A uniform rod of length 2ℓ and mass m is suspended from one end by an inextensible vertical string and other end lies on smooth ground. The angle made by rod with vertical is θ = sin⁻¹(1/√3). If N₁ and N₂ represent the contact force from ground on rod just before and just after cutting the string, then find the ratio of N₁/N₂.
Physics · Mixed Chapter
JEE Mains

Question

A uniform rod of length 2ℓ and mass m is suspended from one end by an inextensible vertical string and other end lies on smooth ground. The angle made by rod with vertical is θ = sin⁻¹(1/√3). If N₁ and N₂ represent the contact force from ground on rod just before and just after cutting the string, then find the ratio of N₁/N₂.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Before cutting, torque equilibrium about the ground contact gives string tension T=mg/2; vertical force balance gives N₁=mg−T=mg/2. Step 2: Immediately after the cut the rod starts rotating with zero initial angular speed; the bottom end remains on the smooth horizontal ground. Step 3: I꜀ₘ=(1/3)mℓ² and τ about CM=N₂ℓ sinθ, so α=3N₂sinθ/(mℓ). The centre-of-mass vertical acceleration is −αℓsinθ. Step 4: N₂−mg=m(−αℓsinθ), hence N₂[1+3sin²θ]=mg. Since sin²θ=1/3, N₂=mg/2. Therefore N₁/N₂=1.