Irodov Problem 1.150

The center of mass of the system will act as if it were a point particle with mass m1+m2 with an initial velocity of subject to the gravitational force g. Thus, at any give time its velocity will be given by,




So its momentum will be,


Its position will be given by,

Irodov Problem 1.149

The velocity of the center of mass of the two discs is given by,

.


The total energy of the system will be conserved. Initially there is no extension in the spring and thus the total energy is equal to the kinetic energy in the discs. The kinetic energy of the system at the initial time instant as seen from the CG is given by,







But since v1 and v2 are perpendicular,

Irodov Problem 1.148

Let the particles in the system have masses , where n is the number of particles in
the system.

The total mass of the system is, so,





Let the velocities of the particles with respect to the center of mass of the system of particles be . The kinetic energy of the system as seen from the center of mass is and so,





With respect to a frame fixed on the center of mass, its velocity is zero. In other words,









As seen from the frame K which moves with a velocity V with respect to the center of mass, the particles have velocities, . Thus, the total kinetic energy as seen of the system of particles as seen from K is given by,

Irodov Problem 1.147

Suppose that the frame K' is one that has a velocity v. Then as seen from K' the velocity vectors of masses m1 and m2 will be v1-v and v2-v respectively. So the total kinetic energy of the two masses as seen from K' will be,



The extrema of E will be at dE/dv = 0, i.e.






Since v is a vectors, to show that this corresponds to the minima, we will have to show that the Hessian matrix of E with respect to v must be positive definite. The Hessian matrix of E is given by (m1 + m2) I which is positive definite. Hence v must be the minima.

The kinetic energy of the system as seen from the frame K' can be obtained by substituting v in E. The kinetic energy is given by,

Irodov Problem 1.146

As in problem 1.145, we can replace the cone by its center of mass to answer the question. As seen by an observer in the rotating frame of the CG, there are four forces acting on the cone (and hence the CG), i) the force of gravity mg, ii) the normal reaction N, and iii) the force of friction F opposing the tendency of the cone to slide downwards and iv) the centrifugal force acting away from the center. As seen from the CG, the CG has no acceleration.

i) Considering the force along the direction of the slope of the cone we have,





ii) Considering the direction normal to the slope we have,





The maximum force of friction that the surface can offer is F = kN. Thus, for the cone not to slip,

Irodov Problem 1.145

No matter how the different parts of the chain move, the dynamics of the center of mass of the chain act as if it were a point particle with the entire mass of the chain m concentrated at the point. The sum of all the forces acting on different parts of the chain then, becomes the force acting on the center of mass. We can conveniently use this fact about the center of mass to solve the problem.

Suppose that the CG is x units off from the axis of rotation. As seen by an observer in the rotating reference frame, there are three forces acting on the center of mass, i) tension in the string T, ii) the force of gravity mg and iii) the centripetal force . In the rotating reference frame, the CG is not accelerating .



In the vertical direction we have,





In the horizontal direction we have,