A long, heavy, flexible rope with mass ρ per unit length is stretched by a constant force F. A sudden movement causes a circular loop to form at one end of the rope. In a manner similar to that in which transverse waves propagate, the loop runs (rolls) along the rope with speed c as shown in the

Question

A long, heavy, flexible rope with mass [latex] ρ [/latex] per unit length is stretched by a constant force [latex] F.[/latex] A sudden movement causes a circular loop to form at one end of the rope. In a manner similar to that in which transverse waves propagate, the loop runs (rolls) along the rope with speed [latex] c [/latex] as shown in the figure.

(i) Calculate the speed [latex] c [/latex] of the loop.
(ii) Determine the energy, momentum and angular momentum carried by a loop of angular frequency [latex] w [/latex]. What is the relationship between these quantities?

Accepted Answer

(i) Examine the motion of the loop in the frame of reference, which moves with the loop at speed [latex] c.[/latex] In this system, the pieces of a circular loop of radius [latex]R [/latex] rotate uniformly. The acceleration of a piece of the rope, which subtends central angle [latex] Δα [/latex] and has mass [latex] \rho R Δα,[/latex] is [latex] c^2 / R, [/latex] whilst the net force due to the tension in the rope is [latex] FΔα [/latex] (see S105). The Newtonian law of motion yields

[latex] F\Delta \alpha =\rho R\Delta \alpha \frac{c^2}{R}, [/latex]

i.e. the 'loop-wave' moves with speed [latex] c = \sqrt{F/ \rho } [/latex],-identical to the speed at which small transverse waves would propagate along the same rope.
(ii) The angular frequency of the rolling loop of radius [latex] R [/latex] is [latex] w = c/ R,[/latex] and the loop has mass, [latex] m = 2πRρ [/latex]. The energy carried by the loop can be expressed in terms of these quantities as

[latex] E=E_{transl}+E_{rot}=\frac{1}{2}(2\pi R\rho )(c^2+R^2 w^2) , [/latex]

which can be written in the form

[latex] E(w)=2\pi Fc\frac{1}{w} =K\frac{1}{w} . [/latex]

The quantity [latex] K = 2πFc [/latex] is a constant, characteristic of the rope and the tension in it. Of course, several [latex] (n) [/latex] loops can be simultaneously excited and their total energy is

[latex] E_n(w)=nK\frac{1}{w} \ \ \ \ \ (n=0,1,2,3,...). [/latex]

The momentum of the loop(s) can be calculated in a similar way,

[latex] P_n(w)=n(2\pi R\rho )c=nK\frac{1}{c \ w} \ \ \ \ \ (n=0,1,2,3,...), [/latex]

as can the angular momentum

[latex] J_n(w)=\pm n(2\pi R\rho )Rc= \pm nK \frac{1}{w^2} \ \ \ \ \ (n=0,1,2,3,...). [/latex]

In this latter formula, the two signs correspond to loops moving 'above' and 'below' the rope, i.e. to the two possible 'polarisations'.
It can be seen that- provided only circular excitations of a single frequency are allowed-the energy, momentum and angular momentum can only assume discrete values, those that can be written as the product of an integer and a basic 'quantum'. The following relationships are valid between these (frequency-dependent) 'quanta':

[latex] E(w)= cP(w)=wJ(w). [/latex]

It is not difficult to recognise that the same relationships are valid for photons

[latex] E_{photon}= \hbar w ; \ \ \ P_{photon}=\hbar w/c; \ \ \ J_{photon}= \pm \hbar . [/latex]

Naturally, this formal analogy must not be taken too seriously, e.g. by thinking of a photon as equivalent to a loop. However, the similarity can be used to show that, even in classical physics, there are objects more complicated than a point mass which are easy to understand, but which can still be excited to many discrete energy, momentum and angular momentum levels.

Question P.106: A long, heavy, flexible rope with mass ρ per unit length is...

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