Irodov Problem 1.260

Since the cylinder if affixed to the mount, due to the application of the force F, both of them will move together and can be treated as one body. If the acceleration of the mount+cylinder system is a then based on Newton's laws,

The force F also induces a torque of Fr on the cylinder since it acts at a distance r from the axis of rotation. Since the moment of intertia of the cylider about its axis of rotation is , we have,

This rotation of the cylinder unwraps the wound thread on the cylinder and releases it, thus point K moves away from the mount+cylinder system with an acceleration of . The net acceleration of point K as seen by a stationary observer is the sum of i) the acceleration of the mount+cylinder system a, and ii) the acceleration of point K as seen by an observer on the mount . Thus, from (1) and (2) we have the acceleration of point K as,

Irodov Problem 1.259

Motion of mass m: There are two forces acting on mass m, i) the force of gravity mg and ii) the tension in the string T1. Let its acceleration be a1. So from Newton's laws we have,

Motion of mass M: There are four forces acting on mass M, i) the force of gravity Mg, ii) the tension in the string connecting to mass m i.e. T2, iii) the forces on the two strings holding mass m of T1 each. Let the acceleration of mass M be a2. So we have,

Since the force of gravity Mg passes through the axis of rotation of mass M, it does not produce any torque in it. However, the rest of the three forces produce torques inducing rotation in the mass. Since the tensions T2 act at a distance of R from the axis, they produce a torque of T2R each in the clockwise direction. The tension T1 acts at a distance of 2R from the axis of rotation of mass M and so it induces a torque of T1(2R) on the mass also in the clockwise direction. Let the angular acceleration of mass M be . Then we have,

Since the threads holding mass M do not slip we have,

From (3) and (4) we can write,

As seen in the figure, suppose that the mass M rotates an angle , each of the two threads unwrap a length of and so the mass M falls down this distance. Meanwhile, the rotation of mass M also releases a length of of the thread holding mass m, making the distance between mass M and mass m to increase by this additional distance. Consequently the total distance moved by mass m is as shown in the figure. From this discussion and (4) it is clear that,

Now using (1), (2), (5) and (6) we can solve for a1 as follows,