The solid S consists of a circular cylinder radius r, height h, with a hemispherical cap at one end. S is placed with the center of the cap on the table and the axis of the cylinder vertical. For some k, equilibrium is stable if r/h > k, unstable if r/h < k and neutral if r/h = k. Find k and show that if r/h = k, then the body is in equilibrium if any point of the cap is in contact with the table.
Solution
Let O be the centre of the base of the hemispherical cap. As the solid rolls, O is always a height r above the table. Consider a point P on the axis a distance d from O and inside the cap. If the solid is rolled to a position where the axis is at an angle θ to the vertical, then P ends up a vertical distance d cos θ below O, so its height has increased. Similarly, a point on the axis a distance d above O would end up a vertical distance d cos θ above O, so its height would decrease. So equilibrium will be stable, neutral or unstable according as the centre of mass is below, at or above O.
We start by finding the position of the centre of mass of the cap. Take the x-axis along the axis with the origin at O. Assume the density is ρ. The moment about an axis perpendicular to the x-axis is ∫0r π(r2 - x2)ρ x dx = π ρ r4/4. The centre of mass of the cylindrical portion of the solid is obviously a distance h/2 from O, so we require (π ρ r2h) h/2 = π ρ r4/4 or r/h = √2 get get the centre of mass at O. Thus k = √2.
If the centre of mass is at O, then the centre of mass is always at the same height above the table, so the solid is in equilibrium however far it is tilted from the vertical.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002