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Rotational inertia

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 I. For a single body such as the tennis ball of mass m (shown in Figure 1), rotating at radius r from the axis of rotation the rotational inertia is
and consequently rotational inertia has SI units of kgm2.
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 m 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 FT to the mass. From Newton’s 2ⁿᵈ law,
this can also be written as
Newton's 2ⁿᵈ law relates force to acceleration. In rotational mechanics torque τ takes the place of force. Multiplying both sides by the radius gives the expression we want.
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 Nm and a maximum rotation speed of 150 rad/s is connected to a flywheel with rotational inertia 0.1 kgm2. What angular acceleration will the flywheel experience as the motor is switched on?
Exercise 1b:
How long will the flywheel take to reach a steady speed if starting from rest?

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.
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?
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?

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 r rotated about a central axis is
and for a hollow cylinder with inner and outer radii ri and ro respectively,
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.
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 o as long as we known the moment of inertia of the shape around its centroid c, mass m and distance d between points o and c.
Exercise 3:
If the shape shown in Figure 5 is made by welding three 10 mm thick metal discs (each with mass 50 kg) to a metal ring with mass 100 kg. If rotated about a central axis (out of the page), what is the rotational inertia of the object?
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.

Want to join the conversation?

  • leaf green style avatar for user Sukanya Sudhiram
    I do not understand how rotational inertia increases with increasing distance of mass?
    isn't it easier to push open a door from a farther point than a point close to its hinges?
    (39 votes)
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    • blobby green style avatar for user Teacher Mackenzie (UK)
      Good question.

      OK; so there are two concepts here and we much keep them seperate.

      The pushing of the door is correct; ie the further away from the hinge, the easier it is to turn. This is the concept of moments. (or torque) and the moment of the force = distance x force. bigger distance = easier to turn.

      Moment of inertia is a different concept. This is about how easy it is to turn a body based on its mass and the distribution of the mass.

      so, if you have a mass of 20kg attached to your door near the hinge and you push the door handle, it will be easy to 'get it moving' or, indeed, to stop it moving. If, however, you now replace the 20kg mass nearer to the edge of the door (far away from the hinge) then it becomes more difficult to get it moving (or stop it moving).

      two concepts: moment of a force
      Moment of inertia.

      make sense??
      (103 votes)
  • duskpin ultimate style avatar for user Alex.Piotrowski
    how do you derive the moment of inertia, using calculus?
    (22 votes)
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    • piceratops tree style avatar for user jrapeur
      You do the surface integral of the object described by its boundaries. It becomes the double integral of d^2.y.dS, where d is the distance of the particles that make the object to the axis, y is the density function, which is usually known and not linear and dS is the area element, which is defined as being the square root of your external product sqrt(Dg1 x Dg2), where g is the parametrization of your surface, Dg1 is the first column of your derivative matrix and Dg2 is the second column of your derivative matrix.
      (25 votes)
  • blobby green style avatar for user ranjit pradhan
    how did you find d in exercise 3?
    (13 votes)
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    • blobby green style avatar for user levgenid
      Although it is not exactly stated it is implied by Figure 5 that the center of each disk is right in the middle between the outer radius (ro=1 meter) and the inner radius (ri=0.75 meters) of the main steel ring. So by d=1/2(1+0.75) he finds the mean of the outer and inner radius.
      (9 votes)
  • leaf green style avatar for user pedro magalhaes
    In the previous video, I think David said that for objects rotating around a point external to them (like the three disks in exercise 3), we should consider them as if all the mass was in the center of mass. However, in the solution of exercise 3, it's calculated the moment of inertia for each object, considering its shape, and then it's used the parallel axis theorem, which gives a different result. How shall I solve this exercises?
    (5 votes)
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    • aqualine ultimate style avatar for user Kadmilos
      Imagine what would happen as the ring gets bigger, and the radius of the disk gets relatively smaller. Inertia is increased more per unit of radius than per unit of mass. In exercise 3, much of the three disks mass was distributed throughout the ring. But as you increase the radius of the ring, the mass of the disks stays close to the ring, and treating those masses as 'point masses' on the ring, and also treating the thickness of the ring as negligible becomes more feasible.

      Just like when we use 'g' for acceleration due to gravity on earth, and for acceleration due to gravity 100m above earth. It's true that there is a difference, but not usually a significant difference. However if Earths radius where only 200m, then that difference would be very significant.
      (7 votes)
  • mr pink orange style avatar for user Jennifer Kopajtic
    who discovered inertia
    (1 vote)
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  • blobby green style avatar for user GABRIELE BARBANTI
    If the cylinder has the center of mass in a different position (not in the object center) how do I calculate the rotational inertia about this reference? For example if I have a cylinder of length 120 cm and the center of mass is at 80 cm from the top how do I calculate the rotational inertia Ix=Iy? Thank you
    (4 votes)
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    • leaf red style avatar for user Bean Jaudrillard
      A cylinder whose centre of mass doesn't coincide with it's geometrical centre has a non-uniform density - the density is variable throughout the object. Without a function for its density it's going to be impossible to know it's rotational inertia, but if you do know the function, you can use calculus to calculate the rotational inertia, specifically by solving the integral I = ∫ r² dm (which is just the calculus counterpart of I = ∑ mr² ). Using a function for density you can relate dm to r and then integrate with respect to r.
      (4 votes)
  • primosaur tree style avatar for user H.SHAH
    Do I need to memorize equations of all shapes to find rotational inertia?
    (4 votes)
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  • aqualine ultimate style avatar for user Halomoan
    What's the idea behind moment of inertia? In my physic book it says that moment of inertia of something comes from the kinetic energy of it (E=1/2(m(v)^2)).

    This is how it is explained:

    -Explain Begin-
    "But because v=rw the equation become this:"
    Linear speed v analog with angular speed w and physicists think that to make the equation c analog with equation a, the mr(r)^2 must be analog with m on equation a. This formula is now called moment of Inertia (noted as I)
    -Explain End-
    I photographed my book (It's in Bahasa Indonesia) if you're curious: https://s30.postimg.org/sm6dwqcm9/20170206_161635.jpg
    Now what he heck is analog with? What is analog? Why should it be analog-ed?
    (3 votes)
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  • blobby green style avatar for user Farhan Naufal
    In exercise 3, why do we must consider the three small discs to rotate about its axis, they also rotate around the center right? Why don't just consider them to rotate around the center?
    (3 votes)
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  • blobby green style avatar for user sali1
    How does frictional torque affect the angular acceleration? Like if a frictional torque is applied, where does that go in the equation to find angular acceleration of a rod/disk?
    (1 vote)
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