A small block slides down from rest at point A on the surface of a smooth circular cylinder, as shown. At point B, the block falls off (leaves) the cylinder. The equation relating the angles θ₁ and θ₂ is given by

Question

G1 Physics · Accepted Answer

For a block sliding on the outside of a smooth sphere, at the leaving point N=0, so v²=gR cosθ₂. Energy from rest at θ₁ gives v²=2gR(cosθ₁-cosθ₂). Equating gives cosθ₂=(2/3)cosθ₁. Hence C.