Irodov Problem 1.246

Let the tensions in the string in the two sides of the cylindrical pulley connecting masses m1 and m2 be T1 and T2. The difference in tension is because of the friction between the string and the cylinder that is responsible for avoiding slippage and for turning the cylinder. Let the acceleration of the masses be a and let be the angular acceleration of the cylinder. Let us assume that mass m1 is moving downwards and that mass m2 is moving upwards (the value of a will be negative if this is wrong).

Since the thread is not slipping over the cylinder,



There are two forces acting on mass m1, i) the tension in the string T1 pulling it upwards and ii) the force of gravity pulling it down. So we have,



There are two forces acting on mass m2, i) the tension in the string T2 pulling it upwards and ii) the force of gravity pulling it down. So we have,



Each of the tensions (T1 and T2) in the two parts of the string on either side of the cylinder induce a torque in the cylinder in opposite directions causing it to rotate. The moment of inertia of solid cylinder about an axis passing through its center parallel to its length is . Hence we have,






Solving, for a from equations (1), (2) and (3) we obtain,









Substituting (4) in (1) we get,






Similarly substituting (4) in (2) we get,






From (5) and (6) we get,

Irodov Problem 1.245

The force is initially applied perpendicular to the rod. This causes the rod to rotate. As the rod rotates, since the direction of F is no longer perpendicular to the rod and so the moment arm decreases. As seen in the diagram, if the rod has rotated an angle , the effective torque on the rod is only - due to the component of the force that is perpendicular to the rod. The moment of inertia of the rod about its axis of rotation (passing through one if its ends perpendicular to the plane) is . So we have,

Irodov Problem 1.244

The arrangement is shown in the figure beside. The force of gravity pulls the Maxwell's disc downwards. The rod is accelerated upwards at a rate a such that, the Maxwell's disc remains in place and does not move downwards and the string unwinds.

Let the tension in each of the strings be T. There are three forces acting on the Maxwell's disc, i) the tensions from each of the string pulling the disc upwards and ii) the force of gravity mg. Since the Maxwell's disc does not move, its acceleration is 0. From Newton's laws we have,


Let the acceleration of the rod be a and the angular acceleration of the Maxwell's disc as the thread unwinds be . Clearly, these two quatities are related as,



since the thread does not slip.

The tension in the strings induces a torque on the Maxwell's disc equal to 2Tr. Given that the moment of inertia of the Maxwell's disc is I, we have,



From (1), (2) and (3) we have,