Question

Question 2 The simple pendulum, discussed in week 4 and lab session 4, has the equation of motion f0/dt2_0? sin θ 9-(g/L)I/2. with The total energy of the pendulum is constant during the motion, and is given by _mgL cos θ, wher dt is the angular speed of the motion in radians per second. Consider the simple pendulum with initial conditions θ(0) and u(0)-wi, 0 i.e. starting from the vertically down position with an initial angular speed w1 (a) Use the energy equation to show analytically that the pendulum will overshoot θ around) if wi>2(g/L)172 π (i.e. whirl (b) Suppose that friction (damping) is present and is proportional to w, so that the equation of motion is where k is the coefficient of friction. By introducing the non-dimensional time t ts -2T(L/g)1/2, show analytically that the equation of motion can be re-written t/ts, where (10) where K tsk, and where a dot denotes differentiation with respect to t. (c) Write a code to solve Eq. (10) using fourth order Runge-Kutta. For k 1.0, determine how many complete revolutions the damped pendulum makes when: i) wi 4(g/L)1/2; (ii) wi - 10(9/L)1/2.

I need help with C

Answer 1

Given data the equation of motion is d^2 theta/dt^2 = -Ohm^2 sin theta with Ohm = (g/L)^1/2 Total energy of the pendulum is constant during motion Energy at starting point when w = w is given by E = mL^2 w^2/2 - mgL consider intial condition as theta (0) = theta w (0) = w_1 Therefore theta = 0, cos theta = 1 To over shoot at theta = pi at the top point The pendulum should over tone potential energy is should have same Kinetic energy Therefore Therefore total energy (initially) should be Greater than potential energy at the top E_initial > potential energy at top > mgL Approch 2 E_kinetic (at the top) > 0 E (pi) = E (0) - 2 mgL \ r \ n = m L^2 w^2/2 - 2 mgL > 0 = m L^2 w^2/2 > 2 mgL Rightarrow w_i > 2 squareroot theta/2 According