Irodov Problem 1.252

There are three forces acting on the sphere, i) the force of gravity mg acting vertically downwards, ii) the normal reaction from the surface N and iii) the force of friction f that acts to resist any slipping between the sphere and the inclined plane acting in the direction of the slope of the plane. Let the acceleration of the sphere down the inclined plane be a.
In the direction of the slope of the inclined plane we have,

In the direction perpendicular to the slope of the plane we have,

Now let us consider the torques acting on the sphere. Since both normal reaction N and the force of gravity pass through the sphere's center of mass, they do not generate any torque on the sphere. The only torque that is responsible for the sphere's rotation is due to the force of friction given by fR. Let the angular acceleration of the sphere be . Since the moment of intertia of a sphere is we have,

If there is no slipping between the surface of the sphere and the inclined plane, then,

From eqn (1), (3) and (4) we have,

The maximum force of friction that any surface can provide is given by Nk, where k is the coefficient of friction. In order for no sliping to be there, the inclined surface must be able to provide the neccessary force of friction f to allow rolling given in (6). From (2) we have,

b) If the velocity of the spehere is v and its angular velocity is , then the total kintetic energy of the sphere is the sum of i) translational kinetic energy and ii) the rotational kinetic energy given by .The velocity of the sphere as a fucntion of time is at and the angular velocity is at/R. So the total kinetic energy of the sphere will be,

Irodov Problem 1.251

At any instant if x is the length of the hanging part of the cord then the mass of the hanging part of the cord is given by and so the force of gravity acting on it is given by . This means that as more and more cord starts to hang (x increases), the force of gravity acting on the cord also increases.
There are two forces acting at the topmost point of the hanging part of the cord, i) the tension from the end point of rest of cord (non haging part) wound on the cylinder T and ii) the force of gravity of the hanging part of the cord. Suppose that the downward acceleration of the cord is given by a. Then from Newton's laws we have,

Now let us consider the rotation of the cylinder. The combined moment of inertia of the cylinder + cord system about its axis of rotation is the sum of moment of intertia of two parts i) the solid cylinder of mass M and radius R given by and ii) the moment of interia of the cord wound on the cylinder. The mass of the wound part (non-hanging part) of the cord is given by and each point of the cord is at a distance R from the axis of rotation of the cylinder. Thus, the moment of inertia of the wound part of the cord is given by . The total moment of interia of the cylinder + cord system is given by,

The tension acting on the end point of the wound part of the cord will induce a torque on the cylinder + cord system and will be equal to TR. Let the angular acceleration of the cylinder + cord system be . Then we have,

Since there is no slipping between the cord and the surface of the cylinder we have,

Now we have all the equations necessary to solve the problem. From (1),(2),(3) and (4) we have,