Concepts: moment of inertia and the axis theorems
System of Particles and Rotational Motion22 min readLesson 4 of 8
4.1 Mass distribution matters
The moment of inertia of a body about an axis is the sum of each mass element multiplied by the square of its distance from that axis. It is the rotational analogue of mass: a body with a larger moment of inertia is harder to start rotating and harder to stop. Unlike mass, the moment of inertia is not a property of the body alone, because the same body has different values about different axes.
Pulling mass away from the axis increases the moment of inertia, which is why a ring has a larger value than a disc of the same mass and radius, and why a hollow sphere has a larger value than a solid one. The radius of gyration is defined so that the body's moment of inertia equals its mass times the square of that radius, and it gives a single length that describes how far the mass is spread from the axis.
The rotational kinetic energy and the angular momentum both depend on the moment of inertia, so every rotational calculation begins by finding the correct value for the chosen axis. The formulas for the ring, disc, rod, sphere and cylinder are the ones used most often and should be known along with the axis each belongs to.
4.2 The parallel and perpendicular axis theorems
The parallel axis theorem compares the moment about any axis with the moment about a parallel axis passing through the centre of mass. The off-centre value exceeds the centre value by the total mass times the square of the gap between the axes. This is why a rod has a larger moment of inertia about its end than about its centre, and the theorem gives the end value as M L squared over twelve plus M L squared over four, which is M L squared over three.
The perpendicular axis theorem applies to a thin flat lamina and says that the moment of inertia about an axis perpendicular to the plane equals the sum of the moments about two perpendicular axes lying in the plane through the same point. A disc about its central axis, for example, has a moment about a diameter equal to half of the central value. The theorem lets a planar result be converted between axes without a new integral.
The two theorems together cover most composite problems. For a rectangular plate about an axis through a corner, use the parallel axis theorem on each of the two in-plane axes; for a ring about a diameter, use the perpendicular axis theorem and symmetry. The theorems only apply about a centre of mass axis in the parallel case, and only to a lamina in the perpendicular case, so check the conditions before using them.
4.3 Standard results worth remembering
A ring of mass M and radius R about its central perpendicular axis has I equal to M R squared. A disc of the same mass and radius has half of that, and about a diameter it has a quarter of M R squared by the perpendicular axis theorem. A solid sphere has two fifths of M R squared about a diameter, and a hollow sphere has two thirds. A solid cylinder about its axis has half of M R squared, and a hollow cylinder with thin walls has M R squared.
A rod of length L about a perpendicular axis through its centre has M L squared over twelve. About a parallel axis through one end it has M L squared over three. About an axis along the rod the value is negligible for a thin rod. A rectangular lamina of sides a and b about an axis through the centre and perpendicular to the plane has M times the sum of a squared and b squared, divided by twelve.
The moment of inertia of a body with a hole can be found by subtracting the moment of the removed part about the same axis, using the parallel axis theorem to shift it to the hole's position. This is the standard method for a disc with a hole or a plate with a circular cut out.
4.4 Worked examples
Find the moment of inertia of a uniform rod of mass 2 kg and length 1 m about an axis through its end and perpendicular to the rod. The centre value is M L squared over twelve, which is 2 times 1 over 12, about 0.167 kg metre squared. The parallel axis theorem adds M times the distance squared, with distance 0.5 m, giving 0.5, so the end value is about 0.667 kg metre squared, which is also M L squared over three.
Find the moment of inertia of a disc of mass 2 kg and radius 0.5 m about a diameter. The central value is half M R squared, 0.25, and a diameter value is half of that, 0.125 kg metre squared. For a point on the rim about an axis through the rim and perpendicular to the disc, the parallel axis theorem gives 0.25 plus 2 times 0.25, which is 0.75 kg metre squared.
A composite body made of a disc and a ring fixed together is handled by adding the moments about the common axis. The calculation is only tedious, not difficult, and a table of masses, moments and axis distances keeps it clean.
Exam
State the axis in every moment of inertia answer. A number without the axis is meaningless, and most of the marks in a moment of inertia derivation are for saying clearly which axis the result belongs to.