Irodov Problem 1.258

Let the tension in the tension in each of the threads be T. This tension is responsible for inducing the spinning in both the cylinders. Let the downwards acceleration of the free cylinder be a. Let the radius of the cylinders be r. Also let the angular accelerations of the two cylinders be b1 and b2 as shown in the figure.

Spinning motion of the fixed cylinder: The net torque acting on the fixed cylinder is 2Tr. The moment of intertia of the cylinder is . So we have,

Motion of the free cylinder: there are three forces acting on the free cylinder, i) the force of gravity mg and ii) the two tensions of value T in each of the threads. So we have,

The force of gravity passes through the axis of rotation of the free cylinder, so it does not induce any torque on the free cylinder. The two tensions however induce a torque of Tr each on the cylinder. So we have,

From (1) and (3) it is clear the angular accelerations of both the cylinders are the same in magintude, in other words,

Figure above shows how various parts of the system move as the free cylinder descends. Consider that initially the point of contact of the thread and the fixed cylinder is A and that with the free cylinder is B. When the fixed cylinder spins an angle , thread of length of is unwrapped from the fixed cylinder and so the point A descends by this length. Similarly when the free cylinder spins an angle , thread of length is unwrapped from the free cylinder. So the total length of string unwrapped is . The the total distance that the free cylinder descends is equal to the total length of thread unwrapped and is given by,

Differentiating (5) twice with respect to time we have,

Using (1), (2) and (6) we have,

From (7) and (1) we have,

Irodov Problem 1.257

If there were no friction, then the spindle would simply be dragged forward due to the horizontal component of the force F. The force would also induce a torque of magnitude Fr on the spindle and cause it to spin in the counter-clockwise direction. Friction between the floor and the spindle however, opposes this with a force f acting in opposition to the force F. This friction force acts on the wider circular part of the spindle and induces a torque fR in the clockwise direction. This frictional torque is eventually responsible for making the cylinder roll without slipping on the floor.

In the Horizontal direction: there are two forces acting on the spindle, i) the horizontal component of the force F and ii) the force of friction f. Let the acceleration of the spindle be a. So we have,

The force of gravity passes through the axis of rotation and does not induce any torque on the system. Let the angular acceleration of the cylinder be b. As mentioned there are two torques acting on the spindle, i) due to the friction of magnitude fR in the clockwise direction and ii) due to the force F of magnitide Fr in the counter-clockwise direction. So we have,

Since there is no slipping between the spindle and surface we have,

Now we can find the acceleration as,

b) Since there is no slipping between the spindle and the surface, friction force does not do any work. All the work done by the force F is converted into the kinetic energy of the spindle. The velocity of the spindle is v = at at some point of time t. The kinteic energy of the spindle has two parts, i) the translational kinteic energy and ii) the rotational kinetic energy . So the total energy of the spindle is given by,