IIT JEE , AIEEE Forum - Physics , Mechanics

amitshivani

hc verma q 33 197 ana q 40

rooney

Since moment of inertia of pulley is not negligible , we cant take tension on both strings to be same.

Let tension in horizontal string (with mass m) be T2
"    "          " vertical    "    ( "       " M)    be T1

Now, considering mass M ,
=>Mg - T1=Ma           -[1]

Considering mass m,
=>T2=ma    -[2]

Consider the pulley,
Net torque = I (Alpha)
Since Alpha = a/r,
Net torque = Ia/r

(T1-T2)r=Ia/r

So, T1-T2 = Ia/r^2

Add [1] and [2] ,
Mg - (T1-T2)=(m+M)A

Put value of T1-T2

Mg - Ia/r^2=(M+m)a

So, a = Mg / (M + m + I/r^2)

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