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You are not logged in. #1 20121010 09:16:47
Rod rotating about fixed axis"One end of a rod of uniform density is attached to the ceiling in such a way that the rod can swing about freely with no resistance. The other end of the rod is held still so that it touches the ceiling as well. Then, the second end is released. If the length of the rod is l metres and gravitational acceleration is g metres per second squared, how fast is the unattached end of the rod moving when the rod is first vertical?" where I is the moment of inertia of the rod about an axis perpendicular to the rod and in the plane of the rod, and w is the angular velocity of the rod. So we need to find the M.I. of the rod. Suppose I_{z} is the M.I. of the rod about an axis through one end of the rod and perpendicular to the rod. By the parallel axis theorem, the M.I. is but we want I_{y}. But I_{x} = 0, so I_{y} = I_{z} by the perpendicular axis theorem. Since Loss in GPE = Gain in KE, then therefore is this correct? #2 20121010 09:25:59
Re: Rod rotating about fixed axisPhewh! For a second there I though it was our admin rotating. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #3 20121010 09:27:16
Re: Rod rotating about fixed axisI let the mass of the rod be m, but it appears that it doesn't matter, since the mass cancels when comparing the GPE and KE. #4 20121010 09:40:18
Re: Rod rotating about fixed axisOk, step by step. How'd you get the loss in GPE formula? The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #5 20121010 09:42:37
Re: Rod rotating about fixed axisIf the rod is at first horizontal, then rotates 90 degrees (when first vertical), then the change in GPE is mgl/2, since the centre of mass of the rod has dropped by a distance of l/2. #6 20121010 09:49:36
Re: Rod rotating about fixed axisBut, the question never states the starting position of the rod. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #7 20121010 09:54:03
Re: Rod rotating about fixed axis
I don't see how else this could be interpreted... clearly the starting position is the rod in its horizontal position. #8 20121010 10:07:08
Re: Rod rotating about fixed axisOops! Me and my careful reading. Didn't see the "also touches the wall" part. Sorry. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #9 20121010 10:13:29
Re: Rod rotating about fixed axisFor a uniform rod of mass m and length 2l, the moment of inertia about an axis perp. to the rod through the centre is ml^{2}/3 (you can prove this by integration but it should be fine to quote it). For a rod of length l, you square (l/2) instead. Then apply the parallel axis theorem, to get the M.I. of the rod about the axis through the end. I sort of wasted time thinking about I_x, I_y and I_z, since it is a rod, not a lamina, and therefore I_y = I_z. #10 20121010 10:23:06
Re: Rod rotating about fixed axisI would have to look at this in the morning. My brain is not functioning properly right now. The limit operator is just an excuse for doing something you know you can't. “It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman “Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment #11 20121010 18:03:29
Re: Rod rotating about fixed axishi zetafunc, You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #12 20121010 18:09:22
Re: Rod rotating about fixed axisThanks. #13 20121010 18:21:40
Re: Rod rotating about fixed axishi zetafunc You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #14 20121010 18:24:49
Re: Rod rotating about fixed axis...... stopped thinking You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #15 20121011 02:57:39
Re: Rod rotating about fixed axisI've studied moments of inertia before (had to do it for my exam) and deriving them via integration  but I know the M.I. of the rod is definitely correct (and therefore the change in KE). Since: as v = rw. But, this question still confuses me... are they asking for the ANGULAR velocity w, or just the instantaneous velocity of the end of the rod? In which case, with r = l; v = rw, using my w from post #1; #16 20121011 05:23:52
Re: Rod rotating about fixed axisOh yes, I'm terrible for only half reading the question. It does ask 'how fast' so I guess they want the velocity. You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei #18 20121012 07:32:37
Re: Rod rotating about fixed axisA maths teacher said I am wrong but never explained why. Confused... 