In the figure `A` and `B` are two blocks of mass `4 kg` and `2 kg` respectively attached to the two ends of light string passing overa disc `C` of mass `40 kg` and radius `0.1 m`. The disc is free to rotate about a fixed horizontal the string does not string does not slip over the disc. Find.
(i) The linear acceleration of mass `B`.
(ii) the number of revolutions made by the disc at the end of `10 sec`. from the start.
(iii) the tension in teh string segment supporting the block `A`.
.

Answer

(i) `(10)/(13) m//s^(2)`,
(ii) `(5000)/(26 p)`
(iii) `(480)/(13 N)`

Explanation

(i) For `A 4g - T_(1) = 4a`….(1)
For `B T_(2) - 2g = 2 a`….(2)
For disc `T_(1) xx R - T_(2) xx R =I alpha`

`(T_(1) - T_(2)) R = ((1)/(2) MR^(2)) (a)/( R)`
`T_(1) - T_(2) = 20 a`l….(3)
adding (1) (2) and (3)
`a = (g)/(13) m//s^(2)`
(ii) `a = R alpha, alpha = (g)/(13) xx (1)/(0.1) = (100)/(13) rad//sec^(2)`
Also `theta =omega t + (1)/(2) alpha t^(2) rArr theta = (1)/(2) xx (100)/(13) xx 10^(2)`
`theta = (5000)/(13) rad`
no of revolution `= (theta)/(2 pi) = (5000)/(26 pi) rev`.
(iii) For `A`
By putting the value of a in equation (1)
`4g - T_(1) = 4a`
`T_(1) = 4g - 4 xx(g)/(13)` or `T = (480)/(13) N`.