A sphere of mass M and radius r rests on a smooth horizontal floor. A small mass m is attached to a fixed point by a string of length √3r and rests on sphere. String is at 30° to vertical. After release, sphere acceleration is √3mg/(am + bM). Find a + b. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A sphere of mass M and radius r rests on a smooth horizontal floor. A small mass m is attached to a fixed point by a string of length √3r and rests on sphere. String is at 30° to vertical. After release, sphere acceleration is √3mg/(am + bM). Find a + b. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Sphere centre O is at y=r, fixed string support at y=3r, and contact point makes an angle 30° between the string and vertical. Geometry gives contact point relative sphere centre (−√3r/2, +r/2). Step 2 — Let sphere accelerate horizontally at a, and particle initial accelerations be aₓ and aᵧ. From rigid sphere contact, −√3(aₓ−a)+aᵧ = 0; from fixed taut string, aₓ+√3aᵧ = 0. Step 3 — Solve accelerations aₓ=3a/4 and aᵧ=−√3a/4. Step 4 — Let contact normal N point outward from sphere and st…