Irodov Problem 1.143

Suppose that at some instant masses m1 and m2 are x and y units below the pulley. The center of mass will be located,. Let the acceleration of the mass be w.

There are two forces acting on m1 , i) the tension pulling the mass up and ii) the force of gravity pulling it down and so we have,



Similarly for mass m2 we can write,


From (1) and (2) we have,





The acceleration of the CG will be,

Irodov Problem 1.142

When the rod rotates at a rate , the mass experiences a centrifugal force (in the rotating reference on the rod) which extends the spring. Suppose that the spring extends by a length . When seen from the rotating reference frame on the rod, there are two forces acting on the mass, i) the tension in the spring and ii) the centrifugal force . There is no acceleration for the mass along the horizontal as seen in the rotating reference frame. So we have,





The work done in getting the system to this rotating state is equal to the total energy stored in the system which is a sum of i) potential energy stored in the spring and ii) the kinetic energy of the mass. Thus, the work done is,

Irodov Problem 1.141

As the plank is pulled at an angle , the thread elongates to a length . Suppose that the spring constant of the thread is . The tension in the thread will be




Let us consider the forces in the vertical direction. There are three forces acting on the bar in this direction, i) the normal reaction N, ii) the component of the tension in the elongated string trying to lift the bar upwards, iii) the force of gravity acting downwards mg. There is no acceleration of the bar in the vertical direction and so we have,





In the horizontal direction, there are two forces acting on bar, i) the component of the tension pulling the bar back and ii) the force of friction f. Before slipping, the bar has no acceleration and so we have,




The maximum force of friction that the bar can be offered is Nk and since the bar starts to slip at an angle , we have from (2) and (3),







From the point of view of the bar, the force of friction is responsible for pulling it and this elongating the thread (as seen in the figure). This means that the entire potential energy stored in the thread is due to the work done by the friction. When the thread is at an angle , the energy stored in the thread is given by,







and this is the work done by friction!