Moment of Inertia by Integration and the Parallel-Axis Theorem
Moment of inertia is an integral of distance squared over the mass distribution, so it depends on the axis as well as the body. The parallel-axis theorem moves a known result to any parallel axis without redoing the integral.
What a learner can do afterwards
- Sets up and evaluates the integral for a rod, a disc or a hoop about a stated axis
- Applies the parallel-axis theorem and says why the moment of inertia is smallest about the centre of mass among parallel axes
- Explains why the same body has different moments of inertia about different axes
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Moment of inertia is built piece by piece: chop the body into tiny bits of mass dm, multiply each by its squared perpendicular distance from the axis, and integrate. Far flung mass counts far more, because distance enters squared. That is why the same body has different moments about different axes: move the same mass outward and I grows, so spin changes meet more resistance.
For a uniform rod of mass M and length L about its central perpendicular axis, the density is M over L, so a bit at position x has dm equals M over L dx, and I equals the integral of x squared dm from minus L over 2 to L over 2, which is one twelfth M L squared kilogram metres squared. About one end the bounds run 0 to L, giving one third M L squared.
Symmetry picks the result. A hoop has every bit at radius R, so I equals M R squared with no integral needed. A uniform disc of mass M and radius R about its central perpendicular axis has density M over pi R squared, so a ring at radius r has dm equals 2M over R squared r dr, and I equals the integral of r squared dm from 0 to R, which is half M R squared kilogram metres squared.
For the same mass M, with the hoop and disc of radius R and a uniform rod of length 2R, all about central perpendicular axes, the values are rod one third M R squared, disc half M R squared, and hoop M R squared, so the hoop is largest. Just as linear kinetic energy is half m v squared, rotational kinetic energy is half I omega squared: with I equal to 3 kilogram metres squared and omega 2 radians per second, the energy is 6 joules.
The parallel-axis theorem moves a known result to any parallel axis without redoing the integral: add M times the squared perpendicular shift distance to the centre of mass value. Among parallel axes, shifting away always raises I and shifting toward lowers it, which is why I is smallest about the centre of mass among parallel axes. For a rod of mass 2 kilograms and length 6 metres, the centre value is 6 kilogram metres squared, and shifting 3 metres to the end adds 2 times 3 squared, giving 24 kilogram metres squared.
Square the distance, integrate over the mass, shift parallel axes with M d squared, and spin energy follows half I omega squared.
Three readings, six flashcards were written for this moment of inertia topic. The videos are linked from their original sources.
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