Analysis of puzzle

Problems such as posed by the 4 cases are solved in elegant fashion applying conservation laws such as pertaining to angular momentum and energy. For A, B and D energy conservation will do nicely, but for case C, when using energy conservation, one arrives perhaps at the right order, even having used an invalid approach.  

Let’s deal first with A, B and C.  Simply applying the conservation  of energy will quickly give an response.

Notice from the relation  above that the mass does not come into account. Hence in A and B the speed V at the bottom is exactly the same.
For case D,  y2 is slightly smaller and hence equally a  smaller speed V at the bottom.

The allusion to shortening the swing radius was purposely introduced to entice people to think it would increase the speed of the mass1. It does not. Hence there is more involved than just shortening the swing radius to increase peripheral speed.

Remains to analyze case C.  Applying simply energy conservation alone is not appropriate here. Quite readily seen with m2 positioned at x=0, vertically in line with the pin. There will be a head-on collision with the pin and, since a perfectly inelastic collision is assumed, energy is not conserved.

We treat the transition as a perfectly inelastic collision between the ensemble and the pin. It is then most appropriate to use conservation of angular momentum.

We distinguish three phases: 1)vertical fall, 2) transition and 3) circular motion.

1 - Linear vertical motion (Fig3a)

2 - Transition from linear to circular motion (Fig3b)

There is an impulsive transfer of angular momentum during the transition. Think of it as an inelastic collision between the hook and the pin. Angular momentum is conserved hence one can write :

m1 V0 x1  + m2 V0 x2  =  m1 V1 x1 + m2 V2 x2

V2   =  V1 (x2/x1),  

Hence one can write for the speed V1 of mass m1 :



Assuming some arbitrary values  
   
m1=1kg,        m2 ={1,10,50,100 kg},             y1=1m,            x1=1m,    

we can plot V1 and V2 as a function of the position x2 of the mass m2 , as shown below in Figs 5a and 5b.

The two set of curves above display respectively the speed of mass m1 and m2 as a function of the distance of m2 from the center of rotation. Notice something very interesting. At the transition from linear to circular motion the speed of m1 always increases and that of m2 always decreases. The proximal mass m2 hence gives up some of its momentum/energy to the distal mass m1.

For those who don’t believe in the kinetic chain action this is something to give some thought. Even for the extremely simple case of a single rigid arm with two masses there is already transfer. How wonderful. It seems that nature just loves it to transfer wherever it can to optimize the peripheral over the proximal.  

Even when treating these very simple models one can see the subtle mechanism at work when dealing with curvilinear motion and mass distribution.  It indicates that when going from large to small radii and there being a large mass close to the center, one can expect velocity multiplication at the periphery. However, this is pefectly true for an ideal mathematical model. In real life it is only approximately true but still there is the same trend.

Since for C there is some speed gain  for mass m1 it is evident that, for C, m1 at the bottom is the fastest in the lot as the circular motion is the same for all four.

3 - Circular motion (Fig3c)