Irodov Problem 1.278

As soon as the discs are put on each other, friction forces start to act between them to eliminate their relative motion. These friction forces generate equal and opposite torques on the two discs. This will decelerate one of the discs and accelerate the other until both spin at the same angular velocity and there is not relative motion between them.

Since friction forces are completely internal to the two disc system, there is no external torque acting on the two disc system. Consequently, the total angular momentum of the two disc system must be conserved at all times. Initially, the total angular momentum of the two disc system is I1w1 + I2w2. The moment of inertia of the two disc system when they are spinning together is I1+I2. Let the final angular velocity be w. Hence, we have,






The work done by friction is entirely responsible for reducing the total kinetic energy of the two disc system and is equal to the final kinetic energy - initial kinetic energy given by,

Irodov Problem 1.277

(a) Since there is no external torque acting on the (man + disc) system with respect to the axis of rotation, its angular momentum about its axis of rotation with will be conserved at all points in time. Since, the initial angular momentum of the (man + disc) system was zero, so it will be at any given time when the man walks on the edge of the disc. suppose that when the man walks with v'(t) relative to the disc, the disc turns with an angular velocity w(t). The net motion of the man is given by v'(t)+Rw(t). Thus, his angular momentum is given by R[v'(t)+Rw(t)]m1. The angular momentum of the disc is given by (see the list of moment of inertias). Since, net angular velocity of the (man + disc) system must be zero we have,






The angle turned can be found by integrating w(t) over time as,








s'(t) is the distance traveled by the man along the edge relative to someone on the disc. Since the person walks along the edge the angle walked by the person along the edge is given by . Hence, we have,




b) As the man walks he imparts force on the disc and turns it around. We know that,

Irodov Problem 1.276

(a) The force F acts through the axis of rotation and so does not impart any torque on the rotating (disc + point mass) system. As a consequence the angular momentum of the (disc + point mass system) is always conserved. As the point mass moves radially inwards, this decreases the moment of inertia of the (disc + point mass system) and thus increases the angular velocity.

The moment of inertia of the (disc + point mass) system about the axis of rotation, when the point mass is at a distance r from the center is given by is given by . Since the initial angular velocity of the disc when r=R is w0, since angular momentum is conserved we have,








In its final state, r=0 since the point mass reaches the center. Thus, setting r=0 in (1) we have,




(b) Since the problem states that the point mass in the radial guide is pulled very slowly, it means that the point mass was not imparted any acceleration. In the rotating reference frame, there are two forces acting on the point mass in the radial direction when it is at a distance r from the center, i) the force F in the thread pulling the point mass radially inwards, and ii) the centrifugal force pulling the point mass outwards. Since there is no radial acceleration, the force F must exactly balance the centrifugal force. In other words,



The work done by the force F is then given by,