Project 2: A bead sliding along a rod Simple Harmonic Motion Free Sliding Bead A is constrained to slide along a rod of length . The rod is rotating in a vertical plane beadwith a constant angular speed, , 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 denote the distance of the bead away from the pivot where can be positive or negative. 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 where is the mass of the bead, is the coefficient of viscous damping, is the constant speed of angular rotation, is the acceleration due to gravity, and is the distance between the pivot and the bead. The rod is initially horizontal, and the initial conditions for the bead are and . Problem 1 Consider the frictionless rod, i.e. . The equation of motion becomes with and a constant angular speed . The rod is initially horizontal, and the initial conditions for the bead are and . A) Analytically solve this initial value problem for B) Consider the initial position to be zero, i.e. . Find the initial velocity, , that results in a solution, , 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 bead to fly off the rod. D) Given displays simple harmonic motion, i.e. part B), find the minimum required length of the rod, , as a function of the angular speed, . E) Suppose , graph the solutions, , for the initial conditions given here: and initial velocities of , and the initial velocity you found in part B). Use Problem 2 Consider the frictionless rod, i.e. . The equation of motion becomes with and a constant angular speed . The rod is initially horizontal, and the initial conditions for the bead are and . You will need to write an Improved Euler Method system solver to find and A) Numerically solve for when , , and . Solve in the time interval . Use step sizes and compare your results. Also, compare your best numerical answers with your analytic answers from Problem 1 part E). B) Numerically solve for when , , and is selected to give simple harmonic motion, i.e. Problem 1 part B. Use small step sizes, e.g. etc. Solve for the longest time interval that provides reasonable values for . Compare your results to the analytic solution that gives simple harmonic motion. What does this demonstrate about numerical solutions? Problem 1. md 2 r mr w 2=mgsin( wt ) 2 dt For natural response: d2r =r w 2 2 dt r ( t ) =A e wt + B ewt For forced response: r ( t ) =Csin ( wt ) Substituting this in equation, we get: ( C= g 1 w2 ) ( wg 1) sin ( wt ) r (t )= Therefore, 2 Total response = forced response + natural response Therefore, r ( t ) =A e wt + B ewt + (A) ( g 1 sin ( wt) 2 w ) (B) r ( 0 )= A+ B=0 e g 1) sin ( wt ) 2 w r ( t )= A ( wt ewt )+( For this to be simple harmonic motion, A should be equal to 0. ( wg 1) sin ( wt ) r (t )= v ( t )= 2 dr g g = 2 1 cos ( wt )= v ( 0 ) = 2 1 dt w w ( ) ( ) (C) If there is any other initial velocity, then the motion would be no longer simple harmonic and there would be exponential terms in the equation, therefore, at some point of time the bead will fly off the rod. Maximum value of r ( t ) is at wt= (D) in part (B) minimum length of the rod=L=2rmax=2 pi g = rmax= 2 1 2 w ( wg 1) 2 (E) For w = 2, v(0) in part(B) = (g/4 -1) = 1.4525 m/s e g ( wt + ewt )+ w cos ( wt ) w v (t ) = A w ( ) ( wg w )= v ( 0) at w=2=4 A+( g2 2) v ( 0 ) =2 A w + v ( 0 ) =4 A+ 2.905 Using this equation: v(0) 1.4525 2.4 2.45 2.5 Plots of r(t) for above values of A and w = 2. A 0 -0.12625 -0.11375 -0.10125 ( ) Problem 1. md 2 r mr w 2=mgsin( wt ) 2 dt For natural response: d2r =r w 2 2 dt r ( t ) =A e wt + B ewt For forced response: r ( t ) =Csin ( wt ) Substituting this in equation, we get: ( C= g 1 w2 ) ( wg 1) sin ( wt ) r (t )= Therefore, 2 Total response = forced response + natural response Therefore, r ( t ) =A e wt + B ewt + (A) ( g 1 sin ( wt) 2 w ) (B) r ( 0 )= A+ B=0 e g 1) sin ( wt ) 2 w r ( t )= A ( wt ewt )+( For this to be simple harmonic motion, A should be equal to 0. ( wg 1) sin ( wt ) r (t )= v ( t )= 2 dr g g = 2 1 w cos ( wt )= v ( 0 )= 2 1 w dt w w ( ) ( ) (C) If there is any other initial velocity, then the motion would be no longer simple harmonic and there would be exponential terms in the equation, therefore, at some point of time the bead will fly off the rod. Maximum value of r ( t ) is at wt= (D) in part (B) minimum length of the rod=L=2rmax=2 pi g = rmax= 2 1 2 w ( wg 1) 2 (E) For w = 2, v(0) in part(B) = (g/2 -2) = 2.905 m/s e g ( wt + ewt )+ w cos ( wt ) w v (t ) = A w ( ) ( wg w )= v ( 0) at w=2=4 A+( g2 2) v ( 0 ) =2 A w + v ( 0 ) =4 A+ 2.905 Using this equation: v(0) 2.905 2.4 2.45 2.5 Plots of r(t) for above values of A and w = 2. A 0 -0.12625 -0.11375 -0.10125 ( ) Problem 1. md 2 r mr w 2=mgsin( wt ) 2 dt For natural response: d2r =r w 2 2 dt r ( t ) =A e wt + B ewt For forced response: r ( t ) =Csin ( wt ) Substituting this in equation, we get: g 2 w2 ( ) C= ( 2gw ) sin (wt ) r (t )= Therefore, 2 Total response = forced response + natural response Therefore, r ( t ) =A e wt + B ewt + (A) g sin ( wt ) 2 2w ( ) (B) r ( 0 )= A+ B=0 e g )sin ( wt) 2 2w r ( t )= A ( wt ewt )+( For this to be simple harmonic motion, A should be equal to 0. ( 2gw ) sin (wt ) r (t )= v ( t )= 2 dr g g = w cos ( wt )= v ( 0 )= 2 dt 2w 2w ( ) ( ) (C) If there is any other initial velocity, then the motion would be no longer simple harmonic and there would be exponential terms in the equation, therefore, at some point of time the bead will fly off the rod. Maximum value of r ( t ) is at wt= (D) in part (B) minimum length of the rod=L=2rmax= pi g = rmax= 2 2 w2 ( wg ) 2 (E) For w = 2, v(0) in part(B) = (g/4) = 2.4525 m/s e ( 2gw )cos ( wt ) ( wt + ewt )+ v ( t ) =A w ( 2gw )= v ( 0) at w=2=4 A + g4 v ( 0 ) =2 A w + v ( 0 ) =4 A+ 2.4525 Using this equation: v(0) 2.4525 2.4 2.45 2.5 Plots of r(t) for above values of A and w = 2. A 0 -0.013125 -0.000625 0.011875 ( ) A) V0 = 2.4 m/s : 1. h = 1/32 2. h = 1/128 3. h = 1/512 V0 = 2.45 1) h = 1/32 2) h = 1/128 3) 1/512 V0 = 2.5 1) h = 1/32 2) h = 1/128 3) h = 1/512 syms y(t); span = 0:1/8192:5; %for different step size just change the span v0 = 2.4525; % for different initial values of v0 change this V = odeToVectorField(diff(y,2) == (4*y - 9.81*sin(2*t))); M = matlabFunction(V,'vars', {'t','Y'}); sol = ode45(M,span,[0 v0]); fplot(@(x)deval(sol,x,1), [0, 5]) The blue graph comes from numerical solution. The red graph comes from analytical solution with step size of 1/8192 which gives the best solution. The comparison shows that after some time, the numerical solution deviates from analytical solution. 2. A *NOTE* Blue graph is numerical solution and red graph is analytical solution. Comparison: a) V0 = 2.4 m/s and step size = 1/512 b) V0 = 2.45 m/s and step size = 1/512 c) V0 = 2.5 m/s and step size = 1/512 The comparison for this shows that the graphs almost match, but they tend to deviate from each other at higher values of t. Green: v0 = 2.5 Blue: v0 = 2.4 Yellow: v0 = 2.45 Red: v0 = 2.4525 \fSolve the problem analytically and program an... My Questions n Solve the problem analytically nismesmisanmred. Tip: Always try to answer questions quickly. Work submitted early oould result in a bonus! f: Subiect: Differential Equations 0 Price: $18.00 @ Deadline:1 week a Assigned to: You a mission2win posted a question May 2?, 201? at 1152pm is). Solve the problem analytically and program an Improved Euler's Method numerical solver. It is required to write the numerical methods in either . Online MATLAB or MS Exoel. Please be as detailed as you can with the explanation. Thank you in advance! Download Attachment: Projectzpptx immanuelsoi suggested a price change May 28, 2017 at 1:1?am W" Price suggested by a Tutor. \fsyms y(t); span = 0:1/8192:5; %for different step size just change the span v0 = 2.4525; % for different initial values of v0 change this V = odeToVectorField(diff(y,2) == (4*y - 9.81*sin(2*t))); M = matlabFunction(V,'vars', {'t','Y'}); sol = ode45(M,span,[0 v0]); fplot(@(x)deval(sol,x,1), [0, 5]) The blue graph comes from numerical solution. The red graph comes from analytical solution with step size of 1/8192 which gives the best solution. The comparison shows that after some time, the numerical solution deviates from analytical solution. 2. A *NOTE* Blue graph is numerical solution and red graph is analytical solution. Comparison: a) V0 = 2.4 m/s and step size = 1/512 b) V0 = 2.45 m/s and step size = 1/512 c) V0 = 2.5 m/s and step size = 1/512 The comparison for this shows that the graphs almost match, but they tend to deviate from each other at higher values of t. Green: v0 = 2.5 Blue: v0 = 2.4 Yellow: v0 = 2.45 Red: v0 = 2.4525 \fSolve the problem analytically and program an... My Questions n Solve the problem analytically nismesmisanmred. Tip: Always try to answer questions quickly. Work submitted early oould result in a bonus! f: Subiect: Differential Equations 0 Price: $18.00 @ Deadline:1 week a Assigned to: You a mission2win posted a question May 2?, 201? at 1152pm is). Solve the problem analytically and program an Improved Euler's Method numerical solver. It is required to write the numerical methods in either . Online MATLAB or MS Exoel. Please be as detailed as you can with the explanation. Thank you in advance! Download Attachment: Projectzpptx immanuelsoi suggested a price change May 28, 2017 at 1:1?am W" Price suggested by a Tutor. \f