A disc shaped pulley of mass `'m'=4 kg` and radius `R=0.5m` can rotate freely about its center. A block of mass `M=2kg` hangs from the pulley through is massless string that is tightly wrapped around the pulley. When the system is released, there is no slipping between pulley and string.

`(a)` Calculate the angular accelerating of the pulley
`(b)` Calculate the tension in the string

Explanation

we choose the puley and mass as two separate systems and draw free body diagrams as shown.

Applying Newton's Second Law in translational form,
`Mg-T=Ma`……..`(1)` [For the block]
Applying Newton's Second Law in Rotational form,
`sumT_(cm)=l_(cm)*alphaimpliesTR=(mR^(2))/(2)alpha`........`(2)` [For the disc]
If point `P` on the pulley rotates by any angle `theta`, the block of mass `M` moves down by a distance `X`, such that `x=Rtheta`
Differentiating this equation twice we get
`(d^(2)x)/(dt^(2))=(Rd^(2))/(dt^(2))`
`a=R alpha`.........`(3)`
Using equations `(1)`, `(2)` and `(3)`, we get
`Mg-T=Ma`
and `T=(ma)/(2)`........`(4)`
Adding the two equations
`a=(Mg)/((M+(m)/(2)))`
`impliesa=(20)/(2+2)=5m//s^(2)`
`impliesalpha=(a)/(R )=(5)/(0.5)=10rad//s^(2)`
Tension `T=(ma)/(2)=5N` [from equation `(4)`]