular momentum, it can be used to describe many fascinating physical phenomena and predict motion in a wide range of situations.

The moment of inertia for an object describes its resistance to angular acceleration, accounting for the distribution of mass around its axis of rotation.

It essentially quantifies how difficult it is to change the speed of an object's rotation, whether that means starting its rotation, stopping it or changing the speed of an already rotating object.

It's sometimes called rotational inertia, and it's useful to think about it as an analogue of mass in Newton's second law: ​_F_net​ = ​ma_​. Here, the mass of an object is often called the inertial mass, and it describes the object's resistance to (linear) motion. Rotational inertia works just like this for rotational motion, and the mathematical definition always includes mass.

The equivalent expression to the second law for rotational motion relates ​torque​ (​τ​, the rotational analogue of force) to angular acceleration ​α​ and moment of inertia ​I​:

\(\tau =I\alpha\)

The same object can have multiple moments of inertia, however, because while a big part of the definition is about the distribution of mass, it also accounts for location of the axis of rotation.

For example, while the moment of inertia for a rod rotating around its center is ​I​ = ​ML​²/12 (where ​M​ is mass and ​L​ is the length of the rod), the same rod rotating around one end has a moment of inertia given by ​I​ = ​ML​²/3.

So a body's moment of inertia depends on its mass ​M​, its radius ​R​ and its axis of rotation.

In some cases, ​R​ is referred to as ​d​, for distance from the axis of rotation, and in others (as with the rod in the previous section) it's replaced by length, ​L​. The symbol ​I​ is used for moment of inertia, and it has units of kg m².

As you might expect based on what you've learned so far, there are many different equations for moment of inertia, and each refers to a specific shape and a specific rotation axis. In all moments of inertia, the term ​MR​² appears, although for different shapes there are different fractions in front of this term, and in some cases there may be multiple terms summed together.

The ​MR​² component is the moment of inertia for a point mass at a distance ​R​ from the axis of rotation, and the equation for a specific rigid body is built up as a sum of point masses, or by integrating an infinite number of small point masses over the object.

While in some cases it may be useful to derive the moment of inertia of an object based on a simple arithmetic sum of point masses or by integrating, in practice there are many results for common shapes and axes of rotation that you can simply use without needing to derive it first:

Solid cylinder (symmetry axis):

\(I = \frac{1}{2} MR^2\)

Solid cylinder (central diameter axis, or the diameter of the circular cross-section in the middle of the cylinder):

\(I = \frac{1}{4} MR^2+\frac{1}{12} ML^2\)

Solid sphere (central axis):

\(I = \frac{2}{5} MR^2\)

Thin spherical shell (central axis):

\(I = \frac{2}{3} MR^2\)

Hoop (symmetry axis, i.e., perpendicularly through the center):

\(I = MR^2\)

Hoop (diameter axis, i.e., across the diameter of the circle formed by the hoop):

\(I = \frac{1}{2} MR^2\)

Rod (center axis, perpendicular to rod length):

\(I = \frac{1}{12} ML^2\)

Rod (rotating about end):

\(I = \frac{1}{3} ML^2\)

Understanding why there are different equations for each axis of rotation is a key step to grasping the concept of a moment of inertia.

Think about a pencil: You can rotate it by spinning it around in the middle, by the end or by twisting it around its central axis. Because the rotational inertia of an object depends on the distribution of mass about the axis of rotation, each of these situations is different and requires a separate equation to describe it.

You can get an instinctive understanding of the concept of moment of inertia if you scale this same argument up to a 30-foot flag pole.

Spinning it end over end would be very difficult – if you could manage it at all – whereas twirling the pole about its central axis would be much easier. This is because torque depends strongly on the distance from the axis of rotation, and in the 30-foot flag pole example, spinning it end over end involves each extreme end 15 feet away from the axis of rotation.

However, if you twirl it around the central axis, everything is quite close to the axis. The situation is much like carrying a heavy object at arm's length vs. holding it close to your body, or operating a lever from the end vs. close to the fulcrum.

This is why you need a different equation to describe the moment of inertia for the same object depending on the rotation axis. The axis you choose affects how far parts of the body are from the axis of rotation, even though the mass of the body remains the same.

The key to calculating the moment of inertia for a rigid body is learning to use and apply the appropriate equations.

Consider the pencil from the previous section, being spun end-over-end around a central point along its length. While it isn't a ​perfect​ rod (the pointed tip breaks this shape, for instance) it can be modeled as such to save you having to go through a full moment of inertia derivation for the object.

So modeling the object as a rod, you would use the following equation to find the moment of inertia, combined with the total mass and length of the pencil:

\(I = \frac{1}{12} ML^2\)

A bigger challenge is finding the moment of inertia for composite objects.

For example, consider two balls connected together by a rod (which we will treat as massless to simplify the problem). Ball one is 2 kg and positioned 2 m away from the axis of rotation, and ball two is 5 kg in mass and 3 m away from the rotation axis.

In this case, you can find the moment of inertia for this composite object by considering each ball to be a point mass and working from the basic definition that:

\(\begin{aligned}
I &= m_1r_1^2 + m_2r_2^2 + m_3r_3^2....\)
\(&= \sum_{\mathclap{i}}m_ir_i^2
\end{aligned}\)

With the subscripts simply differentiating between different objects (i.e., ball 1 and ball 2). The two-ball object would then have:

\(\begin{aligned}
I &= m_1r_1^2 + m_2r_2^2\
&= 2 \;\text{kg} × (2 \;\text{m})^2 + 5 \;\text{kg} × (3 \;\text{m})^2\)
\(&= 8 \;\text{kg m}^2 + 45 \;\text{kg m}^2\)
\(&= 53 \;\text{kg m}^2
\end{aligned}\)

Angular momentum (the rotational analogue for linear momentum) is defined as the product of the rotational inertia (i.e., the moment of inertia, ​I​) of the object and its angular velocity ​ω​), which is measured in degrees/s or rad/s.

You'll undoubtedly be familiar with the law of conservation of linear momentum, and angular momentum is also conserved in the same way. The equation for angular momentum ​L​) is:

\(L = Iω\)

Thinking about what this means in practice explains many physical phenomena, because (in the absence of other forces), the higher an object's rotational inertia, the lower its angular velocity.

Consider an ice skater spinning at a constant angular velocity with arms outstretched, and note that his arms being outstretched increases the radius ​R​ about which his mass is distributed, leading to a greater moment of inertia than if his arms were close to his body.

If ​L​₁ is calculated with his arms outstretched, and ​L​₂, after drawing his arms in must have the same value (because angular momentum is conserved), what happens if he decreases his moment of inertia by drawing in his arms? His angular velocity ​ω​ increases to compensate.

Cats perform similar movements to help them land on their feet when falling.

By outstretching their legs and tail, they increase their moment of inertia and reduce the speed of their rotation, and conversely they can draw in their legs to decrease their moment of inertia and increase their speed of rotation. They use these two strategies – along with other aspects of their "righting reflex" – to ensure their feet land first, and you can see distinct phases of curling up and stretching out in time-lapse photographs of a cat landing.

Continuing the parallels between linear motion and rotational motion, objects also have rotational kinetic energy in the same way they have linear kinetic energy.

Think about a ball rolling across the ground, both rotating about its central axis and moving forward in a linear fashion: The total kinetic energy of the ball is the sum of its linear kinetic energy ​E​_k and its rotational kinetic energy ​E​_rot. The parallels between these two energies are reflected in the equations for both, remembering that an object's moment of inertia is the rotational analogue of mass and its angular velocity is the rotational analogue of linear velocity ​v​):

\(E_k = \frac{1}{2}mv^2\)

\(E_{rot} = \frac{1}{2}Iω^2\)

You can clearly see that both equations have exactly the same form, with the appropriate rotational analogues substituted for the rotational kinetic energy equation.

Of course, to calculate the rotational kinetic energy, you'll need to substitute the appropriate expression for the moment of inertia for the object into the space for ​I​. Considering the ball, and modeling the object as a solid sphere, the equation is this case is:

\(\begin{aligned}
E_{rot} &= \bigg(\frac{2}{5} MR^2\bigg) \frac{1}{2} ω^2\)
\(&= \frac{1}{5}MR^2 ω^2
\end{aligned}\)

The total kinetic energy (​E​_tot) is the sum of this and the ball's kinetic energy, so you can write:

\(\begin{aligned}
E_{tot} &= E_k + E_{rot}\)
\(&= \frac{1}{2}Mv^2 + \frac{1}{5}MR^2 ω^2
\end{aligned}\)

For a 1-kg ball moving at a linear speed of 2 m/s, with a radius of 0.3 m and with an angular velocity of 2π rad/s, the total energy would be:

\(\begin{aligned}
E_{tot} &= \frac{1}{2} 1 \;\text{kg} × (2 \;\text{m/s})^2 + \frac{1}{5}(1 \;\text{kg} × (0.3 \;\text{m})^2 × (2π \;\text{rad/s})^2)\)
\(&= 2 \;\text{J} + 0.71 \;\text{J}\)
\(&= 2.71 \;\text{J}
\end{aligned}\)

Depending on the situation, an object might possess only linear kinetic energy (for example, a ball dropped from a height with no spin imparted onto it) or only rotational kinetic energy (a ball spinning but staying in place).

Remember that it is ​total​ energy that is conserved. If a ball is kicked at a wall with no initial rotation, and it bounces back at a lower speed but with a spin imparted, as well as the energy lost to sound and heat when it made contact, part of the initial kinetic energy has been transferred to rotational kinetic energy, and so it ​can't​ possibly move as fast as it did before bouncing back.