swaha Posted November 20, 2009 Posted November 20, 2009 if the pendulum is kept vertical & the earth rotates the direction of swing of bob is not influenced by the roattion of earth. so the angle the earth rotates in time t should be same as traversed by the pendulum in same time t. why then omega/sin(of latitude) is the time period of oscillation? can anybody please expalin me that cosidering an inertial reference frame? pls reply.
swansont Posted November 20, 2009 Posted November 20, 2009 The plane of oscillation for the pendulum will be constant if the observer is in an inertial frame, at rest with respect to the plane of oscillation. i.e. the plane is constant with respect to the distant stars, which are approximately at rest. The earth is not an inertial frame because of its rotation
swaha Posted November 21, 2009 Author Posted November 21, 2009 where does the sine factor come then? & why?
swansont Posted November 21, 2009 Posted November 21, 2009 where does the sine factor come then? & why? Would a Foucault pendulum work at the equator? Why or why not?
swaha Posted November 22, 2009 Author Posted November 22, 2009 i didnt understand the reason in books why not?
swansont Posted November 22, 2009 Posted November 22, 2009 The Foucault pendulum moves relative to the circle in which it is swinging because the earth rotates. You have to project one onto the other.
swaha Posted December 5, 2009 Author Posted December 5, 2009 complaint for negligience to the moderators. i think i get least no of replies. WHY?
michel123456 Posted December 5, 2009 Posted December 5, 2009 Would a Foucault pendulum work at the equator? Why or why not? "À l'équateur le pendule y oscille dans un plan fixe." from french Wiki.http://fr.wikipedia.org/wiki/Pendule_de_Foucault transalation:" at the equator, pendulum oscillates in a fixed plane" in other words, it does not rotate. Why? see Wikipedia.
mooeypoo Posted December 5, 2009 Posted December 5, 2009 "À l'équateur le pendule y oscille dans un plan fixe." from french Wiki.http://fr.wikipedia.org/wiki/Pendule_de_Foucault transalation:" at the equator, pendulum oscillates in a fixed plane" in other words, it does not rotate. Why? see Wikipedia. If you're going to send people to wikipedia instead of answering their question directly (as opposed of, say, answering and using wikipedia as a source), at the very least send them to the *ENGLISH* version of that page: http://en.wikipedia.org/wiki/Foucault_pendulum That said, I'm not quite sure I understand how what you're saying here answers the poster's question..?
michel123456 Posted December 5, 2009 Posted December 5, 2009 (edited) Sorry That was not intended to frustrate the english language. But you must know that articles in Wiki are not translated from one language to another. Each language has its own article. usually, english Wiki is the best. Articles are more developped and more readible. The sentence from english Wiki is " When a Foucault pendulum is suspended on the equator, the plane of oscillation remains fixed relative to Earth." You are right. i will be more carefull in the future. Edited December 5, 2009 by michel123456
swansont Posted December 5, 2009 Posted December 5, 2009 complaint for negligience to the moderators. i think i get least no of replies. WHY? I'm guessing it's because "I don't understand" doesn't give any feedback on what explanation is needed. If you don't understand some basic principle of a Foucault pendulum, consult the wikipedia article or the many explanations you can find on the web. If you have a specific question, ask it. I suspect there is a language barrier here that makes things more difficult. But you aren't "owed" a response by anyone, least of all in response to a question that is not clear.
swaha Posted December 12, 2009 Author Posted December 12, 2009 i had read the article before. what i dont understand why the pendulum plane should rotate wrt the earth, and if it does why not at the equator. the sine of latitude comes by doing some mathematics considering coriolis force. but why the question of coriolis force comes at all for the rotation occurs at same latitude.(dr/dt=0). can the expalnation be given considering from an inertial frame of reference? if so please give me one for its easier to understand. Merged post follows: Consecutive posts mergedi know i dont owe any response but generally someone tries to answer.
swansont Posted December 12, 2009 Posted December 12, 2009 The pendulum looks like it's rotating, but it isn't — the earth is. Do you understand that the earth, because of its rotation, is not an inertial frame of reference?
swaha Posted December 13, 2009 Author Posted December 13, 2009 yes the earth is non inertial frame of reference so if we look from outside we'll observe the earth roatting alone & not the pendulum? but that also doesnt answer that why the effect of coriolis force is concerned.
D H Posted December 13, 2009 Posted December 13, 2009 The pendulum looks like it's rotating, but it isn't — the earth is. Not true. Or rather, true at only two spots on the Earth: The north and south poles. At any point between the equator and one of the poles, the pendulum assembly rotates with a period of one sidereal day (Earth rotation), the pendulum swings back and forth with a period of [math]2\pi\sqrt{L/g}[/math], and this swing precesses with a period of [math]1/\sin\lambda[/math] sidereal days. With three distinct periods involved, describing the motion from the perspective of an inertial frame is a bit tough. Do you understand that the earth, because of its rotation, is not an inertial frame of reference?[/quote']yes the earth is non inertial frame of reference so if we look from outside we'll observe the earth roatting alone & not the pendulum? but that also doesnt answer that why the effect of coriolis force is concerned. There is no coriolis force when explain the behavior of the Foucault pendulum from the perspective of an inertial frame. Doing that is a bit challenging, however. Explain that behavior from the perspective of someone standing on the surface of the Earth and there is (must be) a coriolis force. It's an inherent part of the equations of motion for a rotating reference frame.
swaha Posted December 13, 2009 Author Posted December 13, 2009 cant the mathematics be done without the coriolis effect or atleat the physical interpretation? but why does the precession occur at the rate of 24/sin(latitude) hrs & not 24hrs?
swaha Posted December 28, 2009 Author Posted December 28, 2009 can anybody atleast explain why the effect of Coriolis force is considered here at all? i think it happens as in case like a bead rolling from the edge towards the centre of a rotating disc.
D H Posted December 28, 2009 Posted December 28, 2009 The coriolis force appears in any rotating frame of reference. The effect on the trajectory of a baseball are very small and can be ignored. The effect on a long-range missile, or a Foucault pendulum are not small.
mooeypoo Posted December 28, 2009 Posted December 28, 2009 The coriolis force appears in any rotating frame of reference. The effect on the trajectory of a baseball are very small and can be ignored. The effect on a long-range missile, or a Foucault pendulum are not small. DH, I'm not sure about this, so I'd love it if you could verify/correct me - the effect is noticeable on a long-range missile because of the large distance (where it "picks up" enough of the effect to actually be noticeable) and it is noticeable on the focault pendulum because of *time* right? That is, the longer the pendulum is moving freely, the more we see the effect. The effect still exists in short-distance or short-duration movements on a rotating frame but if it's too short in time or distance then it's negligible. Is that right?
D H Posted December 28, 2009 Posted December 28, 2009 Add in the fact that those long-range missiles inherently have to be moving very fast (remember that coriolis force is proportional to velocity) and yes, that is right, mooey.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now