Wednesday, April 6, 2011

Teaching of Conservation of angular momentum

In this article, I show how easily the problems of physics are resolved using conservation of angular momentum. From only with an explicit declaration of the conservation of angular momentum allows us to solve difficult problems apparently quite easily. As always, I use the problem solutions to demonstrate my approach.

Once again, limited text editor force capabilities I use some unusual rating. This notation is now summarized in one spot, section "teaching dynamic rotation.

Problem. (Not illustrated) sketch shows a boy of mass m standing at the edge of a cylindrical platform of mass M, moment of inertia Ip and RADIUS R = (M. * 2) / 2. The platform is free to rotate without friction around its central axis. The platform is rotating angular speed us when the boy starts edge (e) platform and walk towards the Centre. What is the platform angular velocity when the boy reaches mid-point (m), distance R/2 from the Centre of the platform? What is the angular velocity when it reaches the Centre (c) platform?

Analysis. (a) we consider rotation around the vertical axis through the Centre of the platform. With boy a distance r from the axis of rotation, moment of inertia of the disk in the boy's I = Ip + m. * 2. Since there is no net torque system around the central axis, angular axis is preserved. First, we calculate moment of inertia of the system at three points of interest:

...................................... EDGE...IE = (M. * 2) / 2 + M. * 2 = ((M_+_m_2) R * 2) / 2

...................................... MIDDLE...IM = (M. * 2) / 2 + m (R/2) * 2 = ((M_+_m/2) R * 2) / 2

.......................................CENTRE...CI = (M. * 2) / 2 + m (0) * 2 = (M. * 2) / 2

Equating angular three-point, we have

.................................................Conservation of angular momentum

..........................................................IeWe = ImWm = bridging

...................................((M_+_m_2) R * 2)US / 2 = ((M_+_m/2) R * 2) Wm/2 = (M. * 2) Wc/2.

These latest equations are easily solved for Wm and Wc in terms of we:

.....................................WM = ((M_+_m_2) / (M + m2)) we, Wc = ((M_+_2_m) M) we.

Problem. (Not illustrated) sketch shows a uniform rod (Ir = Ml?/12) of the mass M = 250 g and length l = 120 cm. The stem is free to turn in a horizontal plane a fixed vertical axis from its Centre. Two pearls of small, each mass m = 25 g are free to move in grooves on the stem. Initially, the stem is rotating at angular speed Wi = 10 rad/sec with maintained beads in place on the sides of the Center by latches located opposite d = 10 cm in the axis of rotation. When locks are released, the pearls slide off unto the ends of the stem. What is the angular velocity Wu stem when beads reach the ends of the stem from? (b) suppose beads to join the ends of the stem and are not be stopped, so they slip off the coast of the stem. What, then, is the angular velocity of the stem?

Analysis. The forces on the system are all vertical and exert no torque about the axis of rotation. As a result, the angular momentum about the vertical rotation axis is preserved. our system is the stem (I = (Ml * 2) / 12) and two beads. We have vertical axis

..............................................Conservation of angular momentum

......................................((Rod) L + L (beads)) I = ((rod) L + L (beads)) u

............................((Ml**2)/12 + take * 2)Wi = ((Ml**2)/12 + 2 m (l/2) * 2) Wu

so.................................Wu = (Ml * 2 + 24md * 2) Wi / (Ml * 2 + 6 ml * 2)

With the given values for the different quantities inserted in this last equation, we find that

...........................................................Wu = 6.4 rad/sec.

(b) it is always 6.4 rad/sec. When Pearl slide out sticks, they carry their speed and therefore their angular with them.

Again, we see the advantage of from each physical problem solution consultant is a fundamental principle in this case, the conservation of angular momentum. Two apparently difficult problems are easily solved with this approach.

Dr. William Moebs is physical retired professor who has taught at both universities: Indiana - Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how stressed the fundamental principles by visiting physical support.

Article Source: http://EzineArticles.com/?expert=William_Moebs

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