Inertia II - the ring and disk | Scoins.net | DJS

## Inertia II - the ring and disk

A simple ring

A uniform ring  of mass M and radius r has negligible thickness. Each element of the ring has mass m and thus inertia mr². The whole ring must have mass M and inertia Mr².

1  A hollow cylinder spinning on its longitudinal axis is made up of many rings. If the mass of the whole cylinder is M and its radius is r, what is the inertia?

2  Consider a uniform disk of mass M and radius r spinning about an axis perpendicular to the disk and through its centre. The axis is perpendicular to the plane of the disk. Each element of width dr is approximately a ring, and has mass of m (kg/m²) so its moment of inertia is 2πr (dx) m ρ². Sum these for all x (integrate) to get an expression; recognise that M=πr²m and therefore the inertia is ½ Mr².

3  Find the moment of inertia of a uniform hollow cone of base radius r and height h and mass M when rotated about its axis. I think ½ Mr² is an obvious result.

4  Find the moment of inertia of a uniform solid cylinder of base radius r and length h and mass M when rotated about its axis. I think ½ Mr² is another obvious result.

5  Show that the moment of inertia of a uniform solid cone of base radius r and height h and mass M is, when rotated about its axis, 3/10 Mr².

6  Consider a hollow sphere. Draw a diagram to help you show that the elemental moment of inertia is (2πmr² cos ø dø) (rcosø)² and use this to show that the moment of inertia of a hollow sphere is ⅔ Mr².

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