Irodov Problem 1.199

As soon as the mass m is given the velocity the system will start to spin about the system's CG. The CG itself will also start to move in linear motion with a velocity (m.v0+m.0)/(2m) = v0/2.

If we fix an observer at the CG however, he will see that the upper mass and lower mass are moving with a speed v0/2 in the opposite directions as shown in the figure. In other words the observer at the CG will feel as though the entire system were imparted an initial angular velocity of .

From this point onwards we shall solve the problem in the frame fixed to the CG. As the system spins around the CG, the centrifugal force will extend the spring. The moment of inertia of the system will also change as the spring extends since the masses will move further apart. Suppose that at some instant the spring is extended by a length x, then each of the masses would be at a distance x/2 from the CG. The moment of inertia of the system as a function of x will then be,




Initially since x=0, the initial angular momentum of the system as seen from the CG is given by,



Since there are no external torques acting on the system as seen from the CG, this angular momentum must be conserved at all times. Thus we can compute the angular velocity of the system as seen by the CG as a function of x as,









As seen from the CG the masses will be subject to two kinds of motion i) rotational (the spinning) w and ii) the radial motion (masses moving towards or away from CG) vr shown in the figure. The radial motion of both masses will be equal and opposite as seen from CG since the linear momentum of the system as seen from CG must be 0, this is indicated in the figure. At maximally stretched position, however, vr will be zero.

Let the maximum stretch of the spring be x=s. The energy of the system will also be conserved at all times and so considering the initial and maximally stretched positions we have,











Substituting in (5) we have,









(6) is a quadratic equation and can be solved, however, as stated in the problem is very small in comparison to 1 and so 2+ = 2 and 1+=1. Thus we have,

Irodov Problem 1.198

After collision two different motions will be induced in the dumbbell, i) a linear velocity v2 with which the entire dumbbell (dumbbell's center of mass) will move and ii) a spin with which the dumbbell will spin about its center of mass, say with an angular velocity w. This is shown in the figure. Suppose that the mass m continues to move with a speed v1 after the collision.

Since this is a completely elastic collision and no external forces or external torques are involved, momentum, angular momentum and energy will be conserved for the entire system.

Conservation of Momentum



Note that here we have considered the entire dumbbell as a point mass m concentrated at the center of mass of the dumbbell.

The moment of inertia of the dumbbell about its CG is given by,





Conservation of Angular Momentum
Let us consider the angular momentum of the system about the stationary point O, that was collocated with the CG of the dumbbell before collision. The angular momentum of the system before collision is given by which is the angular momentum of mass m w.r.t point O. After collision the mass contributes an angular momentum of since it moves with a velocity v1. The dumbbell's total angular momentum is the sum of the angular momentum of the CG (due to its linear motion) and the angular momentum of the dumbbell about it CG (due to its spin). Since we have chosen to evaluate the angular momenta about point O which is the initial position of the CG of the dumbbell, the moment arm of the velocity vector of the CG of the dumbbell about O will be 0. Thus the only contribution of the dumbbell's angular momentum comes from its spin about its CG given by Iw. So we have,







Conservation of Energy The total energy of the dumbbell after collision is the sum of the kinetic energy of the CG and the kinetic energy due to its spin about the CG. So We have,





Now its just a matter of solving (1), (3) and (4) to get w since angular momentum of the dumbbell about its CG is Iw.