Physics 325 HW 10 Solutions

# Physics 325 HW 10 Solutions - Physics 325 Homework 10...

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Physics 325 Homework 10 Solutions – Fall 2010 1) Taylor: Problem 7.52, p 291. Consider a mass m that hangs from a string, the other end of which is wound several times around a wheel (radius R , moment of inertia I ) mounted on a frictionless horizontal axle. Use as coordinates for the mass and the wheel the co-ordinates x , the distance fallen by the mass, and ! , the angle through which the wheel has turned (both measured from some convenient reference position). a) Write down the modified Lagrange equations for these two variables and solve them (together with the constraint equation) for !! x , the second time derivative of double dot”, and the Lagrange multiplier, " . b) Write down Newton’s second law for the mass and wheel, and use these to check your answers for !! x and “ double dot”. Show that " f / " x is the tension force on the mass. Comment on the quantity " f / " # Solution : Let us point the x-axis downwards. As the string unwinds, it is clear that x = R , so the constraint equation is f = x ! R = 0 . (1.1) The potential energy is U = ! mgx The kinetic energy of the mass and the wheel is T = 1 2 m ! x 2 + 1 2 I ! 2 The Lagrangian becomes L = 1 2 m ! x 2 + 1 2 I ! 2 + mgx and the two modified Lagrange equations, that include the Lagrange Undetermined Multiplier, ! , are ! L ! x + ! f ! x = d dt ! L ! ! x ! mg + = m !! x , and (1.2) ! L ! + ! f ! = d dt ! L ! ! ! 0 ! R = I !! . (1.3) Solving these three equations, the constraint equation and the two modified Lagrange equations, ((1.1), (1.2) and (1.3)) we find that !! x = gm m + I R 2 = g mR 2 mR 2 + I , !! = !! x R = 1 R gm m + I R 2 = g mR mR 2 + I and = " mg I R 2 m + I R 2

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If you write down Newton’s second law as applied to the mass and the wheel, you should get two equations with exactly the form of Eqs.(1.2) and Eqs.(1.3) except that ! is replaced by ! T (minus the tension in the string). Naturally these give the same answer for !! x and !! . The simplest way to identify is to compare the Lagrange equation (1.2) with the Newtonian equation to give = " T . Since the constraint function is f = x ! R " , we see that " f " x = # T , as it should ( " f " x - is a projection of the constraint force on x- axis, since tension is pointed upwards, we have this minus). On the other hand, " f
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## This note was uploaded on 10/06/2011 for the course PHYS 325 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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Physics 325 HW 10 Solutions - Physics 325 Homework 10...

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