Irodov Problem 1.249

As soon as the disc is placed flatly on the floor, each point on the disc starts to experience a force of friction dF that opposes its motion relative to the floor. While the friction force dF experienced at each point on the disc is the same (since the problem states that the disc applies equal pressure from all points), the resistive torque r x dF generated by the friction force depends on its distance from the center of the disc. Points of the disc that are far from the center generate more resistive torque than those that are closer to the center.

Let us consider an infinitesimally small section of the disc that is at a distance r from the center and has a thickness dr and at an angle and subtends an angle from the center of the disc. The total area of this infinitesimally small part of the disc is given by . The mass dm of this small area of the disc is given by the area spanned by the disc multiplied by the mass per unit area of the disc,





There are two forces acting on this small piece in the vertical direction the normal reaction from the surface dN and the force of gravity (dm)g acting downwards, and its acceleration in this direction is 0. In other words,



The force of friction dF experienced by this piece will be equal to kdN and will act tangentially to the circle of radius r since its motion is along a cirlce of radius r as the disc spins. The resistive torque experienced by the piece will now be







The net torque is given by,












Since the moment of inertia of the disc about its axis of rotation is given by , the angular retardation (deceleration) experienced by the disc is given by,






Given that the disc was originally spun at an angular velocity of , the time taken by it to stop will be given by,

Irodov Problem 1.248

There are two surfaces of contact for the cylinder, i) the vertical wall and ii) the floor. Both these surfaces offer a force of friction to resist the slipping of the cylinder relative to them. Since the cylinder is spinning in the anti-clockwise direction when it was initially placed, the force of friction on each of these surfaces will oppose its relative motion and so the friction force from vertical wall fy will act upwards while the friction force fx from the surface will act to the left as shown in the figure. Let Nx and Ny be the normal reactions acting from the vertical and horizontal wall respectively. Let the mass of the cylinder be m.

There are three forces acting on the body in the vertical direction, i) the force of gravity mg acting downwards, ii) the normal reaction from the floor Ny acting upwards and iii) the force of friction Ny from the vertical wall. Since there is no accelaration for the cylinder in the upward direction we have,



In the horizontal direction there are two forces acting on the cylinder, i) the force of friction fx from the floor and ii) the normal reaction from the vertical wall Nx. Since the cylinder is not accelerating in the horizontal direction we have,



Now from our basic understanding of friction force is Nk and so we have,





From (1), (2), (3) and (4) we obtain,















The two friction forces from the wall and the floor result in a resistive torque slows the rotating cylinder and this torque is simply equal to (fx + fy)R. Since the moment of inertia of the cylinder about its axis of rotation is its angular deceleration is given by,







If at some instant of time the angular velocity of the cylinder is , and it has turned an angle then we have,



Since in the end the cylinder stops completely its final angular velocity is 0, so from (10) we have,







Since each complete rotation means that the cylinder has turned by radians. If the cylinder turns an angle radians then the number of rotations its undergoes is . Hence the total number of rotations that the cylinder will undergo before it stops will be,

Irodov Problem 1.247

Let us consider the forces acting on each body one by one.
Mass m1:
On mass m1 there are two forces acting in the vertical direction i) the force of gravity m1g pulling it downwards and ii) the normal reaction N from the surface stopping it from moving down into the surface. Since this mass has no acceleration in the vertical direction we have,



In the horizontal direction, there are two forces acting on the mass i) the tension in the string T1 pulling it forward and ii) the force of friction between the surface and the mass, kN, resisting its motion on the surface. Let the acceleration of the mass be a. Then we have,



Mass m2: There are two forces acting on this mass, i) the force of gravity m2g pulling it down and ii) the tension in the string T2 pulling it upwards. Its acceleration will be a the same as that of mass m1. So we have,



The Cylindrical Pulley: Since the thread does not slip over the cylinder, each of the ends of the string induces a torque on it at its two ends namely T1R and T2R resulting in a net resultant torque of (T2-T1)R that is responsible for turning the cylinder and inducing angular acceleration in it. Since there is no slipping between the thread and the cylinder we have,



Since the moment of inertia of the cylinder is we have,






Solving (1),(2),(3) and (4) we get,





The work performed by friction force is simply -Nkx, where x is the distance traveled by mass m1 (the negative sign comes because the force of friction is in the opposite direction of that of the direction of motion of the mass m1). Now, since mass m1 moves with a constant acceleration a we have,