Question

The pulleys in figure are identical, each having radius R and moment of inertia I. Find the acceleration of the block M.?

Answer 1

Given in the question :-

Assume, Acceleration = a

ang. acceleration α = a/R

for M , tension = T

and For m, tension = T'

Now, the tension b/w the pulley = T''

Therefore ,

$Mg-T =Ma$

$T = M(g-a)$

Now ,$T'-mg = ma$

$T' = m(g+a)$

Now For the pulleys,

$\alpha = a/R$

$\alpha = (T-T") R/I$

or

$a =(T-T")R^2/I$

Since $a/R =(T"-T)R/I$

Therefore $a = (T"-T') R^2/I$

Now solving these two values of a , we get

2a = (T-T')R²/I

Now, put the value of T & T'

$2a =\frac{ (Mg-Ma -mg-ma)R^2}{I}$

$2al =(M-m)gR^2 -(M+m)aR^2$

or

$a (2I+MR^2+mR^2) = (M-m)gR^2$

$a = \frac{(M-m)gR^2}{2I} +MR^2+mR^2$

$a = \frac{(M-m)g}{M+m} +2I/R^2$

Hence the accelration of the block at M is $a = \frac{(M-m)g}{M+m} +2I/R^2$

Hope it Helps :-)