A uniform equilateral triangular plate ABC rotates freely about vertex A in a vertical plane. Initially AB is horizontal and plate is released. Determine XY for the prescribed formula for acceleration of centre of mass. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A uniform equilateral triangular plate ABC rotates freely about vertex A in a vertical plane. Initially AB is horizontal and plate is released. Determine XY for the prescribed formula for acceleration of centre of mass. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Use gravitational torque about the pivot A: τ = mg x₍CM₎, where x₍CM₎ is horizontal distance of the triangular-plate centre of mass from A. Step 2 — Angular acceleration at release α = τ/Iₐ, with Iₐ moment of inertia of plate about pivot A. Step 3 — Initial angular velocity is zero, so centre-of-mass acceleration is purely tangential: a₍CM₎ = αr₍CM₎. Step 4 — Substitute the equilateral-triangle centroid r₍CM₎ and source Iₐ into requested parametrization. The printed answer is 6. QUALIT…