Two wooden discs, one with radius 2 cm and mass 1 kg and the other with radius 4 cm and mass 2 kg, are welded together coaxially and mounted on a frictionless axis through their common centre. A light string is wrapped around the edge of the smaller disc, and a 3 kg block is suspended from the free end of the string. The acceleration of the block after it is released is (2/3)a m s⁻²; then what is the value of a? Assume that string does not slip on the discs.
Physics · Mixed Chapter
JEE Mains

Question

Two wooden discs, one with radius 2 cm and mass 1 kg and the other with radius 4 cm and mass 2 kg, are welded together coaxially and mounted on a frictionless axis through their common centre. A light string is wrapped around the edge of the smaller disc, and a 3 kg block is suspended from the free end of the string. The acceleration of the block after it is released is (2/3)a m s⁻²; then what is the value of a? Assume that string does not slip on the discs.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Combined moment of inertia of two coaxial solid discs is I=(1/2)m₁r₁²+(1/2)m₂r₂²=(1/2)(1)(0.02²)+(1/2)(2)(0.04²)=0.0018 kg·m². Step 2: For the hanging 3 kg block, mg−T=ma. No slipping gives α=a/0.02, and torque on discs is T(0.02)=Iα. Step 3: Eliminate T: a=mg/[m+I/r₁²]=30/[3+0.0018/0.0004]=30/7.5=4 m/s². Step 4: Given actual acceleration=(2/3)aₛ, we have 4=(2/3)aₛ and aₛ=6.