[Vo]:resend2/pendulum and spinning bob
Hi steven, Please excuse the brief reply, because my webmail account makes formatting difficult.(This is also a resend as my last message was apparently truncated.) Harry wrote:>> It also worthwhile to note that there is a theorem which states that the>> period of a pendulum depends on the distance from the centre of mass of> >the bob to the pivot point and is independent of the total mass of the> >bob. I think you need to consider this in your analysis.Steven wrote: > In any case I can see how it's very tempting to believe that the > pendulum arm's periodicity will not be affected regardless of whether > the break is applied or not applied, but unfortunately that's not the > case. I had, in fact, toyed with this possibility myself for a brief > period of time, until Mr. Lawrence correctly showed the flaw in my > reasoning. The pendulum arm's periodicity WILL definitely be affected > depending on the application of the break to the spin-bob wheel. When > the break is ON, the length of pendulum arm is effectively lon! ger causi This is inconsistent with the theorem. The length of the "arm" is the distance from the centre of mass of the bob to the pivot point.The Theorem says it is THIS distance which governs the period.The radius of the bob and the state of the brakes do not change this distance, so they do not affect the period. > This is precisely why this apparatus does not make additional energy. > > Again, I recommend increasing the size of the spin-bob wheel to around > three times the length of the pendulum arm. Hopefully, this will make > the experiment more easy to grasp as the ramifications of the theorem > you state, specifically the changes in periodicity due to changes in > length, should be more obvious visually. As I said above, the parameter that counts is the distance from the pivot point to the centre of mass of the bob. > Better yet, forget about all these theoretical conjectures. ! Get down
Re: [Vo]:resend2/pendulum and spinning bob
Harry Veeder wrote: Hi steven, Please excuse the brief reply, because my webmail account makes formatting difficult.(This is also a resend as my last message was apparently truncated.) Harry wrote: > It also worthwhile to note that there is a theorem which states that the > period of a pendulum depends on the distance from the centre of mass of >the bob to the pivot point and is independent of the total mass of the >bob. I think you need to consider this in your analysis. Steven wrote: In any case I can see how it's very tempting to believe that the pendulum arm's periodicity will not be affected regardless of whether the break is applied or not applied, but unfortunately that's not the case. I had, in fact, toyed with this possibility myself for a brief period of time, until Mr. Lawrence correctly showed the flaw in my reasoning. The pendulum arm's periodicity WILL definitely be affected depending on the application of the break to the spin-bob wheel. When the break is ON, the length of pendulum arm is effectively lon! ger causi This is inconsistent with the theorem. But it's not. The theorem, if correctly stated, just says the period is not affected by the total mass of the bob. However, the period _is_ affected by the size and shape of the bob. (The common statement that the period depends only on the length of the pendulum is not actually correct, save in certain limiting cases.) The theorem is actually included in the analysis I posted earlier, implicit in the solutions given in the post. Furthermore, the oft stated but not quite correct "theorem" that the period depends only on the shaft length can be seen in the solution for the free-spinning bob, or in the "locked" case when the bob is very small. In the latter case, whether it's locked or not makes no significant difference, because the rate at which it turns is small, the moment of inertia of a small bob is very small, and the amount of energy going into turning the bob is ignorably small. None the less, as a standard pendulum is like your pendulum with the brake LOCKED, a claim that the period depends /only/ on the shaft length would be, at best, an approximation. The solution for a bob of fairly small radius, with the brake locked, is, quoting from my "part II" post, theta = theta_0 * Cos(sqrt((g/l)(1 - (1/2)(R/l)^2)) * t) and the period, which is also the moment when the argument to the Cos(...) function is exactly 2pi, is: t_0 = 2 pi sqrt(l/g) / sqrt(1 - (1/2)(R/l)^2) This is, indeed, independent of the mass of the bob (and the length of the arc), but it's _NOT_ independent of the RADIUS of the bob. Again, the common statement that the period of a pendulum depends "only on the length of the arm" is made with the assumption that the bob is very small relative to the arm, so we can ignore the term (R/l)^2. In a grandfather's clock, with a 4 foot long pendulum and a 3" diameter disk on the end, (R/l)^2 = (1/256), and the effect due to the bob not being a point mass will be to lengthen the period by about 1 part in 1000. That's about 4 seconds per hour, or close to 2 minutes a day. It's significant -- but on the other hand, with a 48" shaft, it's equivalent to changing the shaft length by about 1/10 of an inch. That's well within the normal range of adjustment on such a pendulum. The standard textbook (freshman physics) solution for a simple pendulum will be the one for the case where the brake is UNLOCKED (or the bob is very small), which is just theta = theta_1 * cos(sqrt(g/l) * t) Again, this is independent of the mass of the bob, and in fact in this case it depends /only/ on the length of the pendulum. In particular, it does not depend on the length of the arc, which is one reason pendulum clocks are good timekeepers, once adjusted. On the other hand, the period does depend quite strongly on the value for "g", the acceleration due to gravity. Again, the period, assuming a small bob or free-spinning bob (and assuming I did the math right), is t_0 = 2 pi sqrt(l) / sqrt(g) The observed value for "g" is rather smaller at the equator than at the poles, due to the "centrifugal force" caused by the Earth's rotation (the equator moves at about 1000 mph). Consequently, a pendulum clock which kept correct time at the poles would run slow at the equator. By the way, the independence of the period on the length of the arc is one of the things that's only true for small angles -- for a pendulum swinging through a large arc the period depends strongly, and in complex fashion, on the length of the arc. The length of the "arm" is the distance from the centre of mass of the bob to the pivot point.The Theorem says it is THIS distance which governs the period.The radius of the bob and the state of the brakes do not change this distance, so they do not affect the period. This is precisely why this apparatus does not make additional energy. A
Re: [Vo]:resend2/pendulum and spinning bob
Howdy Vorts, Stephen Lawrence has posted some good pendulum info. A careful look at how he has approached the subject shows he is more than conversant with the subject. Back in Houston and Dallas in the late 1940's and 50's we saw a rise of technical interest in pendulums for use in seismic work for oil exploration. Surprisingly, much of the electronics industry had it's origin in some of these firms( like SIE) from which emerged TI. et.al. The industry awaits the next generation of " doodle buggers" that have insight into the workings of a pendulum that can apply that insight into a new generation of analytical and measuring instrumentation. I believe Stephen has a glimpse and should pursue it. For his good work posted I offer a thanks and a clue laser generated pendulum whereas the light undulates following a similar math. In otherwords.. light can now be made to bend, stop, accelerate and spring return. Hmm.. the measureable features of such light properties are astounding. For example.. a fiber optic section should gain weight under certain induced light properties and react under induced undulations which could possibly allow certain measurement in pressure and temperature now thought to be impossible. Richard
Re: [Vo]:resend2/pendulum and spinning bob
Hello Harry. Let's make sure I'm on the same page with you on this pendulum matter. Are you claiming the pendulum arm is going to swing back-and-forth with the exact same periodicity frequency regardless of whether the spin-bob wheel has the breaks on or off? > Steven wrote: > >> In any case I can see how it's very tempting to believe >> that the pendulum arm's periodicity will not be affected >> regardless of whether the break is applied or not applied, >> but unfortunately that's not the case. I had, in fact, >> toyed with this possibility myself for a brief period of >> time, until Mr. Lawrence correctly showed the flaw in my >> reasoning. The pendulum arm's periodicity WILL definitely >> be affected depending on the application of the break to >> the spin-bob wheel. When the break is ON, the length of >> pendulum arm is effectively longer causi > This is inconsistent with the theorem. > The length of the "arm" is the distance from the centre of > mass of the bob to the pivot point. The Theorem says it is > THIS distance which governs the period.The radius of the > bob and the state of the brakes do not change this distance, > so they do not affect the period. I believe this is where the problem (or inconsistency) lies. I believe you are selectively reinterpreting portions of the theorem to suit your diagram particularly in regards to when the break is applied. Unfortunately when the breaks are applied this DOES effectively change the pendulum arm's distance, and as such, the periodicity of the pendulum swing. When the break is applied to the spin-bob wheel you must add its inertial mass to the inertial mass of the pendulum arm. By the sacred laws of Newtonian physics this must slow things down a bit, or else the physics books will have to be rewritten - followed cats and dogs soon living together. This is why I have consistently suggested re-envisioning your diagram with the spin-bob wheel three times the size of the pendulum arm as this should make the change in periodicity more obvious depending on whether the breaks are applied and when not applied. >> This is precisely why this apparatus does not make >> additional energy. >> >> Again, I recommend increasing the size of the spin-bob >> wheel to around three times the length of the pendulum arm. >> Hopefully, this will make the experiment more easy to grasp >> as the ramifications of the theorem you state, specifically >> the changes in periodicity due to changes in length, should >> be more obvious visually. > As I said above, the parameter that counts is the distance > from the pivot point to the centre of mass of the bob. Indeed that is the case when the breaks are not applied - pendulum arm swings faster. And that is NOT the case when the breaks are applied - pendulum arm swings slower. I suggest another visual exercise particularly since you are very good with Adobe Illustrator. Create two distinct diagrams where both models possess the exact overall 3-D dimensions and mass distribution. (1) The first model contains two distinct parts that move independently of each other, a pendulum arm and a spin-bob wheel where they are attached through a frictionless bearing. IOW, the spin-bob wheel spins freely and independently of the pendulum arm's swing. (2) Next create another model with the exact same dimension of the first model but where the pendulum arm and the so-called spin-bob wheel are actually physically attached together as one solid unit. As previously stated, it's imperative that both models possess the exact same mass proportioned in the exact same locations. Now, swing back-and-forth both devices, in your head. What do you think the periodicity is going to be for both models? The same or different? Regards, Steven Vincent Johnson www.OrionWorks.com
Re: [Vo]:resend2/pendulum and spinning bob
Steven and Stephen, - Original Message - From: OrionWorks <[EMAIL PROTECTED]> Date: Monday, September 3, 2007 10:47 am Subject: Re: [Vo]:resend2/pendulum and spinning bob > Hello Harry. > > Let's make sure I'm on the same page with you on this pendulum matter. > > Are you claiming the pendulum arm is going to swing back-and-forth > with the > exact same periodicity frequency regardless of whether the spin- > bob wheel > has the breaks on or off? Not anymore. Stephen has made that clear. I should know better to not expect any surprises to emerge from systems involving purely mechanical linkages or models of systems based on Newton's laws of motion. > > Steven wrote: > > > >> In any case I can see how it's very tempting to believe > >> that the pendulum arm's periodicity will not be affected > >> regardless of whether the break is applied or not applied, > >> but unfortunately that's not the case. I had, in fact, > >> toyed with this possibility myself for a brief period of > >> time, until Mr. Lawrence correctly showed the flaw in my > >> reasoning. The pendulum arm's periodicity WILL definitely > >> be affected depending on the application of the break to > >> the spin-bob wheel. When the break is ON, the length of > >> pendulum arm is effectively longer causi > > > This is inconsistent with the theorem. > > The length of the "arm" is the distance from the centre of > > mass of the bob to the pivot point. The Theorem says ! it is yup. > Unfortunately when the breaks are applied this DOES effectively > change the > pendulum arm's distance, and as such, the periodicity of the > pendulum swing. > When the break is applied to the spin-bob wheel you must add its > inertialmass to the inertial mass of the pendulum arm. By the > sacred laws of > Newtonian physics this must slow things down a bit, or else the > physicsbooks will have to be rewritten - followed cats and dogs > soon living > together. This is why I have consistently suggested re-envisioning > yourdiagram with the spin-bob wheel three times the size of the > pendulum arm as > this should make the change in periodicity more obvious depending > on whether > the breaks are applied and when not applied. > > > >> This is precisely why this apparatus does not make > >> additional energy. > >> > & I 'm afraid I was never very good at performing mechano-logical simulations in my head. Harry
