Two uniform cylinders, each if mass `m = 10 kg` and radius `r = 150 mm`, are connected by a rought belt as shown. If the system is released from rest. Determine.
(a) the velocity of the centre of cylider `A` after it has moved through `1.2 m` and
(b) the tension in the portion of the belt connecting the two cylinders.
.

Answer

(a) `4 sqrt((3)/(7)) m//s`,
(b) `(200)/(7) N`

Explanation

`m = 10 kg, r = 150 mm`
(a) `W_(g) = Delta KE`
`mgh = ((1)/(2) mv^(2) + (1)/(2) I omega^(2)) + (1)/(2) I omega^(, 2)`
`mgh =(1)/(2) mv^(2) +(1)/(2) ((1)/(2) mr^(2)) ((v)/(r))^(2) +(1)/(2) ((1)/(2) mr^(2))((2v)/(r))^(2)`
`mgh = (7)/(4) mv^(2)`
`v = sqrt((4)/(7) xx 10 xx 1.2)`
`v = 4 sqrt((3)/(7)) m//s`
(b) for translational motion of `C.M` of `A`
`mg -(T_(1) + T_(2)) = ma` ....(1)
For rotational motion of `B`
`T_(2) xx r = I alpha`
`T_(2) xx r = ((1)/(2) mr^(2)).((2a)/(r)) rArr T_(2) = ma`....(2)
For rotational motion of `A`
`(T_(1) - T_(2)) r = ((1)/(2) mr^(2)) (a)/( r)`
`T_(1) -T_(2) = (ma)/(2)` ....(3)
`T_(1) = (3ma)/(2)`
from (2) & (2)
`T_(1) = (2)/(2) T_(2)` ....(4)
solving (1) (2) and (4)
`T_(2) = (2)/(7) mg = (200)/(7) N`.
.