Irodov Problem 1.256

There six forces acting on the cylinder, i) the two normal reaction from each of the planks of N each, ii) the two forces of friction f between the planks and the cylinder surface that resists any slipping, iii) the force F with which the cylinder is being pulled downwards and iv) the force of gravity mg acting downwards. If the acceleration of the cylinder is a then, applying Newton's laws in the vertical and horizontal directions we have,

The forces of normal reaction and that due to gravity pass through the axis of rotation of the cylinder. The torque on the cylinder is generated by three forces that act tangentially on the cylinder's surface at a distance r (the radius of the cylinder) from the axis, i) the force F and ii) the two forces of friction f each from each of the planks. The torque due to the forces of friction is opposed to that due to F. The moment of intertia of the cylinder about its axis of rotation is . Assuming that the cylinder's angular acceleration is we have,

If there is no slipping between the cylinder and the planks then,

From (4) and (3) we have,

From (5) and (2) we have,

The maximum friction force f that the planks can offer is Nk from each of the planks, in other words,

Irodov Problem 1.255

There are three forces acting on the spindle at any point of time, i) the force of gravity mg acting downwards, ii) the normal reaction from the surface N and iii) the tension in the string T. Since there is no friction between the surface of the plane and the spool, the spool will only slip, it will never roll. Further there will be no force of friction to resist the slipping of the spool downwards. Let the acceleration of the spool be a then considering the forces in the direction of the inclined plane we have,

Since both the normal reaction N and the force of gravity mg pass through the axis of rotation of the spool they do not cause any torque on the spool. However, the tension in the string acts at a distance r from the axis and so it will induce a torque Tr on the spool and cause it to spin. Let the angular acceleration of the spool be , then we have,

Now the only way for the spindle to move down is if the thread unwraps. If the thread spins an angle in the anti-clockwise direction as shown in the figure, (point A has moved along the circumference of the spindle's narrow part) then a length AA' equal to of of the thread will unwrap causing the spindle to move down a distance of

Differentiating this twice with respect to time we will have,

Now having all the necessary equations we can solve the problem