A uniform solid cylinder (1 kg) and a thin plank (0.5 kg) move on a smooth plane with no slipping at contacts. Horizontal applied force is 0.25 N. Find N from the source expression for plank acceleration. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A uniform solid cylinder (1 kg) and a thin plank (0.5 kg) move on a smooth plane with no slipping at contacts. Horizontal applied force is 0.25 N. Find N from the source expression for plank acceleration. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Use rolling-no-slip constraints at ground–cylinder and cylinder–plank contacts to link their velocities. Step 2 — Apply Newton’s second law to plank and cylinder, and torque equation Iα = fR to cylinder (I = (1/2)m₁R²). Step 3 — Eliminate contact friction forces to determine plank acceleration from applied F = 0.25 N, m₁ = 1 kg, m₂ = 0.5 kg. Step 4 — The printed result uses dimensionless N = 7 in its specified acceleration expression. QUALITY FLAG — The scaling/formula for N is not leg…