Free Sliding Bead A bead is constrained to slide along a rod of length L. The rod is rotating in a vertical plane with a constant angular speed, a), about a pivot in the middle of the rod. The pivot allows the bead to freely slide along the rod, i.e. the pivot does not impede the movement of the bead. Let r03) denote the distance of the bead away from the pivot where Ht) can be positive or negative. W {6' Equation of Motion Applying Newton's second law provides a balance of forces due to gravity, friction, centripetal acceleration, and linear acceleration. The equation resulting from these forces is d2?\" dr 2 in? + ,8 E mm 1' = mg sin(wr) where m is the mass ofthe bead, [3 is the coefficient ofviscous damping, w is the constant speed of angular rotation, g : 9.81 'm/s2 is the acceleration due to gravity, and r is the distance between the pivot and the bead. The rod is initially horizontal, and the initial conditions for the bead are r(0) : re and r'(0) : v0. \f(C) for any other velocity , the centrepetal force and the tension is not balanced. D The minimum agoized leath of rode. fam part B 70) = 0 + 9 coswt = D aut 20# Cosutt = Vo 37- wt Cosut = Ve = - Lyth is minimum when cut is minimum.O Equation of Motion m dy - myw = - mg sin ( wt ) dt im = Constant w= Constant oh ( 2 - 8w" ) = - g sin (wt ) D'r - Yar = - g sin (wt ) solution have Complementary integral and particulary integra CI O-W) = 0 m - a =0 9 m = +w wt - WE 9 =de = Ae+ Be PI = - g sin(wt ) - g sin ( wt ) smat D- WY D'a" g fin ( wt ) = ginat 2 W 2 - a tal Now = Sihat 9 = detyp - 201 4. = Ae + Be + 8 sin ( wt ) 1= Ac fnewt gsin(wt ) Now 8 (0 ) = 70 Y' ( 0 ) = UDProblem 1 Consider the frictionless rod, i.e. ,8 = 0. The equation of motion becomes d2?" 2 _ my mo) 5" = mg s1n(wt) with g = 9.81 1/11/32 and a constant angular speed to. The rod is initially horizontal, and the initial conditions for the bead are T(U) = 3'0 and r'(0) = 190. A) Analytically solve this initial value problem for r(t) B) Consider the initial position to be zero, i.e. To = 0. Find the initial velocity, :20, that results in a solution, r(t), which displays simple harmonic motion, i_.e, a solution that does not tend toward infinity. C) Explain why any initial velocity besides the one you found in part B) causes the head to fly offthe rod. D) Given r(t) displays simple harmonic motion, i.e. part B), find the minimum required length of the rod, L, as a function ofthe angular speed, to. E) Suppose m = 2, graph the solutions, t), for the initial conditions given here: r0 = 0 and initial velocities ofvc, = 2.40, 2.45, 2.50, and the initial velocity you found in part B). Use 0 E t E 5 Problem 2 Consider the frictionless rod, i.e. ,8 = 0. The equation ofmotion becomes d2?\" 2 _ new mm 5" = mg sm(wt) with g = 9.81 mfs2 and a constant angular speed to. The rod is initially horizontal, and the initial conditions for the bead are r03) = To and r'(0) = 120. You will need to write an Improved Euler Method system solver to find r(t) and v(t) A] Numerically solve for ?'(t) when co = 2, To = U, and v0 = 2.40. 2.45, 2.50. Solve in the time interval 1 1 1 t E [0.5]. Use step sizes h = E' E' E and compare your results. Also, compare your best numerical answers with your analytic answers From Problem 1 part E). B) Numerically solve for T0) when co = 2, To = U, and 190 is selected to give simple harmonic motion. i.e. l 1 1 Problem 1 part B. Use small step sizes, e.g. h = , , . etc. Solve for the longest time interval 512 2043 3192 that provides reasonable values for r(t). Compare your results to the analytic solution that gives simple harmonic motion. What does this demonstrate about numerical solutions