In figure mass `m_(1)` slides without friction on the horizontal surface, the frictionless pulley is in the form of a cylinder of mass `M` and radius `R`, and a string turns the pulley without slipping. Find the acceleration of each mass, and tension in each part of the string.

Explanation

Step I: Analysis of motion: In this case the string rotates the pulley, i.e, the string does niot slide. The puley will have angular acceleration,. Hence, the tension on the string on both the sides of the pulley will not be equal themotion of `m_(1)` and `m_(2)` will be translational. Positive direction of angular acceleration `alpha` is taken along the corresponding acceleration `a_(1)` which is taken positive in the direction of not force.
Step II: Equation of motion: Let these be `T_(1)` and `T_(2)`. The equation of motion respectivley, of masses `m_(1)` and `m_(2)` is
`T_(1)=m_(1)a`..........i
and `m_(2)g-T_(2)=m_(2)a` ........ii

Step III: Application of torque equation
The torque equation for the pulley.
`T_(2)R-T_(1)R=Ialpha`.......iii
`(N_(2)` constitutes no torque about the axis of rotation)
There is no slipping of the string over the pulley.
`a=alphaRimpliesalpha=a/R` ............iv
From eqwn iii and iv we get
`:. T_(2)R-T_(1)R=Ia/R`
or `T_(2)-T_(1)=(Ia)/R^(2)`...........v
Step V: Solving equations:
Solving the above equation we get `a=(m_(2)g)/((m_(1)+m_(2)+I/R^(2)))`
Here `I=(MR^(2))/2`
`T_(1)=(m_(1)m_(2)g)/((m_(1)+m_(2)+I/(R^(2))))` and `T_(2)=((m_(1)+I/(R^(2)))m_(2)g)/((m_(1)+m_(2)+L/R^(2)))`