Question

A uniform solid cylinder of mass M and radius R rotates on a horizontal. friction less axle. Two masses hung from light cords wrapped around the cylinder. If the system is released from rest, the tension m each cord is-

• A. $\frac{Mmg}{(M+m})$
• B. $\frac{Mmg}{(M+2m})$
• C. $\frac{Mmg}{(M+3m})$
• D. $\frac{Mmg}{(M+4m})$

Answer 1

The correct option is D $\frac{Mmg}{(M+4m})$

$mg-T=ma$

$2TR=I\propto$

$T=\frac{MR\propto }{4}$

$a=R\propto$

$mg=ma+\frac{ma}{4}$

$T=\frac{M}{4}\frac{4mg}{4m+M}$

$T=\frac{Mmg}{M+4m}$