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.

Answer

`(4M_(1)M_(2)g)/(4M_(1)M_(2)+M_(0)(M_(1)+M_(2))`

Explanation

If `a_(0)a_(1)` and `a_(2)` are accelerations of `M_(0)M_(1)`, and `M_(2)`, respectively, then

`a_(0)=(a_(1)-a_(2))/(2) implies 2a_(0)=a_(1)-a_(2)` ..(i)
For motion of `M_(0), 2T=M_(0)a_(0)` ..(ii)
For motion of mass `M_(1), M_(1)g-T=M_(1)a_(1)` ...(iii)
For motion of mass `M_(2), T-M_(2)g=M_(2)a_(2)` ..(iv)
Substituting value of `a_(0), a_(1)` and `a_(2)` From Eqs. (ii), (iii) and (iv) in (i) 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_(2)g)/(4M_(1)M_(2)+M_(0)(M_(1)+M_(2)))` ...(v)
Acceleration of the pulley of A from Eq. (ii) is
`a_(0)=(2T)/(M_0) = (4M_(1)M_(2)g)/(4M_(1)M_(2)+M_(0)(M_(1)+M_(2)))`.