---------------------- Show that if a bar magnet suspended in a uniform magnetic
ID: 2101535 • Letter: #
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Show that if a bar magnet suspended in a uniform magnetic field B is displaced slightly from its equilibrium position, it will oscillate with a period T = 2 pi where mu is the dipole moment of the magnet and I is the moment of inertia of the bar. I can be determined by weighing and measuring the bar and realizing that I = m/12 (l2 + a2) where l and a are the length and width of the bar and m is its mass. HINT: This problem is the rotational analog of the simple pendulum covered in your text in which Newton's second law for rotational motion t = alpha = I d2theta/dt2 is used to find the period of the simple harmonic motion. Since the torque on a magnetic dipole is given by t = mu Times B the magnitude of the torque is t = -mu Bsin theta = -mu B theta where we have used the small angle approximation. Note that this is a restoring torque. In our experiment, the magnet will be suspended from its midpoint by a thin wire so that only the horizontal component of the earth's magnetic field BH will cause a torque. Thus we will determine the product mu BH.Explanation / Answer
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 )
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