Irodov Problem 1.254

This problem is identical in all respects to problem 1.253 except for the fact this is inside an accelerating elevator. Using the famous Einstien's principle of equivalence, with respect to an observer in the accelerating elevator, it will be as if the net gravitational attraction inside the elevator has become g-wo. Thus, the solution for this problem is exactly the same as that of 1.253 except that the acceleration due to gravity g will be replaced by g-wo.

Thus, based on eqn (5) in 1.253, in this problem the acceleration as seen by the observer in the elevator will be,

The force F exerted on the elevator the combined tensions in the two strings 2T. So replacing g by g-wo in eqn (6) we have,

Irodov Problem 1.253

There are three forces acting on the cylinder as it unrolls downwards, i) the force of gravity mg acting through its center of mass and ii) the tensions T each in the two pieces of string that hold it. Let the acceleration of the cylinder be a. Then we have,

Since the force of gravity passes through the axis of rotation it does not induce any torque on the clinder, however, the tension in each of the strings induces a net torque of 2TR on the cylinder causing it to rotate. Let the angular acceleration of the cylinder be . Since the moment of intertia of the cylinder is , we have,

Since there is no slipping between the string and the cylinder we have,

From, (2) and (3) we can write,

From, (4) and (1) we obtain,

From (5) and (4) we have,

b) At some instant of time if the cylinder is falling at a speed v, its angualr velocity will be v/R.The total energy of the cylinder at any given time is the sum of its translational kinetic energy and its rotational kinetic energy . At any given time the velocity of the cylinder if given by at. Hence we have,

Since all the power is generated by gravity, this must be the power generated by gravity.