Two masses `2m` and `m` are attached to the ends of the card which passes over a pulley of mass `m` and radius `R`. Calculate the tension in the wire on both sides of the pulley. Assume no slipping.

Explanation

`2mg-T_(1)=2ma` (`i`)
`T_(2)-mg=ma` (`ii`)
pulley:
`tau_(0)=(T_(1)-T_(2))R=I alpha` (`iii`)
`I=(1)/(2)mR^(2)` (`iv`)
` a=Ralpha` (`v`)
From (`iii`),(`iv`) and (`v`),
`(T_(1)-T_(2))R=(1)/(2)mR^(2)((a)/(R))`
`T_(1)-T_(2) =(1)/(2)ma` (`vi`)
Adding (`i`). (`ii`) and (`vi`), we get
`mg=(7)/(2)ma implies a =(2g)/(7)`
`2mg-T_(1)=2mxx(2g)/(7)=(4mg)/(7) implies T_(1)=(10mg)/(7)`
`T_(2)=m(g+a)=m(g+(2g)/(7))=(9mg)/(7)`
Alternatively: considering (blocks `+` pulley) as system and taking torque about `O`
`tau_(0)=2 mgR-mgR=mgR`
`I_(0)=2mR^(2)+mR^(2)+(1)/(2)mR^(2)=(7)/(2)mR^(2)`
`alpha=(tau_(0))/(I_(0))=(mgR)/((7)/(2)mR^(2))=(2g)/(7R)`
`a=Ralpha=(2g)/(7)`