Firstly pack it in with the attitude. There's only so long I or anyone else will let it slide. Secondly the assumption of no friction is everywhere in introductory mechanics, just like no air resistance or perfectly flat surfaces or any other mathematical realisation. Where are you bringing it up, you've ready enough physics books haven't you?
Another question about the s/t diagram: given there are four beads each released at the same time (say t[sub]0[/sub]), when does the bead at the highest point (the blue one) reach the next lowest bead's initial position along s? Does the next lowest (the red one) reach the inital position of the bead below it (the green one) at the same time? What does it mean if these descent times (not the total descent times) are equal or not?
In 1673 Christiaan Huygens constructed the first pendulum clock with a isochronous pendulum by forcing the pendulum to swing in an arc of a cycloid. Unfortunately, friction along the two arcs causes a greater error than that corrected by the cycloidal path [1]. It wasn't until 100 years AFTER Huygens that John Harrison, an English clockmaker, without any formal education, solved the problem [2]. You see, it was the FRICTION that killed the cycloidal clock, rendering it a non-starter. [1] . Gardner, M. The Sixth Book of Mathematical Games from Scientific American. Chicago, IL: University of Chicago Press, pp. 129-130, (1984). [2] http://en.wikipedia.org/wiki/John_Harrison
That really doesn't look like it has much bearing on this thread. Friction and a rolling body only mean the body will eventually stop rolling if no external force is applied.
It shows what happens when you think you are "solving" stuff and you neglect to describe the problem completely : you end up with a non-starter, exactly the way Huygens ended.
Huygens thought he could use a cycloidal pendulum to build a clock. Is that what you think I'm trying to do? What "problem" do you think I've failed to describe completely?
...and, since he considered an idealised case, he failed to get any practical solution. There is a lesson to be learned from this. For one, you will need to chuck away the book you've been learning from, there are much better books to learn mechanics from. For example, the solution to the problem has never been one of kinematics, it is one of dynamics, you need to consider the forces (or the energies). Finding the equations of motion for the cycloidal pendulum. Deriving its period. Finding the equations of motion for a steel ball rolling down a cycloid.
\(m\frac{d^2s}{dt^2}=\Sigma {F}\) Your book should have that at the beginning of the chapter on dynamics.
That's an equation. Give a definition in English, no more than two or three sentences should suffice. You're allowed to include mathematical equations, as long as their meaning is clear. I.e. don't assume you're addressing anyone with more than a 1st year grasp of the "difficulties" you've been so clear about. For instance, I can assume without you clearing up what your equation means, that it means a constant mass times its acceleration along a path, s, is equal to the sum of forces acting on m. I also am assuming that since we've been discussing kinematics of rigid bodies, that m is a rigid body. Let's be careful about those assumptions.
Yes, the type of stuff that you were supposed to do for your homework but you managed to get prometheus to do for you. After you learn how to form the ODE's describing the equations of motion you need to learn how to solve them. I suspect that this is a different class that you needed to take before you started the mechanics class.
Well, I'm actually posing questions that you, being an expert, should be able to answer OTH. But here you are, deflecting. One more try: would you agree that e.o.m. correspond to prediction, of the future evolution of a system, that is, the equations predict the positions and momenta of systems of particles of which a rigid body is a special case?
Sure. How does any of this metaphysical stuff bring you any closer to being able to put together and solve any of these ODEs? This is what mechanics is all about. So far, you just posted pretty pictures lifted off the internet, when will you get down to doing actual physics? there isn't always going to be one of us solving your homework for you, you know.
Still got that "I can't let an idea go" problem? I'd go and get that seen to, if I were you. You've been wrong about several things pertaining to the behaviour of the system I first described (somewhat loosely) in the OP. Why should I, or anyone, think you're right about "doing actual physics" ?? How about you post the equations that show if the ball rolls down the track instead of sliding, the amplitude has to be small? You remember claiming that, don't you?
That is very easy, the non-realistic case of zero friction has the solution: \(T=\sqrt{\frac{0.7}{g}} \int_{-\alpha_0}^{+\alpha_0} \sqrt{\frac{a^2+(a-r)^2-2a(a-r) cos(\alpha)}{a-(a-r) cos(\alpha)}} d \alpha\) where: the "track" has the equation: \(x=a(\alpha-sin(\alpha))\) \(y=a(1-cos(\alpha))\) \(-\pi<\alpha<\pi\) \(r\) is the radius of the steel ball \(\alpha_0\) is the initial value for the parametric angle \(\alpha\) As you can see, the solution is an elliptic integral that depends on \(\alpha\) . For the realistic case when there is friction, the above integral gets a lot more complicated. Deriving it is left as an exercise for you, you haven't done any math/physics outside cutting and pasting pictures off the web anyway, so this would be a good time for you to get started. And, yes, I derived the above from base principles. And, no, you aren't going to con me to spoon feed the derivation to you. PS: the case of the frictionless pendulum is obtained as a particular case of the above by replacing \(\sqrt{\frac{0.7}{g}\) with \(\sqrt{\frac{0.5}{g}\) and by setting \(r=0, a=4R\).
Perhaps another way to say it so that you won't sharply disagree with me (snigger) would be "the overall length of the string is constant but the position of the point from which the string is fixed constantly moves, hence the effective length of the string that is actually in motion is not constant." That's a lot more effort to type so I didn't to start with but since you have an enormous pedantry issue, I have had to type it here. If you aren't careful and continually insist on raising points that are not relevant I shall start undeleting posts of yours, which would be very amusing indeed. Please Register or Log in to view the hidden image!
Ohhh, I was unaware that mods could read pre-edited posts! How exactly does one become a mod, again? Please Register or Log in to view the hidden image!
I confess I don't understand how this: is related to this: Have I missed something? Well (goodness gracious, etc), here is another analysis based on the total energy of the 'system of particles'. If instead of a curve, the particle rolls or slides down an inclined plane (i.e. instead of a changing slope the slope is constant), we have to account, or not, for the rotational kinetic energy. This energy will depend on the shape of the rolling body; the rotational velocity is related to its moment of inertia by the equations: \( E_k\; =\; \frac {1} {2} (\sum_i m_i R_i^2)\omega^2\; =\; \frac {1} {2} I \omega^2\) ...........(i) My reference states: equation (i) "... is correct for any axis, even if it is not a principal axis, because the magnitude of the velocity of each particle of the [rigid] body is always \( v_i\; =\; \omega R_i \)." Also: "When the rotation [of each particle] is about a principal axis, we may write \( E_k\; =\; \frac {L^2} {2I}\;\;\; \) ".............(ii) Since the particles in a rigid body are at rest relative to each other, the internal energy must remain constant, so the total energy for a system of particles reduces, in the case of a rigid body, to \( E\; =\; E_k\; +\; E_p \) which is equal to \( \frac {1} {2} m v^2\; +\; \frac {1} {2} I \omega^2\; +\; E_p \) We can solve for \( v \), by substituting for \( \omega\; =\; v/R \), and rewriting the moment of inertia \( I\; =\; m K^2 \) where \( K \) is the radius of gyration. So we get: \( E\; =\; \frac {1} {2} m ( 1\; +\; \frac {K^2} {R^2} ) v^2\; +\; mgy \) since \( mgy_0 \) is the initial total energy--i.e. the 'force' \( mg \) times a horizontal 'displacement' \( y_0 \). Therefore: \( v^2\; =\; \frac {2g(y_0\; -\; y)} {1\; +\; K^2/R^2} \) But if the rigid body slides instead of rolling, the term \( K^2/R^2 \) goes to zero, so then we have: \( v^2\; =\; 2g(y_0\; -\; y) \)
