Moment of Inertia

Question

A cheerleader has a batton consisting of light rod of 1m and an point mass of 1/2kg at each end.Find the moment of inertia about

1.an axis perpendicular to the rod and passing through the centre,

2.an axis perpendicular to the rod and passing through one of the masses

3.an axis parallel to the rod and passing through both masses

4.an axis at 45° to the rod and passing through the centre

Answer 1

1. Both point mass are 0.5m form the axis, so total moment of inertia is 2 * 0.5 * 0.5² = 0.25 kgm²
2. One point mass is 0 m from the axis, while the other is 1m, so the total moment of inertia is 0.5 * 1² = 0.5 kgm²
3. Both point mass are 0 m from the axis, so the total moment of inertia is 0 kgm²
4. Both point mass are 0.5 *sin45 from the axis, so the total moment of inertia is 2 * 0.5 * (0.5 *sqrt(2)/2)² = 0.125 kgm²