A uniform rod initially inclined at 60° to horizontal is supported by a vertical string at its top and a rough floor at bottom. String is suddenly cut; bottom end does not slip. Given mg = 16/√3 N, find initial friction force on rod (N). [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A uniform rod initially inclined at 60° to horizontal is supported by a vertical string at its top and a rough floor at bottom. String is suddenly cut; bottom end does not slip. Given mg = 16/√3 N, find initial friction force on rod (N). [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Immediately after cutting the string, lower rod tip is stationary and rod begins angular acceleration about this tip. Step 2 — Moment of inertia Iₚ=mℓ²/3; torque τ=mg(ℓ/2)cos60°, so α = 3g cos60°/(2ℓ). Step 3 — Centre horizontal acceleration magnitude aₓ=(ℓ/2)α sin60° = (3g/4)sin60°cos60°. Step 4 — Friction f=maₓ=(3mg/8)sin120°. With mg=16/√3 N, sin120°=√3/2: f=(3/8)(16/√3)(√3/2)=3 N.