Physics 325 Discussion 7 Solution3/9/2015Problem 1The dynamics of the given problem is done using the rotational analogue of Newton’s second law:Xτ = Iα (1.1)This requires summing the torques on the mass M (due to both gravity and the pseudoforce). Note that themoment of inertia, I, for a point mass is mr2, and the term α is the angular acceleration, which can also beexpressed as¨θ. We can calculate the torques as:Xτ = −MgL sin θ − M ¨yL cos θ (1.2)where the torques are calculated as r ×F = rF sin φ, where here φ is the angle between r and F. Care shouldbe taken in defining the sign of this cross product based on the right hand rule from r to F. For example,for the pseudotorque, the angle φ =π2− θ, however, the right hand rule dictates the sign be negative, andso we have −sin(π2− θ) = −cos θ. Using this with Equation (1.1) gives:−MgL sin θ − M ¨yL cos θ = ML2¨θ (1.3)To linear order in θ, sin θ ≈ θ and cos θ ≈ 1, such that the above equation can be written:M¨θ +MgLθ = −ML¨y (1.4)For an oscillating ceiling, y(t) = Y0exp(iωt), studying the steady state response θ(t) = Θ0exp(iωt), thisdifferential equation can be evaluated as:−ω2Θ0eiωt+gLΘ0eiωt=ω2LY0eiωt(1.5)1 −g/Lω2Θ0= −Y0L(1.6)to first order, we can ignore the small term, such that Θ0= −Y0L. Thus:θ(t) = −Y0Leiωt= −y(t)L(1.7)Note that given this, in the lab frame, the back and forth acceleration of the ceiling is canceled by the motionof the pendulum in the frame of the ceiling (the horizontal displacement of the pendulum in its frame isL sin θ(t) ≈ Lθ(t) = −y(t)). Thus the pendulum will only bob up and down.1Problem 2Setting x(t) = D exp(−st) in the given differential equation, we get:ms2De−st+ kDe−st= F0e−st(2.1)⇒ D =F0ms2+ k(2.2)Then constructing the full solution using the homogeneous solution, the expression for x(t) is given as:x(t) = De−st+ A cos ωt + B sin ωt (2.3)Solving for the coefficients A and B such that x(t) is quiescent at t = 0 (i.e. x(0) = 0 and ˙x(0) = 0):0 = De0+ A cos(0) + B sin(0) = D + A ⇒ A = −D (2.4)0 = −sDe0− ωA sin(0) + ωB cos(0) = −sD + ωB ⇒ B =sDω(2.5)giving the full solution for x(t) given the initial conditions as:x(t) = De−st− cos ωt +sωsin ωt(2.6)At long times, t s−1, the particular solution dies off and we are left with just the homogeneous solution:x(t s−1) = D−cos ωt +sωsin ωt(2.7)which can be written as a simple oscillation as:x(t s−1) =D√s2+ ω2ωsinωt −
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