Irodov Problem 1.283

Consider a stationary observer observing the top's motion. The figure depicts the picture of a top spinning while undergoing precession. Under precession the spin angular velocity vector w rotates as shown in the figure. Suppose that angular velocity of precession is w', this means that the angular velocity vector w rotates with an angular velocity w'. In other words the angular momentum vector of the top can be written as,







About point O (point of contact of the top with the ground), there is only one torque acting on the top namely due to the force of gravity mg acting at the CG. As shown in the figure the CG is at a distance from point O, the torque due to gravity is given by . The net torque must be equal to the rate of change of angular momentum of the top. Thus we have,






(b) As seen by an observer located at the CG of the top, there are two forces acting in the horizontal direction i) the centrifugal force as the CG rotates in a circle of radius at a rate w' given by and ii) the horizontal component of the reaction from point of contact NH. Since from the observers point of view there is no acceleration for the CG, we have,



Similarly, it is clear that the vertical component of the reaction from floor at the point of contact must be mg to balance the force of gravity.

Irodov Problem 1.282

Consider an infinitesimally small piece of the rod at a distance x along the rod from point C of length dx. If the linear density (mass per unit length) of the rod is p then the mass of the piece will be pdx. The piece spins at a radius of at a rate w and so has a linear velocity of . The momentum of the infinitesimally small piece of rod is given by and is directed perpendicular to the plane of axis and the rod directed into the plane.

Angular momentum w.r.t to C: The angular momentum of the piece with respect to C is then given by . The total angular momentum of the rod with respect to point C is given by,



Angular momentum w.r.t to axis of rotation: The infinitesimally small piece is at a distance from the axis and so the angular momentum of this piece w.r.t to the axis of rotation is given by . The angular momentum of the entire rod w.r.t the axis of rotation is then given by,








(b)








The angular momentum vectors at two extremes of the rod's rotation are depicted in the figure beside. The net change in the angular momentum vector is given by,





(c)










While the magnitude of the angular momentum remains the same as the rod rotates, its direction keeps rotating as shown in the figure. Thus we have,


The change in angular momentum vector is due to the torque provided on the axis given by,