Question

Figure shows two blocks of masses m and M connected by a string passing over a pulley. The horizontal table over which the mass m slides is smooth, The pulley has a radius r and moment of inertia I about its axis and it can freely rotate about this axis. Find the acceleration of the mass M assuming that the string does not slip on the pulley.?

Answer 1

Given in the question :-

Two blocks m and M are here.

So, assume that the acceleration of these two blocks = a

Tension = T & T'

Now for the first case :- It means block M.

$Mg-T =Ma$

$T = M(g-a)$

Now for second case i.e block m.

$T' = ma$

We know that the Net torque acting here

$= Tr-T'r$

or $=(T-T')r$

Now ,Let if angular acceleration = α

Hence,

a =α. r

$\alpha = a/r$

Here for finding the torque of the pulley we have,

T =Iα

Now put the value of α and T

$(T-T')r = I. a/r$

$M(g-a)-ma =Ia/r^2$

or ,

$Mg = Ma+ma+Ia/r^2$

Hence, $a = \frac{Mg}{(M+m+I/r^2)}$

Acceleration of the mass M when string doesn't slip on the pulley is $a = \frac{Mg}{(M+m+I/r^2)}$

Hope it Helps :-)

Answer 2

a force of 20 newton is applied on a body of mass 5 kg resting on horizontal plane the body gainer kinetic energy of 10 joule after moves a distance to metre the frictional force is