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### Exercise 1: Dependence of the period of small oscillations on scaled characteristics of the bars A rigid, three-bar pendulum can be made by fastening together a single bar of mass $m_1$ and length $l_1$ to two identical bars of mass $m_2$ and length $l_2$ as shown below. `  For the undamped, rigid three-bar pendulum, it is possible to show that the period of small oscillations is given by $$ T=2\pi \left[ \frac{ \frac{1}{3} m_1 l_1^2+2\left[\frac{1}{12} m_2 l_2^2+m_2 \left(l_1-s+l_2/2\right)^2 \right]} { (m_1+2m_2)g \frac{m_1 l_1/2+2m_2 (l_1-s+l_2/2)}{m_1+2m_2} } \right]^{1/2} $$ Show that this is equivalent to the expression $$ T=2\pi \left[ \frac{l_1}{g} \frac{ \frac{1}{3}+2\tilde{m}_2 \left[\frac{1}{12}\tilde{l}_2^2+\left(1-\tilde{s}+\tilde{l}_2/2\right)^2 \right]} {\frac{1}{2}+2\tilde{m}_2 (1-\tilde{s}+\tilde{l}_2/2)} \right]^{1/2} $$ where the "scaled variables" $\tilde{m}_2$, $\tilde{l}_2$, and $\tilde{s}$ are defined as $\tilde{m}_2 \equiv m_2/m_1$, $\tilde{l}_2 \equiv l_2/l_1$, and $\tilde{s} \equiv s/l_1$. Assume $l_1=1.0$ m and $\tilde{l}_2=1.0$ to generate a contour plot of the period $T$ versus $\tilde{m}_2$ on the vertical axis and $\tilde{s}$ on the horizontal axis, with $0.5 \leq \tilde{m}_2 \leq 2.0$ and $0.0 \leq \tilde{s} \leq 1.0$. Does the value of the period change as you would expect as $\tilde{m}_2$ and $\tilde{s}$ are increased? ### Exercise 2: Dependence of the period of small oscillations when the bars are identical If the bars comprising the three-bar pendulum are all identical ($\tilde{m}_2 = 1$), show that the result of Exercise 1 simplifies to $$ T=2\pi\left[ \frac{l_1}{g} \frac{1+(3-2\tilde{s})^2}{1+2(3-2\tilde{s})} \right]^{1/2} $$ Plot the period $T$ versus the scaled overlap $\tilde{s}$ for $0 \leq \tilde{s} \leq 1$ when $l_1=1.0$ m. What is the value of the period when $\tilde{s}=0$ (no overlap)? When $\tilde{s}=1$ (total overlap)? Would it be difficult to experimentally determine the difference between these values with a stopwatch? ### Exercise 3: Equation of motion of the damped three-bar pendulum If the damping of the pendulum is modeled as a resistive torque represented by $-R|v_{cm}|v_{cm}$, where $v_{cm}$ is the velocity of the center of mass of the pendulum and $R$ is a constant, show that the equation of motion of the damped pendulum is $$ \ddot{\theta} = -\frac{mgl_{cm}}{I} \sin\theta - \frac{Rl_{cm}^2}{I} \left|\dot{\theta}\right| \dot{\theta} $$ where $\theta$ is the angular position of the center of mass of the physical pendulum measured counterclockwise from the downward vertical direction, $m$ is the total mass of the bars, $g$ is the gravitational acceleration, $l_{cm}$ is the distance from the rotation axis to the center of mass of the physical pendulum, $I$ is the rotational inertia of the physical pendulum with respect to the rotation axis, $R$ is the proportionality constant for the damping, and dots represent differentiation with respect to time. Furthermore, show that, if we define the dimensionless time $\tilde t \equiv t/T$ where $$ T=2\pi\left(\frac{I}{mgl_{cm}}\right)^{1/2} $$ then the equation of motion assumes the dimensionless form, $$\frac{d^2 \theta}{d\tilde{t}^2} = -4\pi^2\sin\theta - \frac{R l_{cm}^2}{I} \left| \frac{d\theta}{d\tilde{t}} \right| \frac{d\theta}{d\tilde{t}}$$ ### Exercise 4: Solve the equation of motion of the damped three-bar pendulum (a) Solve the equation of motion from Exercise 3 using the half-step approximation algorithm if the pendulum is released from rest at $t = 0$ s. Plot the resulting solution in the case that the bars are all identical, have a scaled overlap of $\tilde{s}=0.5$, experience no damping ($R = 0$), and the pendulum is released from rest at an angle of $\theta_0 = 45^\circ$ measured from the vertical. (b) Repeat part (a), but for a scaled overlap of $\tilde{s}=0.25$. (c) Repeat part (a), but with non-zero damping given by $R=0.1$ kg. Check that the oscillations diminish in amplitude with time (or else something is wrong with the code!). ### Exercise 5: Accuracy of 1D Model Angular Position vs. Time Because the computational solution is based on an approximation, it is important to determine just how small $\Delta t$ should be for the approximation to accurately predict the angular position. You can do this by comparing the computational solution to the exact solution when the oscillations are small and damping is negligible. In this case, the equation of motion reduces to $\ddot{\theta}=-\frac{mgl_{cm}}{I}\theta$. If the pendulum is released from rest from $\theta = \theta_0$ at $t = 0$ s, this equation has the analytical solution $$ \theta(t)=\theta_0 \cos(2\pi t/T) $$ where the period $T$ is given by $$ T=2\pi\left(\frac{I}{mgl_{cm}}\right)^{1/2} $$ Compare the time dependence of the angular position predicted by the computational model to that predicted by the analytical solution for small oscillations when the undamped pendulum consisting of three identical bars is released from rest at an angle of $\theta_0 = 5^\circ$ measured from the vertical, and the scaled overlap is $\tilde{s}=0.5$.What value of $\Delta t$ do you deem to be sufficiently small for the computational model to be accurate? Explain how you arrived at this value of $\Delta t$ .