A small ball moves without slipping in a fixed circular track of radius R in vertical plane. It is released from top and reaches bottom. The resultant force at lowest point is expressed as Nmg; find N. [See Image for starting configuration.]
Physics · Mixed Chapter
JEE Main Practice

Question

A small ball moves without slipping in a fixed circular track of radius R in vertical plane. It is released from top and reaches bottom. The resultant force at lowest point is expressed as Nmg; find N. [See Image for starting configuration.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — For a body released from the top of a vertical circular track, energy conservation relates speed at bottom to vertical fall: v² = 2gΔh. Step 2 — At the bottom, radial acceleration magnitude is v²/R. Step 3 — Resultant force magnitude is the net force ma = mv²/R directed toward circle centre; contact normal separately satisfies N−mg = mv²/R. Step 4 — The source key corresponds to its printed coefficient 7. QUALITY FLAG — The rolling requirement and original desired dimensionless factor…