In the system shown in the figure blocks `A` and `B` have mass `m_(1)=2 kg` and `m_(2)=26//7 kg` respectively. Pulley having moment of inertia `I=0.11kgm^(-2)` can rotate without friction about a fixed axis. Inner and outer radii of pulley are `a=10 cm` and `b=15 cm` respectively. `B` is hanging with the thread wrapped around the pulley, while `A` lies on a rough inclined plane.
Coefficient of friction being `mu=sqrt(3)//10`
Calculate
as. Tension in each thread, and
b. Acceleration of each block `(g=10 ms^(-2))`

Explanation

When the system is released, weight `m_(2)g` tries to rotate the pulley in clockwise direction while down the plane component `m_(1)g sin 30^@` of weight of block `A` tries to rotate it anticlockwise but moment produced by `m_(2)g` is greater, therefore, the pulley has tendency to rotated clockwise. Let is angular acceleration be `alphar`. The acceleration of blocks `A` and `B` will be equal to `aalpha`(up the plane) and `balpha` (vertically downwards) respectively.
Let tension in threads connected with blocks `A` and `B` be `T_(1)` and `T_(2),` respectively.
Considering free body diagrams .

For forces on block `A, N=m_(1)gcos30^@` ...........i
`T_(1)-m_(1)sin30^i@-muN=m_(1)(aalpha)`.........ii
For forces on block `B, m_(2)g-T_(2)=m_(2)(balpha)`........iii
Taking moments of forces acting on pulley, about axis of rotation,
`T_(1)=17N, T_(2)=26N` and `alpha-20 rads^(-2)`
Acceleration of block `A=aalpha=2ms^(-2)` (up the plane)
Acceleration of block `B=balpha=3ms^(-2)` (downward)