A uniform disc of mass m and radius R is on a rough horizontal surface (with circular surface in contact). Coefficient of static friction between disc and the rough surface is μ. The minimum amount of torque required to cause rotation of the disc about an axis passing through its centre and perpendicular to its plane is (n/3) μmgR. The value of n is ...
Physics · Mixed Chapter
JEE Mains

Question

A uniform disc of mass m and radius R is on a rough horizontal surface (with circular surface in contact). Coefficient of static friction between disc and the rough surface is μ. The minimum amount of torque required to cause rotation of the disc about an axis passing through its centre and perpendicular to its plane is (n/3) μmgR. The value of n is ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: At limiting static friction, the shear force on an annular ring of radius r and area 2πr dr is dF=μ(mg/πR²)2πr dr. Step 2: Its torque about the centre is dτ=r dF. Integrating from 0 to R gives τ=∫₀ᴿ[2μmg/R²]r² dr=(2/3)μmgR. Step 3: Compare with (n/3)μmgR to obtain n=2.