a light string passes over two identical pulleys of moment of inertia I and radius R each. two blocks of masses M and m are attached to the two ends of the string. find the acceleration of the block M

 

 

 

 

 

 

 

 

 

 

 

 

There is a cute method in many of these problems that reduces the no. of equations required. Assume that M>m. Suppose the masses were both at the same height initially. After a time t, M has moved down by y and m moved up by the same distance y. Their speeds are equal to say v and the magnitudes of acceleration are say a.

 

v=R  and a = R (for the pulleys)

 

Now the energy of the system is

 

E = -Mgy+mgy+0.5Mv²+0.5mv²+ Iv²/R² (there are two pulleys)

 

We must have dE/dt = 0 (as total energy is a constant).

 

dE/dt = -Mgv+mgv+Mva+mva+2Iva/R² = 0

dividing by v throughout

-Mg+mg+Ma+ma+2Ia/R² = 0

 

or a = g(M-m)/(M+m+2I/R²).

I'll tell u a shortcut for these types of problems:

 

 

We know that, Acceleration of the masses= Net Pulling force , right

                                                                Total mass

 

So what is the                 net pulling force= (M-m)g

     what is                     total mass=M+m+(no of pulleys)/r²

 

So substituting we get ,

 

 a=(M-m)g / (m+M+2I/r²⁾

 

 

This works for all types of these problems, see hcv from 30-40 all will work.

 

This is in a way a valid method as we have to account for the mass of pulley which we do by the I/r²      term