A thin uniform rod of length R*sqrt(2) rests inside a hollow uniform sphere of inner radius R, with both ends touching the inner wall. The rod and the sphere have equal mass m. The system is released from rest on a frictionless floor (and the sphere-rod contact is frictionless). By the tim

Question

A thin uniform rod of length R*sqrt(2) rests inside a hollow uniform sphere of inner radius R, with both ends touching the inner wall. The rod and the sphere have equal mass m. The system is released from rest on a frictionless floor (and the sphere-rod contact is frictionless). By the time the rod swings to a horizontal orientation, how far has the sphere moved horizontally relative to the ground?

Accepted Answer

R*sqrt(2)/4. Because equal masses share a fixed centre of mass, the sphere shifts by half of the rod's horizontal shift relative to the sphere; that shift equals the perpendicular distance R/sqrt(2) from centre to the chord.

Solution

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