"Yes" , you are desperately trying to get others to solve your homework. Given the way you behaved: nya, nya.
Hope I'm not pushing the envelope here, and I realise I need to be careful about what I say in the presence of intellectual giants, this question reproduced verbatim, appears at the end of the 5th chapter of the aforementioned textbook: A wheel of radius \( R \) rolls with constant velocity \( v_0 \) along a horizontal plane. Prove that the position of any point on its edge is given by the equations \( x\; =\; R(\omega t\; -\; sin\; \omega t),\; y\; =\; R(1\; -\; cos\; \omega t) \) where \( \omega\; =\; v_0 / R \) is the angular velocity of the wheel and \( t \) is measured from the instant when the point is initially in contact with the plane. Also find the components of the velocity and the acceleration of the point. (my italic)
Without writing down any equations, we can see that since the wheel moves (say, left to right) at constant velocity, the point's x component does too, whereas the y component accelerates (say, up and down). By 'inverting' the context from a rolling wheel to a pendulum, this also inverts the above relation between x and y. In the pendulum's case (with or without rolling), we can imagine a generating circle moving left to right, then right to left, which accelerates in the x direction, and doesn't accelerate (is isochronous) in the y direction. So we know the position vector has a constant velocity plus a changing velocity component, in both cases. Now we need to choose an origin (x[sub]0[/sub],y[sub]0[/sub]) for both cases.
This is not true, because of the angular velocity of the wheel. You can see that by considering the point when it just touches the ground - it is instantaneously stationary so it must accelerate. You can also see it by taking your equations for the x position and differentiating it wrt time - you will find it is non zero.
The easiest way to attack this problem is to consider the wheel rotating in the rest frame of the axle. You can choose a point at a radius R and use basic trig to write down it's x and y position (the origin should be at the bottom of the wheel, so the axle is at (0,R)). Once you've done that, apply a Galilean transformation to get the answer. Notes: remember that the wheel is moving with a speed v, which is equivalent to a frame moving at a speed -v relative to the axle. Since there is no motion relative to the axle in the y direction, the equation for y follows from basic trig only.
Are you saying the point accelerates in both the x and y directions, relative to the plane? I think I got it wrong, the point does appear to accelerate in the x direction, reaching the greatest velocity when at the highest (resp. lowest) point. But does it accelerate in the y direction?
It has to accelerate in both directions because it's rotating (and it will accelerate in any frame - even the rest frame of the axle). If you look at the wheel end on, the point on the edge of the wheel will appear to be moving in simple harmonic motion, correct?
To recap, we are still yet to answer the final question, viz I actually rather like this problem, and if I can be bothered to type up the solution I will unless someone else that's trying to do it comes back and tells me not to spoil the surprise! I have some notes, including some for people that have been following this thread from the beginning: The derivation of the equation of motion relies on making a crucial observation regarding the angle: By basic geometry you may note that the angle of the normal of the cycloid surface to the vertical is exactly half of the parameter used in the definition of the cycloid. Tach's claim that the equation of motion for the cycloidal pendulum is different to the equation of a bead on a cycloidal track is wrong, as is his claim that the equation of motion simplifies for small angles / amplitudes. The whole point of a cycloid is that the period does not depend on the amplitude of the motion. Tach's other claim about the kinetic energy of the bead on a cycloidal track is also wrong. Quite comically so if I may say so.
I think it's interesting that some of the initial suggestions for finding a solution suggested that finding one would be difficult, and yet there's a question at the end of an early chapter of an undergrad text, which suggests all you need is (the chapter on) kinematics. Handily, the question appears at the end of this chapter, which is the first 'real' physics in the book. The first 4 chapters are more your geometry and measurement stuff, i.e. math.
Your counter can be easily proven wrong since the ball rolling along the cycloid has a different kinetic energy due to its rotation about its own axis. arfa brane got this, I wonder how you could miss it. He's waiting for someone to do his homework, I see that you are chomping at the bit to prove me wrong rather than seeking the truth, so , I take it that you will not give the above a second thought and jump straight into showing off and solving his homework. ......for an ideal, not physically realizable situation. For realistic situations , this is not true. He who laughs last, laughs best.
I got the idea, by looking closely at the animations for the cycloid and the tautochrone, that the y velocity is constant--looking at the wheel's edge in the direction of translation, means a mark on a transparent wheel's perimeter will appear to oscillate left and right at a constant velocity, like a sawtooth wave. That is, the y acceleration is "nulled out" by the translation of the axis, observing in the x direction. Maybe I have this wrong, but it looks a lot like the bead on a wire scenario has constant velocity for the y direction, the bead accelerates 'sideways'. I'm also assuming the animations are accurate physically. Have another look: Please Register or Log in to view the hidden image! Please Register or Log in to view the hidden image!
In the lower animation, the graph on the right is the curve ds/dt for each colored bead. We know that at each point on any curve, the relation (ds)[sup]2[/sup] = (dx)[sup]2[/sup] + (dy)[sup]2[/sup] holds (for the curved 'rail' on the left, the physical path for each bead). The animation shows what isochronous curves look like. The four small black arrows demonstrate the time rate of change (in something or other) along s is equal for each bead.
I have to wonder why you're approaching the problem in this way. You have the equation for the y position as a function of time: \(y = R \left(1- \cos(\omega t) \right)\). You can work out the y component of velocity by differentiating this formula with respect to time, and from that you can see that it is certainly not a constant.
While if the problem was to study a ball rolling on a cycloidal track I would agree with you, the original problem was not this, and afra brane was referring to the rolling wheel defining the cycloid when he was talking about angular momentum (post #4). I have to say, I find it rather surprising that you have failed to understand this. Scraping the bottom of the barrel for objections now Tach? If we are taking the surface of the track to be frictionless, the equations of motion are the same for both scenarios. For the pendulum there will be some air resistance that will lead to a similar term in the equation of motion (with a different coefficient) if you want to be completely "realistic." I think you just want to be pedantic, and it's nothing more than trolling. yeesh, a far better version of this phrase is "He who laughs last, laughs longest."
Looks like you have joined the trolling cheerleader's squad. Too bad, try reading (for comprehension, this time) the post: Subsequent exchanges with arfa show that he clearly understood the issue, why is it taking you so long? First off, "frictionless" is not realistic. Second off, even in the "frictionless" case there is a difference between the two cases, you have not written any equation, you have just made unsupported assertions.
Yeah? I'm not disputing the situation of the rolling ball on a cycloidal track is different from a frictionlessly sliding one (I'm not completely sure the linear motion will be any different from the sliding case if you assume no energy losses, but hey). It's quite natural that the thread participants will discuss something like a rolling ball on the track after it is brought up. My point is that it was you that brought it up, not arfa. In these sorts of classical mechanics problems assuming the surface is frictionless is very common. Again, I never said it was realistic (or not). Well, I would very much like to see your proof that a mass sliding frictionlessly on a cycloidal track has a different equation of motion to a cycloidal pendulum. I also note that you've not provided a shred of evidence either, so I am taking your remark about unsupported assertions as trolling. Lets be clear Tach, you've already had a warning on this thread and it's not the only one you've got active. You accusing me of trolling is pretty irrelevant but when I am accusing you of trolling you're very close to a holiday (the usual disclaimer goes here about if you have a problem with moderation take it to a super-mod or admin). Where we go from here is down to you.
Good. That was the point. There is a clear difference between the two cases. If you do the math, you will also find out where the differences lie.
Derivation of the equations of motion for a bead on a frictionless cycloidal track. The equations defining the cylcoid are \(x = R\left(\alpha + \sin \alpha \right), \quad y = R\left(1 - \cos \alpha \right)\). We will compute the equation of motion of \(s(t)\) defined by the Euclidean line element: \(ds^2 = dx^2 + dy^2\) We can show from the equations of the cycloid \(dx = R\left(1 + \cos \alpha\right) d \alpha, \quad dy = R \sin \alpha d \alpha \) hence \(ds^2 = 2 R^2 (1 + \cos \alpha) d\alpha^2\) using the trig identity \(\cos^2 \left(\frac{\theta}{2}\right) = \frac{1 + \cos \theta}{2}\) we can show \(ds = 2 R \cos \frac{\alpha}{2} d \alpha\\ \Rightarrow s = 4 R \sin \frac{\alpha}{2}\) here is a diagram of the setup (ignore the fact that this is a pendulum - I couldn't find a more appropriate picture and I was too lazy to draw one. The important point is the particle has a weight mg, the tangential component of which is \(mg\sin\frac{\alpha}{2}\) and the angle of the normal to the vertical can be worked out by basic geometry to be half of the cycloid parameter \(\alpha\), labeled \(\theta\) in the diagram.) Please Register or Log in to view the hidden image! Consider using \(F = m a = m \frac{d^2 s}{dt^2}\) where the forces are at a tangent to the cycloid. we have \(-mg \sin\frac{\alpha}{2} = m \frac{d^2 s}{dt^2}\). Using the relationship between \(\sin\frac{\alpha}{2}\) and s we can get an equation in terms of s only: \( \frac{d^2 s}{dt^2} + \frac{g}{4R} s = 0\). The solution is clearly oscilliatory, but we don't even need to solve it to get the period: \(\omega^2 = \frac{g}{4R}\) and \(T = \frac{2 \pi}{\omega}\) so \(T = 4 \pi \sqrt{\frac{R}{g}}\). Notes: We have made no assumption about the amplitude of any oscillation. The period does not depend on the amplitude. The only inputs were the path followed by the particle and the fact it is massive. There is nothing regarding a pendulum or a track.
Correction: I was the one who initially suggested that the equations of motion for a sliding or rolling ball in a cycloidal track should look the same. Tach objected and claimed that for a rolling ball the angle of displacement had to be small. I claimed that the only difference is a constant factor because the radius of gyration for the rolling ball is constant, so the equations will not be "different" as far as physics is concerned. However the period will be longer than \( 4\pi \sqrt { \frac {R} {g}} \), in that case. I believe that equation applies only to a body sliding without friction, and my claim might not be true (although I can't see why not). Then again, I should have realised that ds/dt isn't constant for a sliding bead, and the animated plot of s against t in the second animation shows this. What else it shows is that each ds/dt reaches the same point--1/4 of a period--at the same time, independently of the velocity or the initial height of each bead. We know that \( x\; =\; \pi R \) at the point each curve intersects in the s/t graph.
