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Natural Science Forum / Physics / Relativity / April 2006An IITJEE problem
I have a SHM problem. A rod of length L is placed on two circular dics. The co-efficient of friction between rod and disks is k. The rod is displaced by a small distance x. Determine the time period of the oscillations. _______________________ O O <----disks <-------------- L -----------------> The answer is t = 2(L/2kg) [g- acc due to gravity) But I want the solution. I have a SHM problem. A rod of length(> L )is placed on two circular dics. The co-efficient of friction between rod and disks is k. The rod is displaced by a small distance x. Determine the time period of the oscillations. __________ O O <----disks <-------L-----> (There might be some problems in the diagram, but L is actually the distance between the centre of disks. Length of rod is greater than L) The answer is t = 2(L/2kg) [g- acc due to gravity) But I want the solution. ? - pie v- under root | ? - pie Hmm... I like pie. Pizza, shepherd's pie, apple pie, pork pie, they are all good. Perhaps you mean pi. Androcles | v- under root I have a SHM problem. A rod of length(> L )is placed on two circular dics. The co-efficient of friction between rod and disks is k. The rod is displaced by a small distance x. Determine the time period of the oscillations. __________ O O <----disks <-------L-----> (There might be some problems in the diagram, but L is actually the distance between the centre of disks. Length of rod is greater than L) The answer is t = 2(L/2kg) [g- acc due to gravity) But I want the solution. See the correct question below I have a SHM problem. A rod of length L is placed on two circular dics. The co-efficient of friction between rod and disks is k. The rod is displaced by a small distance x. Determine the time period of the oscillations. _______________________ O O <----disks <-------------- L -----------------> The answer is t = 2??(L/2kg) [g- acc due to gravity) But I want the solution. SHM = Sado Hyper Masochism? Please explain. Or you may attempt trolling elsewhere. ;-) I have a SHM problem. A rod of length L is placed on two circular dics. The co-efficient of friction between rod and disks is k. The rod is displaced by a small distance x. Determine the time period of the oscillations. _______________________ O O <----disks <-------------- L -----------------> The answer is t = 2??(L/2kg) [g- acc due to gravity) But I want the solution. First show us your workings - or do you have no idea even how to start? BTW I always hated these problems so I am probably not the best person to answer - but if you make an attempt myself and I suspect others like David smith who is a mechanical engineer will help. Thanks Bill Dear Nishu: > I have a SHM problem. "Simple Harmonic Motion" > A rod of length L is placed on two circular dics. > The co-efficient of friction between rod and disks [quoted text clipped - 7 lines] > The answer is t = 2??(L/2kg) [g- acc due to gravity) > But I want the solution. The problem setup is lacking much. I assume the disks are acting like wheels, so there would be no SHM. Your formula gets trashed if you do not use ASCII. Is this t = 2 . pi . sqrt( L / (2 . k . g) ) ? David A. Smith Yes it is t = 2 . pi . sqrt( L / (2 . k . g) ). I have a little idea that some kind of torque will act about COM of rod by normal reaction forces by wheels. And SHM will surely take place. Let me again remember you that the L is distancebetween centre of wheels and may not be length of rod itself. N:dlzc D:aol T:com (dlzc): > Dear Nishu: > [quoted text clipped - 15 lines] > > The problem setup is lacking much. Try this figure: ================================ (o) L (o) A B > I assume the disks are acting like wheels, so there > would be no SHM. If the wheels, A and B, are *counter rotating* (driven at the same angular speed) the rod will move (slide) to and fro. > Your formula gets trashed if you do not use ASCII. > Is this t = 2 . pi . sqrt( L / (2 . k . g) ) ? Seems to be correct. > David A. Smith mL David you seem to be a genious. I forgot to add that wheels are counter rotating. Now please solve the question > David you seem to be a genious. I forgot to add that wheels are > counter rotating. Now please solve the question So you want others to do the work for you? Show us your attempt. Bill Dear Bill Hobba: >> David you seem to be a genious. I forgot to add >> that wheels are counter rotating. Now please solve >> the question > > So you want others to do the work for you? Show > us your attempt. Easy. Language barrier, Bill. If you saw the movie "The Princess Bride"... "I don't think that word means what you think it means." Inconceivable! David A. Smith > Dear Bill Hobba: > [quoted text clipped - 13 lines] > > David A. Smith Quite possibly. I will try to be a little more tolerant of it in future. BTW nice solution. Thanks Bill Dear mL: > N:dlzc D:aol T:com (dlzc): >> Dear Nishu: [quoted text clipped - 29 lines] > (driven at the same angular speed) the rod will > move (slide) to and fro. Thanks mL.. Counter rotating with A rotating clockwise, and B rotating counterclockwise. L is given as the length of the rod. The distance between A and B needs to be greater than x (for any x) and less than L. >> Your formula gets trashed if you do not use ASCII. >> Is this t = 2 . pi . sqrt( L / (2 . k . g) ) ? > > Seems to be correct. The period of oscillation is a function of wheel speed. So any result will have to be expressed in terms of that. From the result given above, it looks like it might be 1 revolution per second (2 . pi radians). Define the midpoint a, to be at L/2. Assume the member is homogeneous. Define the separation between wheel centers as d, knowing that it will likely fall out of the result... Define a mass per unit length of m_l. Look at the diagram, and note that the rod does not spin in the plane of the page. So the sum of the moments (force times distances) about each of the two points must equal zero. The normal force at B will be (given x positive to the right): The normal force at A will be: I'll fill in the blanks after work... David A. Smith It worked. Really Thanks David. You seems to be a genius in physics. Can you please tell something about yourself Dear Nishu: > It worked. Really Thanks David. No issues. > You seems to be a genius in physics. Not at all. > Can you please tell something about yourself Just some guy. The first 100+ hits on Google even with "David A. Smith" aren't me. Good luck. David A. Smith Hi David, > Dear mL: > [quoted text clipped - 32 lines] > The distance between A and B needs to be greater than x (for any > x) and less than L. To satisfy the given answer, L has to be the distance between the wheel axes (as shown in my figure). >>> Your formula gets trashed if you do not use ASCII. >>> Is this t = 2 . pi . sqrt( L / (2 . k . g) ) ? [quoted text clipped - 4 lines] > result given above, it looks like it might be 1 revolution per > second (2 . pi radians). As it turns out, the period doesn't depend on the angular wheel speed. [...] Mel Dear mL: > Hi David, > [quoted text clipped - 38 lines] > To satisfy the given answer, L has to be the distance > between the wheel axes (as shown in my figure). L is given in the problem statement as the length of the rod. The wheel spacing is inconsequential, as long as d + x < L. >>>> Your formula gets trashed if you do not use ASCII. >>>> Is this t = 2 . pi . sqrt( L / (2 . k . g) ) ? [quoted text clipped - 7 lines] > As it turns out, the period doesn't depend on the angular > wheel speed. It *must*. Lets say I set the wheels rotating at 1 revolution per day. What is the period of the oscillation? Physically, the period will be on the order of a day, and likely the rod will dampen to a stop without ever swinging to significant "negative" values of x, with those wheel speeds. Think about it like this, what are the units of sqrt( L / g ) (since 2 . k is dimensionless)? The "2 . pi" out front carries other units with it. David A. Smith Hi again, N:dlzc D:aol T:com (dlzc): > Dear mL: > [quoted text clipped - 37 lines] >> To satisfy the given answer, L has to be the distance >> between the wheel axes (as shown in my figure). . > L is given in the problem statement as the length of the rod. > The wheel spacing is inconsequential, as long as d + x < L. Nishu made a correction downthreads: "Let me again remember you that the L is distance between centre of wheels and may not be length of rod itself." >>>>> Your formula gets trashed if you do not use ASCII. >>>>> Is this t = 2 . pi . sqrt( L / (2 . k . g) ) ? [quoted text clipped - 5 lines] >> As it turns out, the period doesn't depend on the angular >> wheel speed. . > It *must*. Lets say I set the wheels rotating at 1 revolution > per day. What is the period of the oscillation? Physically, the > period will be on the order of a day, and likely the rod will > dampen to a stop without ever swinging to significant "negative" > values of x, with those wheel speeds. The rod is slipping against the wheels during the entire cycle of motion - and, the velocity of the rod varies continuously as it should do in every SHM. > Think about it like this, what are the units of sqrt( L / g ) > (since 2 . k is dimensionless)? To solve the problem ... - draw a free body diagram of the *rod* to identify the forces acting on it: weight mg, normal forces N_A and N_B, and friction forces kN_A and kN_B - set up the required equations (e.g. two force eqns and one torque eqn) - eliminate redundant quantities (m, N_A, N_B), and simplify to get the wanted eqn of motion: x" + w^2 x = 0, with w^2 = 2k g/L Thus, the period of motion is T = 2pi/w = 2pi sqrt(L/(2k g)) Mel Dear mL: > Hi again, > > N:dlzc D:aol T:com (dlzc): ... >> L is given in the problem statement as the length >> of the rod. The wheel spacing is inconsequential, [quoted text clipped - 5 lines] > distance between centre of wheels and may not > be length of rod itself." Missed it. OK. ... >> It *must*. Lets say I set the wheels rotating >> at 1 revolution per day. What is the period of [quoted text clipped - 7 lines] > entire cycle of motion - and, the velocity of the rod > varies continuously as it should do in every SHM. I accept your explanation, but I do not "believe" it. Not your problem, mine. >> Think about it like this, what are the units of >> sqrt( L / g ) (since 2 . k is dimensionless)? [quoted text clipped - 17 lines] > > T = 2pi/w = 2pi sqrt(L/(2k g)) Thanks, Mel. David A. Smith |
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