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ing_it
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Homework Statement
We have a rod (length L, mass m) suspended at a point whose distance from the center of mass is a.
1) prove that (generally) there exist two values of a (a1, a2) for which the pendulum oscillates with the same period.
2) derive and explain: [tex] T = 2\pi\sqrt{\frac{a_1+ a_2}{g}}[/tex]
3) with what value of a do we get the smallest period?
2) derive and explain: [tex] T = 2\pi\sqrt{\frac{a_1+ a_2}{g}}[/tex]
3) with what value of a do we get the smallest period?
Homework Equations
[tex] E = 1/2 I w^2 [/tex]
[tex]J= J^* + ma^2[/tex]
[tex]J= 1/12 mL^2 + ma^2 [/tex]
[tex] T = 2\pi\sqrt{\frac{J}{mga}}[/tex]
[tex] T = 2\pi\sqrt{\frac{L}{g}}[/tex]
The Attempt at a Solution
1) I thought that two periods being the same would mean T1 = T2
[tex] 2\pi\sqrt{\frac{J_1}{mga}}= 2\pi\sqrt{\frac{J_2}{mga}}[/tex]
[tex]J^* + ma_1^2 =J^* + ma_2^2[/tex]
[tex]a_1^2 =a_2^2[/tex]
[tex]a_1 = +/- a_2[/tex]
and since a1 = a2 would be the same location a1 = - a2 which would mean that the only way to get the same period for two different values of a, would be to have the a be the same but in different directions from the center of mass.
2) I noticed that [tex] T = 2\pi\sqrt{\frac{a_1+ a_2}{g}}[/tex] looks a lot like [tex] T = 2\pi\sqrt{\frac{L}{g}}[/tex]
So I thought that a1 + a2 = L
and that that would mean that the rod would be suspended on one end and the mass would be concentrated on the other end of the rod? I guess it sounds silly but I just couldn't figure out anything else.
3) [tex] \lim_{x\rightarrow 0} T = 2\pi\sqrt{\frac{J}{mga}} = 2\pi\sqrt{\frac{1/12 L^2 + a^2}{ga}} [/tex]
since [tex] 2\pi\sqrt{\frac{1}{g}} [/tex] is a constant I'll derive the rest because a limit is just a special type of derivation (I think?)
[tex] \frac{1/2 L^2 + a^2}{a}[/tex] derived is:
[tex] La + a^2 + {\frac {L^2}{2}}[/tex]
and from that I can see that the smaller the a the smaller the period of the oscillation so we would get the shortest period if the pendulum was hung from its center of mass.I was told my solution is wrong. I don't know which parts, I don't know why and I don't know how to correct it. If you could help me out somehow I'd be really grateful.
[tex] 2\pi\sqrt{\frac{J_1}{mga}}= 2\pi\sqrt{\frac{J_2}{mga}}[/tex]
[tex]J^* + ma_1^2 =J^* + ma_2^2[/tex]
[tex]a_1^2 =a_2^2[/tex]
[tex]a_1 = +/- a_2[/tex]
and since a1 = a2 would be the same location a1 = - a2 which would mean that the only way to get the same period for two different values of a, would be to have the a be the same but in different directions from the center of mass.
2) I noticed that [tex] T = 2\pi\sqrt{\frac{a_1+ a_2}{g}}[/tex] looks a lot like [tex] T = 2\pi\sqrt{\frac{L}{g}}[/tex]
So I thought that a1 + a2 = L
and that that would mean that the rod would be suspended on one end and the mass would be concentrated on the other end of the rod? I guess it sounds silly but I just couldn't figure out anything else.
3) [tex] \lim_{x\rightarrow 0} T = 2\pi\sqrt{\frac{J}{mga}} = 2\pi\sqrt{\frac{1/12 L^2 + a^2}{ga}} [/tex]
since [tex] 2\pi\sqrt{\frac{1}{g}} [/tex] is a constant I'll derive the rest because a limit is just a special type of derivation (I think?)
[tex] \frac{1/2 L^2 + a^2}{a}[/tex] derived is:
[tex] La + a^2 + {\frac {L^2}{2}}[/tex]
and from that I can see that the smaller the a the smaller the period of the oscillation so we would get the shortest period if the pendulum was hung from its center of mass.I was told my solution is wrong. I don't know which parts, I don't know why and I don't know how to correct it. If you could help me out somehow I'd be really grateful.
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