Two masses `m_(1)` and `m_(2)` are joined with the help of massless string passing over a pulley of Radius `R` and mass `M`. Assume that string doesn't slip over the pulley. Take `m_(1)=10kg`, `m_(2)=5kg`, `M=10 kg`, `R=1m`.

`(a)` Calculate the acceleration of each book.
`(b)` Calculate the tensions in the thread.

Explanation

we choose `m_(1)`, `m_(2)` and pulley as separate systems and draw their `FBD`.
`(a)`
`m_(1)g-T_(1)=m_(1)a`………`(1)` [from FBD of block of mass `m_(1)`]
`T_(2)-m_(2)g=m_(2)a`………..`(2)` [from FBD of block of mass `m_(2)`]
`T_(1)R-R_(2)R=l alpha` [from FBD of the pulley]
`impliesT_(1)R-T_(2)R=(MR^(2))/(2)alpha`...........`(3)`
From equation `(3)` we can conculde that tension in a single thread will be different on two sides of the pulley, otherwise it will not be possible for the pulley to accelerate.
As we discussed in previous problem, the relation between the acceleration of the blocks and the angular acceleration of the pulley will be,
`a=alphaR`..........`(4)`
Using equation `(1)`, `(2)`, `(3)` & `(4)`, we get
`m_(1)g-T_(1)=m_(1)a`
`T_(2)-m_(2)g=m_(2)a`
`(T_(1)-T_(2)=(Ma)/(2))/((m_(1)-m_(2))g=(m_(1)+m_(2)+(M)/(2))a)` (adding all the equations)
`impliesa=((m_(1)-m_(2))g)/((m_(1)+m_(2)+(M)/(2)))`
`impliesa=(50)/(15+5)=2.5m//s^(2)`
`(b)` Tensions `T_(1)=m_(1)g-m_(1)a`
`=m_(1)(g-a)`
`=10(10-2.5)`
`=75N`
`T_(2)=m_(2)(g+a)`
`=5(10+2.5)`
`=62.5N`.