Two beads of masses `m_(1)` and `m_(2)` are connected to each other by massless string. Both beads are free to mover over fixed smooth circular loop in vertical plane. At any instant they are in the position shown and subtends angle `90^(@)` at centre. Find tangential accelerations of both beads and tension in the string if string remains taut.

Explanation

Let `a_(1)` and `a_(2)` be the tangential accelerations of two beads respectively. Then for bead `A`
`m_(1)g sin30^(@)+T sin45^(@)=m_(1)a_(1)`…..(`i`)
For bead `B`
`m_(2)gsin60^(@)-Tsin45^(@)=m_(2)a_(2)`…..(`ii`)
since string remains taut
`:. a_(1)cos45^(@)=a_(2)cos45^(@)`
(Acceleration along string remains same)
`rArr a_(1)=a_(2)=a`
`:. ` from (`i`) and (`ii`)
`a=(g(m_(1)+sqrt(3)m_(2)))/(2(m_(1)+m_(2)))`
`(T)/(sqrt(2))=(m_(1)g(m_(1)+sqrt(3)m_(2)))/(2(m_(1)+m_(2)))-(m_(1)g)/(2)`
`T=((sqrt(3)-1)(m_(1)m_(2)g))/(sqrt(2)(m_(1)+m_(2)))`