Shown in the figure below is a simple pendulum with a base motion given by \( X_{0} \sin (\omega t)[m] \). Let \( l=1[\mathrm{~m}], X_{0}=1[\mathrm{~m}] \), and the initial condition \( \left(\theta_{0}, 0\right) \). Note that this initial condition corresponds to moving the pendulum by an angle \( \theta_{0} \) from the vertical and then release from rest. Assume the bar is inextensible and its mass is negligible. Analyze the dynamics of the pendulum and use MATLAB to simulate the dynamics of \( \theta(t) \). (1) [4\%] Show the free-body-diagrams of the pendulum mass. (2) [20\%] Let \( \vec{a}_{B}=a_{B t} \vec{e}_{t}+a_{B n} \vec{e}_{n} \) and \( \vec{e}_{t}, \vec{e}_{n} \) are the tangential and normal basis vectors, respectively. Find \( a_{B t} \) and \( a_{B n} \) in terms of \( l \) and \( \theta \). (3) \( [18 \%] \) Derive the equation of motion of the pendulum mass \( B \) in the tangential and normal directions using the Newton's second law. (4) \( [10 \%] \) Show the first-order-form equation used in the pendulum simulation in the \( \vec{e}_{t} \) direction. Use the equation of motion in the direction to simulate the dynamics of the pendulum and answer the following two questions. (5) \( [28 \%] \) Let \( \omega=1.5 \mathrm{rad} / \mathrm{s} \). Find the maximum initial \( \theta_{0} \) so that the pendulum cannot complete one full revolution about the support \( A \) in the first 50 seconds. (6) \( [20 \%] \) Repeat (5) for \( \omega \) in \( [0.5,3.0] \) and plot the maximum initial \( \theta_{0} \) as a function of \( \omega \). Note there must at least be \( 20 \omega \) values distributed evenly in the range \( [0.5,3] \). HINT: (i) For (2), consider \( \vec{a}_{B}=\vec{a}_{B / A}+\vec{a}_{A} \). (ii) The motion of the pendulum falls on the circle of radius \( l \) and center \( A \). Once the angle \( \theta(t) \) is obtained from the numerical simulation, the position of the mass can be found by \( \vec{r}_{B}=\vec{r}_{B / A}+\vec{r}_{A} \) where \( \vec{r}_{B / A}=l \sin (\theta) \vec{\imath}-l \cos (\theta) \vec{\jmath} \). (ii) For (6), you could implement a repetition loop over the \( \omega \) range, or you could run a selected set of \( \omega \) repeatedly using the provided code. To do a loop, one can modify Lab1.m and enter the \( \omega, \theta_{0} \) values from the input line of \( \mathrm{Lab} 1 . \mathrm{m} \).
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There are two forces acting on the mass B: gravitational force (mg) acting downward and tension force (T) acting along the rod towards the pivot point A. (2) Finding \(a_{Bt}\) and \(a_{Bn}\): We know that \( \vec{a}_{B}=\vec{a}_{B/A}+\vec{a}_{A} \). The Show more…
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69. To What Angle? A small metal ball of mass $m$ hangs from a pivot by a rigid, light metal rod of length $R .$ The ball is swinging back and forth with an amplitude that remains small throughout its motion, $\theta^{\max } \leq 5^{\circ} .$ Ignore all damping. (a) The equation of motion of this ideal pendulum can be derived in a variety of ways and is $$ \frac{d^{2} \theta}{d t^{2}}=-\frac{g}{R} \sin \theta $$ For small angles, show how this can be replaced by an approximate equation of motion that can be solved more easily than the one given. (b) Write a general solution for the approximate equation of motion you obtained in (a) that works for any starting angle and angular velocity (as long as the angles stay in the range where the approximation is OK). Demonstrate that what you have written is a solution and show that at a time $t=0$ your solution can have any given starting position and velocity. (c) If the length of the rod is $0.3 \mathrm{~m}$, the mass of the ball is $0.2 \mathrm{~kg}$, and the clock is started at a time when the ball is passing through the center $(\theta=0)$ and is moving with an angular speed of $0.1 \mathrm{rad} / \mathrm{s}$, find the maximum angle your solution says the ball will reach. Can you use the approximate equation of motion for this motion? If the starting angle is not small, you cannot easily solve the equation of motion without a computer. But there are still things you can do. (d) Derive the energy conservation equation for the motion of the pendulum. (Do not use the small-amplitude approximation.) (e) If the pendulum is released from a starting angle of $\theta_{1}$, what will be the maximum speed it travels at any point on its swing?
Pendulum with Varying Length. A pendulum is formed by a mass m attached to the end of a wire that is attached to the ceiling. Assume that the length $I(t)$ of the wire varies with time in some predetermined fashion. If $\theta(t)$ is the angle in radians between the pendulum and the vertical, then the motion of the pendulum is governed for small angles by the initial value problem $$\begin{array}{l}{l^{2}(t) \theta^{\prime \prime}(t)+2 l(t) l^{\prime}(t) \theta^{\prime}(t)+g l(t) \sin (\theta(t))=0} \\ {\theta(0)=\theta_{0}, \quad \theta^{\prime}(0)=\theta_{1}}\end{array}$$ where g is the acceleration due to gravity. Assume that $$l(t)=I_{0}+l_{1} \cos (\omega t-\phi)$$ where $l_{1}$ is much smaller than $l_{0}$ l0. (This might be a model for a person on a swing, where the $pumping$ action changes the distance from the center of mass of the swing to the point where the swing is attached.) To simplify the computations, take $g=1$Using the Runge-Kutta algorithm with $h=0.1$ study the motion of the pendulum when $\theta_{0}=0.05, \theta_{1}=0, \quad l_{0}=1, l_{1}=0.1$ $\omega=1,$ and $\phi=0.02 .$ In particular, does the pendulum ever attain an angle greater in absolute value than the initial angle $\theta_{0} ?$
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A cylinder of side $L$ meters lies one quarter submerged and upright in a certain fluid. At $t=0,$ the cylinder is pushed down a distance of $L / 2$ meters and released from rest. Show that the resulting motion is simple harmonic, and determine the circular frequency and period of the motion. A simple pendulum consists of a mass, $m$ kilograms, attached to the end of a light rod of length $L$ meters, whose other end is fixed. (See Figure $8.5 .11 .)$ If we let $\theta$ radians denote the angle the rod is displaced from the vertical at time $t,$ then the component of the velocity in the direction of motion is $v=L \cdot \frac{d \theta}{d t},$ so that the component of the acceleration in the direction of motion is $L \cdot \frac{d^{2} \theta}{d t^{2}} .$ Further, the tangential component of the force is $F_{T}=-m g \sin \theta,$ so that, from Newton's second law, the equation of motion of the pendulum is $$ m L \frac{d^{2} \theta}{d t^{2}}=-m g \sin \theta $$ That is, $$ \frac{d^{2} \theta}{d t^{2}}+\frac{g}{L} \sin \theta=0 $$ This is a nonlinear differential equation. However, if we recall the Maclaurin expansion for $\sin \theta,$ namely, $$ \sin \theta=\theta-\frac{1}{3 !} \theta^{3}+\frac{1}{5 !} \theta^{5}-\cdots $$ it follows that for small oscillations, we can approximate $\sin \theta$ by $\theta .$ Then Equation $(8.5 .27)$ can be replaced to reasonable accuracy by the simple linear differential equation $$ \frac{d^{2} \theta}{d t^{2}}+\frac{g}{L} \theta=0 $$
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