Irodov Problem 1.263

Let the tangential velocity of the smaller sphere (its center) be v and its angular velocity about its own center be w. As the sphere rolls down down it looses gravitational potential energy. This loss in potential energy is completely converted into its kinetic energy (since the sphere rolls no energy is lost due to friction). Its kinetic energy is the sum of its translational energy () and its rotational kinetic energy (). When the smaller sphere's CG is at an angle , the CG of the sphere would have come down a height of as shown in the figure. Since the moment of inertia of the smaller sphere about its axis of rotation is we have

Since the sphere is rolling, we have,

From (1) and (2) we have,

Now let us consider the forces acting on the sphere in the direction normal to the surface of the sphere. There are two forces acting in this direction, i) the component of force of gravity and ii) the normal reaction from the surface of the larger sphere N. The acceleration of the smaller sphere in the normal direction is the centripetal acceleration due to the rotation of its CG about the center of the larger sphere directed inwards towards the center. So we have,

When the smaller sphere breaks off of the surface of the larger sphere, the normal reaction N will becomes zero (since there is no contact anymore). Thus, from (2) and (4) and setting N=0 we have,

Now substituting (5) in (3) we have,

Irodov Problem 1.262

As soon as the spinning cylinder (say in the clockwise direction) is placed on the surface it initially slips (the point of contact of the cylinder with the surface - the lowest point of the cylinder, is moving backwards at a speed w0r relative to the surface). The force of friction f acts on it to resist this slipping between the lowest point of the cylinder (the point of contact of the cylinder with the surface) and the surface and as a result accelerates the cylinder forward. The friction force f also generates a retarding torque that opposes the spinning motion and reduces the cylinder's angular velocity. If the velocity of the cylinder in the forward direction at any point of time is v and its angular velocity due to spinning motion is w then the velocity of the lowest point of the cylinder is wr - v backwards. As v increases (due to the force f) and w decreases (due to the resitive torque fr) at some point of time when wr-v = 0, the net velocity of the lowest point of the cylinder becomes 0. At this point there is no relative motion between the surface and the point of contact of the cylinder and the cylinder begins to roll and the force of friction ceases to act.

The only force acting on the mass in the horizontal direction is f and if the acceleration of the cylinder is a then,

Suppose that the angular acceleration of the cylinder is b, then we have,

There are two forces acting acting in the vertical direction, i) the normal reaction N from the surface and ii) the force of gravity mg acting downwards. There is no acceleration for the cylinder in the vertical direction and so we have,

The maximum friction force that can be offered is Nk and so this is the friction force f that is generated on the cylinder. So using (3) we have,

From (1) and (4) we have,

From (2) and (4) we have,

From (5) and (6) we have,

Applying the condition for rolling wr = v as discussed above we have,

The friction force is the only force responsible for reducing the energy of the cylinder. Thus the difference between final and initial kinetic energies is the work done by the firction. Initially the cylinder has no translational motion, it only spins and its kinetic energy is purely due to rotational motion given by . Finally however, the cylinder has both translational and rotational motions and thus the total kinetic energy is the sum of both these parts given by . The work done by friction is thus given by,