Kleppner p-77says that a pendulum hanging vertically and rotating at angular speed w is in an equilibrium unstable w.r.t. another identical one which moves along a circle. In first case semi-vertical angle tends to 0 while in the 2nd case it has some value >0 but <90 deg.I saw that in the 1 st case, K.E. is smsller than in the 2nd case as MI is greater in the 2nd case.Then if the total energy be conserved the PE of 1st config. may be greater and it may be unstable.I am not sure.Please help.
What do you mean by the 1st pendulum's equilibrium's "being unstable with regard to" the 2nd's? As in, you couldn't tie them together? Or they're not the same pendulum? Please Register or Log in to view the hidden image!
I am sorry as I could not give the meaning.What Kleppner says, that if the pendulum of 1st config. rotating quickly but hanging vertically, be perturbed a little, it will move on a circle with a bigger radius.This is not so-called eqilbm. However, I used the term rather loosely.
it is a conical one...you may visualise it as a simple pendulum is made to rotate in a circular trajectory.
Hi neelakash, The second case will have greater PE than the first (difference = m.g.L(1-cosA)). But the second case will also have a lower angular speed than the first, so that energy is conserved. Assuming a spherical bob or radius r and mass m, and a massless rod of length L (L >> r), you should be able to calculate the angular speed and semi-vertical angle of the stable pendulum.
your solution is not accceptable.Because,angular freq. w is the same for both-for this I happened to call them identical. w=2(pi)/T or,T=2pi/w also, T=2(pi)r/v =2(pi)r/wr =2pi/w so be it angular freq. or, angular velocity, they are same for both.
I agree.but I got a soln. When you impart a small perturbation to the bob the semi-vertical angle will be different from 0.We had two soln. in hand -1.semi-vertical angle 0 deg. 2.cosine inverse of (g/lw^2).However, previously, I did not give these.Once the angle gets changed from 0 the 2nd soln. must dominate.Energies will be adjusted in its way.One must supply it from outside as you say.
The energy required to alter the semi-vertical angle by a tiny amount (let's say 0.0000001 degrees) is effectively nothing. But the angle is enough that the pendulum will be pulled out further centrifugally. The energy to do this comes from the bob's kinetic energy, so the frequency decreases. You could also address this problem from the conservation of angular momentum. A spherical bob of mass M, radius r, spinning with angular velocity w has angular momentum 2Mr<sup>2</sup>w/5. The same bob swinging in a circle of radius R (>>r) and angular velocity W has angular momentum MR<sup>2</sup>W. Equating the two: 2M.r<sup>2</sup>w/5 = M.R<sup>2</sup>W w/W = 5R<sup>2</sup>/2r<sup>2</sup> Clearly, if R >> r, then w > W. We can determine an exact solution by examing the forces involved. I'm bored, so I'll do it now: L= length of string A = semivertical angle g = acceleration due to gravity T = tension in the string Find the horizontal and vertical components of the string tension, and arrange in terms of T: T cos(A) = M.g (vertical component, must balance gravity) T = M.g/cos(A) T sin(A) = M.R.W<sup>2</sup> (horizontal component, provides centripetal force) T = M.R.W<sup>2</sup>/sin(A) Equate the two: M.g/cos(A) = M.R.W<sup>2</sup>/sin(A) g/cos(A) = R.W<sup>2</sup>/sin(A) g<sup>2</sup>/cos<sup>2</sup>(A) = R<sup>2</sup>.W<sup>4</sup>/sin<sup>2</sup>(A) Find sin<sup>2</sup>(A) and cos<sup>2</sup>(A) through the relationship of R and L: R = L sin(A) sin<sup>2</sup>(A) = R<sup>2</sup>/L<sup>2</sup> cos<sup>2</sup>(A) = 1 - R<sup>2</sup>/L<sup>2</sup> Substitute back into the forces equation: g<sup>2</sup>/(1 - R<sup>2</sup>/L<sup>2</sup>) = R<sup>2</sup>.W<sup>4</sup>/(R<sup>2</sup>/L<sup>2</sup>) Isolate R<sup>2</sup>: g<sup>2</sup>/(1 - R<sup>2</sup>/L<sup>2</sup>) = W<sup>4</sup>L<sup>2</sup> g<sup>2</sup> = W<sup>4</sup>L<sup>2</sup>(1 - R<sup>2</sup>/L<sup>2</sup>) g<sup>2</sup> = W<sup>4</sup>(L<sup>2</sup> - R<sup>2</sup>) R<sup>2</sup> = L<sup>2</sup> - g<sup>2</sup>/W<sup>4</sup> Substitute back in to the angular momentum equation: w/W = 5R<sup>2</sup>/2r<sup>2</sup> w/W = 5(L<sup>2</sup> - g<sup>2</sup>/W<sup>4</sup>)/2r<sup>2</sup> Rearrange as a polynomial of W: 2w.r<sup>2</sup>/5 = W(L<sup>2</sup> - g<sup>2</sup>/W<sup>4</sup>) 2w.r<sup>2</sup>/5 = W.L<sup>2</sup> - g<sup>2</sup>/W<sup>3</sup>) W<sup>4</sup> - (2w.r<sup>2</sup>/5L<sup>2</sup>)W<sup>3</sup> - g<sup>2</sup>/L<sup>2</sup> = 0 Yuck! I'll stop there. Finding roots of quartics is too much hassle.
The 1st part of your suggestion assumes that once you impart the small perturbation, the bob will fly off to a larger radius in the expense of its own kinetic energy. Have you considered that the perturbation itself will impart energy to the bob to fly centrifugally off-centre? What you told was not the thing I wanted to know. My question was what configuration is more stable? For A=0, we get a motion somewhat like a spin. For A not =0, the bob rotates. But the basic criterion is the unique value of w in both cases. It’s peculiar, I agree. You have to neglect mass of the string. Otherwise, torsion will come into the picture. Kleppner draws this diagm. casA vs w. The vertical on w axis corresponds to w1=[g/l]^0.5. With w<w1, we have only one soln. With w>w1, we have two soln. One is cosA=1 anther is cosA=[g/lw2]. The problem is that you have to explain the instability of the horizontal portion of the curve where w>w1
LIke I said, even a tiny perturbation is enough. That's why the vertical spin configuration is unstable. The circular motion is stable, because a small perturbation will not cause a large change in motion.
neelakash, I don't have my copy of Kleppner on hand, so please correct me if this is wrong, but the physical situation is that you have a pendulum which is being rotated at a constant angular velocity w. Since the pendulum is being driven (with fixed angular velocity in this case), energy is not conserved. If w goes to zero, if the pendulum is not being driven, then you know that hanging vertically is a stable equilibrium. Furthermore, if w is small, if pendulum is being driven sufficiently slowly, then the hanging vertical equilibrium remains stable. However, if you spin the pendulum fast enough, the hanging vertical equilibrium ceases to be stable, and a new equilibrium appears where the bob hangs at some definite angle. Your job is two fold. First, you must find all the equilibrium points of your system. You know that hanging vertically is one. You must determine when a second equilibrium exists and what the equilibrium angle is. Second, you must determine the stability of each equilibrium point. To do this, consider small deviations from the equilibrium point and calculate the force. If the force tries to push you back to equilibrium, then the equilibrium is stable. If the force pushes you further away from equilibrium, then the equilibrium is unstable.
What don't you know? Do you know how to write down the forces? Do you know how to find an equilibrium point given the forces? Do you know how to study small oscillations given the equilibrium points?
