The principle of reversible Kater's pendulum and why its rod's ends are pointed


by hypernova90
Tags: ends, kater, pendulum, pointed, principle, reversible
hypernova90
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#1
Nov30-08, 02:34 AM
P: 2
I am not able think of convincing answers to the following questions-
1.Why the distance between knife edges (when the time periods about them is approx same) is considered equal to the length of an equivalent simple pendulum ?? It will be great if you can derive it....
2. Why the ends of the rod of Kater's pendulum are sharpened (pointed) ??
3. Why the rod itself is so heavy ?

I don't have a clue to question 1's answer....

For question 2, I think it is only so that we can better see when the pendulum passes a certain reference point, thus reducing the error in measurement of time period....but this is not satisfying because a bar pendulum does not have such ends (the rod in bar pendulum is cylindrical with flat ends)....

For question 3, I think it is to reduce the effect of buoyancy due to air ?? I am not sure...

Any help will be appreciated.
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jeebs
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#2
Nov30-08, 12:47 PM
P: 326
i would have thought the blades were sharpened so that it could rock back and forth smoothly, so it doesnt waste any energy. i am curious about why the distance between the blades can be treated as the effective length of a simple pendulum though, i have been scouring the net looking for a derivation myself for some work im doing and i cant find one.
for question three i would imagine that its helpful to have a heavy pendulum so that air resistance is not such a problem.
hypernova90
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#3
Nov30-08, 10:47 PM
P: 2
Jeebs ,
I am not talking about the sharpening of the blades of the knife edges, that is done so that the "point" (actually line) of suspension of the rod remains the same, otherwise a blunt end would cause the rod to swing in a slightly 'distorted' fashion as the position of point of suspension itself will not be constant....
I am talking about the 'sharpened' (maybe this word is causing confusion, but i dont have any other word for it) ends of the rod itself . If u see the rod(not the knife edges) in Kater's pendulum, the ends are made like the head of a missile.....

And why will air drag decrease with increase in mass of rod ?? I thought resistance due to a fluid depends entirely on volume......is the answer decrease in air resistance OR decrease in buoyancy due to air..? OR are both the same thing.....??

jeebs
jeebs is offline
#4
Dec1-08, 02:02 AM
P: 326

The principle of reversible Kater's pendulum and why its rod's ends are pointed


ohh sorry. Well, the reason that our Kater's pendulum had very thin tips was to allow it to fit through the light gate that measured the period.

what i meant by making it more heavy was, compare the flight of a shuttlecock (badminton) to the flight of a tennis ball, you can belt a shuttlecock as hard as you want but the air resistance effect is still massive, whereas a tennis ball is effected a lot less, and its more down to the different masses of the two objects rather than the shapes of them.

if you still dont know what i mean, imagine your pendulum, and then imagine it completely hollow inside. which one is going to be affected by air resistance more?
adriank
adriank is offline
#5
Dec1-08, 03:31 AM
P: 534
Quote Quote by hypernova90 View Post
1.Why the distance between knife edges (when the time periods about them is approx same) is considered equal to the length of an equivalent simple pendulum ?? It will be great if you can derive it....
This result is not obvious.

In general, given a pendulum (with small oscillations) with moment of inertia [tex]I[/tex] about the axis of rotation and d the distance to the centre of mass, recall that the period of oscillation is [tex]T = 2\pi\sqrt{I/mgd}[/tex]. For a simple pendulum, [tex]I = md^2[/tex], so [tex]T = 2\pi\sqrt{d/g}[/tex].

Say the mass of Kater's pendulum is [tex]m[/tex], the distances from the centre of mass to each of the knife edges are [tex]a, b[/tex], and say the moment of inertia about the centre of mass (along the axis of rotation) is [tex]I_0[/tex]. Then the moments of inertia about the two pivots are [tex]I_a = I_0 + ma^2[/tex] and [tex]I_b = I_0 + mb^2[/tex]. From above, their periods are [tex]T_a = 2\pi\sqrt{(I_0 + ma^2)/mga}[/tex] and [tex]T_b = 2\pi\sqrt{(I_0 + mb^2)/mgb}[/tex]; setting them equal and solving gives [tex]I_0 = mab[/tex]. In that case, the common period is [tex]T = T_a = 2\pi\sqrt{(mab + ma^2)/mga} = 2\pi\sqrt{(a + b)/g}[/tex], which is the period of a simple pendulum with length [tex]a + b[/tex].


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