A particle of mass m is attached to an end of a uniform rod of mass M = 2m and length `l_(1)` which is suspended through its mid point by an inextensible string as shown initially the rod is in horizontal position and at rest. The system is released from this position Just after the release

Options

A. The angular acceleration of the system is `(6g)/(5l)`
B. The angular acceleration of the system is `(2g)/(5l)`
C. The Tension in the string is `(12)/(5) mg`
D. The Tension in the string is `(2)/(5)` mg

Answer

A, C

Explanation

`(mgl)/(2)=((2ml^(2))/(12)+(ml^(2))/(4))alpha`
`alpha=(6g)/(5l)rArr a_(cm)=(alphal)/(6)=(g)/(5)`
`3mg -T=(3mg)/(5)rArr T=(12mg)/(5)`
Alternate

The FBD of the rod and the ball are shown. Applying `tai = 1 alpha` about the CM of the rod we have,
`N((l)/(2))=((Ml^(2))/(12))alpha`
Writting newtons lind law in the vertical direction no the CM of the rod we have
`T-N-2mg=0`
and writting newtons `II` law in the vertical direction on the ball we have
`mg -N = m ((l)/(2))alpha`
Solve these equations.