Irodov Problem 1.274

a) Since the collision between the plate and the point mass is elastic, no energy is lost in the collision. Further, since there is no external torque acting on the (point mass + plate) system, the angular momentum of the (point mass + plate) system will be conserved.

Due to the symmetry of the collision it is clear that the point mass will continue to move along its initial direction of motion (either in the same direction or exactly opposite) after the collision. Suppose that the velocity of the point mass after collision be v'. Further suppose that the collision also results in the plate acquiring an angular velocity of w. The moment of inertia of the plate about axis AB will be (see this list). Conservation of angular momentum about the axis AB gives us,






Conserving initial and final energies we have,



From (1) and (2) we have,








One trivial solution for (3) is v=v' which implies that the point mass simply passed through and through the plate without doing anything to it. Clearly this is no possible so we neglect it. The other solution gives us,







Now using (1) and (4) we can find w as,





Let the width of the plate be b and its surface density (mass/unit area) be p. Consider a infinitesimally thin vertical slice of the plate of thickness dx at a distance x from the axis. The mass of this slice is bpdx. Since it rotates at a distance x and with a angular velocity w it pulls the rest of the plate with a force radially outwards. The total force on the axis is found by integrating over all such slices as,






From (5) and (6) we have,






Since the plate is not flying out radially, this must be exactly euqal to the force exterted by the axis to support the rotating plate.

Irodov Problem 1.273

Impulse is a force that acts for a very short duration and cause momentum to change almost instantaneously. The change in momentum of the rod will thus be J. In other words, the CG of the rod will start to move at a speed v=J/m.

The impulse also imparts an impulsive angular momentum of Jl/2 about an axis perpendicular to the plane of the rod passing through its CG (since it acts at a distance l/2 to the CG) to the rod sending it into a spin. Let the final angular velocity of the rod about its CG be w. Then we have,



Consider an infinitesimally small piece of the rod of length dx at a distance x from the axis along the length of the rod. If the linear density (mass per unit length) of the rod is p, then the mass of this piece is pdx. The rod spins about the CG at a rate w and so it exerts a centrifugal force of . The net force exterted by one half of the rod on the other is the integration of dF over the half length of the rod as,

Irodov Problem 1.272

The centrifugal force acting on the sleeve pushes it radially outwards along the rod. Since there are no torques acting on the system about the axis of rotation, the total angular momentum of (sleeve + rod) system must be conserved at all times. As the sleeve moves outwards, the moment of inertia of the (sleeve + rod) system increases. This means that as the sleeve moves out the angular velocity of the spinning rod will decrease. This is similar to a skaters spinning slowly as they extend their arms outwards.

Suppose that at some instant of time the mass m is at a distance x from the the axis of rotation. Further, suppose that at this instant, the angular velocity of the spinning rod is w and that the outward radial acceleration of the sleeve be a. The moment of inertia of the (sleeve + rod) system is given by (see the list). The moment of inertia of the (sleeve + rod) system will be . Initially x=0 and the angular velocity at this time was wo. Since, angular velocity must be conserved, we have,










There is only one force acting on the sleeve - the centrifugal force (the sleeve is at a distance x from the axis of rotation) and this results in the radial acceleration a. So we have,





Here, v is the radial outward velocity. From (1) and (2) we have,