Period of Small OscillationDate: 08/21/2003 at 16:45:00 From: Amelia Subject: Period of small oscillation Hello! I'm desperately stuck on this question...I started working it out as if it were a simple pendulum question but then realised that this is not correct since the trapdoor has a mass as opposed to the string of a pendulum. Anyway, the question is as follows. A square trapdoor, modelled by a uniform rigid lamina of side b and mass 5m, is smoothly hinged at one horizontal edge and has a catch of mass m in the centre of the opposite edge. Find the trapdoor's period of small oscillation about downward hanging equilibrium. (Ignore air resistance and take gravitational acceleration g to be constant.) Since I have only studied simple pendulums in depth, I am unsure as to where to start. Cheers for any suggestions! Amelia Date: 08/22/2003 at 01:17:41 From: Doctor Warren Subject: Re: Period of small oscillation Hi Amelia, Thanks for writing to Dr. Math. You're dealing with an example of what are known as "physical pendula," which are a generalization of the simple pendula you've studied. The first thing to think about is how the force of gravity affects a rigid mass, like this trapdoor. Think first about holding this trapdoor in your hands, disconnected from its hinge, and trying to balance it - say, on your head. (It sounds silly, I know, but stick with me.) Where would you have to put it on your head to make it balance? You'd have to rest its center of mass on your head, of course. If you think about this a bit, it'll help you with your physical intuition. If you can apply an upward force at the trapdoor at exactly one point and balance it (i.e. make it stop moving), it means that gravity acts on the trapdoor as if all of its mass is concentrated at its center of mass. The same thing applies to any mass - gravity acts on the center of mass. In the case of a physical pendulum, you have a mass suspended somehow, and gravity acts on its center of mass. When the center of mass is directly beneath the point of suspension, the pendulum is in equilibrium, and will not move. When the center of mass is not directly under the point of suspension, gravity pulls on the center of mass and exerts a torque around the point of suspension, causing the pendulum to swing. Let's make this precise. Torque is defined by: tau = R x F = r f sin(theta) where R and F are the position vector (from the point of suspension to the location where the force is applied) and the force vector, respectively; r and f are the magnitudes of R and F, respectively; and theta is the angle between R and F. r is the distance between the point of suspension (the hinge) and the center of mass. f, the magnitude of the force, is just Mg, where M is the mass of the pendulum (M = 6m). Let's draw a picture now, showing these quantities. Because the pendulum is suspended all along one edge and is symmetrical about its middle, we can exploit the symmetry to treat the problem as being two dimensional. We'll just visualize the pendulum edge-on: * \ \ \ R, ||R|| = r \ c |\ | \ F |o \ ||F|| = Mg | @ v In this diagram, r is the distance from the hinge (*) to the center of mass (c). o is theta, the angle between R and F. @ is the catch on the edge opposite the suspended edge. Now we have an expression for the torque of a physical pendulum about its suspension point: tau = r mg sin(theta) This torque is a "restoring torque" that always tries to push theta to zero. Since your problem specifies that you're only concerned about small oscillations, we can take the small-angle approximation: sin(theta) approximately equals theta for small theta. tau = r mg theta This equation is essentially Hooke's law, as you'd use for a spring-and-mass oscillator, in angular form. F = k x becomes tau = k theta because torque is the angular analogue of force, and theta is the angular analogue of displacement. You can see immediately that k, the proportionality constant between the angular displacement and the experienced torque, is k = mgr Now, for any simple harmonic oscillator, the angular frequency is given by: w = sqrt( k / m ) and the period by: T = (2 pi / w) = 2 pi sqrt( m / k ) where m is the mass (intertia) of the system. The angular analogue of mass is rotational inertia, I. Thus, in terms of angular quantites, T = 2 pi sqrt( I / k ) or T = 2 pi sqrt( I / Mgr ) This is the final relation that relates quantities concerning the construction of the physical pendulum (its rotational inertia about its point of suspension, its total mass, and the location of its center of mass) to its period of oscillation. To finish this problem, only two things are thus needed: 1) r, the distance between the suspension point and the center of mass, which requires us to calculate the center of mass. 2) the rotational inertia of the trapdoor about its center of mass. Let's see about finding these quantities. 1) We're only concerned with one coordinate of the center of mass - its distance from the point of suspension. When finding the center of mass of a composite object, you can treat each of the components' mass as being concentrated at its center of mass. The lamina of the trapdoor has a center of mass a distance b/2 from the point of suspension by symmetry, so its contribution to the center of mass is (5m)(b/2). The catch on the opposite edge contributes (m)(b), so the total center of mass is: r = (1/6m) * (5mb/2 + mb) = (7/12) b m^2 2) The rotational inertia about the point of suspension is a very similar calculation. While the center of mass is r = (m1 x1 + m2 x2) / (m1 + m2), the rotational inertia is I = (m1 x1^2 + m2 x2^2) = (5m (b/2)^2 + m b^2) = (9/4) b^2 m Substituting these results into the expression for T indicated above will give you the desired period of oscillation. Let me know if you have any more questions. - Doctor Warren, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/