A smooth pulley A of mass `M_(0)` is lying on a frictionless table. A massless rope passes round the pulley and has masses `M_(1)` and `M_(2)` tied to its ends, the two portions of the string being perpendicular to the edge of the table so that the masses hang vertically. Find the acceleration of the pulley.

Explanation

Let the length of portions of string on the table be `l_0 , l_1 and l_2` as shown in figure .
Let mass `M_0` move to the right by x on the table , `M_1` goes down by `y_1 and M_2` goes up by `y_2` .
Then `l_0` becomes `(l_0 -x),l_1` becomes ` (l_1 + y_1) and l_2` becomes ` (l_2 - y_2)`.
As the length of string remains unchanged
` 2l_0 + l_1 + l_2 = 2(l_0 -x) + l_1 + y_1 + l_2 - y_2`
or ` 2 x = y_1 -y_2`
Diff. twice w.r. to t,
we get ` 2(d^(2)x)/(dt^(2)) = (d^(2) y_1)/(dt^(2)) - (d^(2)y_2)/(dt^(2))`
If ` a_0 , a_1 and a_2` are acceleration of

`M_0 , M_1 and M_2 ` respectively , then
` 2a_0 = a_1 - a_2` ......... (1)
For motion of `M_0 , 2T= M_0a_0` ....... (2)
`M_1g-T=M_1 a_1`....... (3)
For motion of mass `M_2` ,
` T-m_2g = M_2a_2` .......... (4)
Substituting values of ` a_0 , a_1 and a_2` from (2) , (3) and (4) in equation (1) , we get
` 2((2T)/(M_0))=(g-(T)/(M_1)) -((T)/(M_2)-g)`,
`(4T)/(M_0)= 2g-T((1)/(M_1) + (1)/(M_2)) i.e., ((4)/(M_0) + (1)/(M_1) + (1)/(M_2))T=2g `
This gives `T=(2M_0 M_1 M_2g)/(4M_1M_2+ M_0 (M_1 + M_2))`........ (5)
`:.` Acceleration of puelly A from (2)
`a_0 = (2T)/(M_0) = (4M_1M_2 g)/( 4M_1 M_2 + M_0(M_1 + M_2))` ............ (6)