Teaching dynamic rotation

Then, I describe how I teach problem solutions for dynamic rotation in terms of basic principles. As always, the statement of principles will be the first line of the solution to the problem. I chose two problems to illustrate my approach. They will require the application of the second law of Newton, angular velocity versus the angular position and the equation of motion of rotation.

In previous articles, I described the unusual scoring, I used. It is necessary due to the limitations of the text editor. Notation is dispersed on four different articles, which makes accessing very annoying. To overcome this difficulty, I summarize this notation here:

1. Most of the kinematic variables are represented by letters. For example, v is represented by the speed and acceleration by a.

2 Indices are designated using lowercase letters. For example, initial speed V-naught"is Vo.

3 Powers are represented by *. For example, "time squared" is T * 2.

4. An amount is represented by the sum (the conditions to be summarized).

5 Ft is represented by FT.

6 Static and kinetic friction coefficients are represented by the United Kingdom and United States, respectively.

7 Angles are represented with a script italic examples, I represent "Theta" by th and "phi" by phi.

8. Rotational quantities are represented with a script in italics. Specifically, the angular position is represented by Q, angular speed W, angular acceleration by A net external forces by text, torque moment of inertia by I.

Examine problems and their solutions.

Problem. Vertical grinder (I = 15 kgm * 2) has a RADIUS R = 80 cm. The wheel turns 80 rev/min when a friction brake is pushed against the rim. The wheel and then stops after five additional revolutions. The coefficient of kinetic friction between wheel and brake pad is Uk = 0.70. What is the normal force n applied to the brake?

Analysis. We have the angular velocity at the beginning and the end of five revolutions. Therefore, if we relate speed angular angular position, we should be able to determine the angular acceleration. Then, we can use the rotation of the motion equation to determine the force of friction and the normal force.

The wheel is supported to the axis of rotation (Group B) bearings, thrust by brake (normal force n and friction force f = UkN), and the acts of its weight to its centre of mass (which is also in the axis of rotation). Only the friction force exercises a couple around the central axis of rotation of the wheel.

Firstly, we must obtain 80.0 rev/min and 5 revolutions in appropriate units:

............................80 rev/min = (80 rev/min) (are rad/rev) (1 min / 60 s) = 8.38 rad/s.

... .rev 5 = (rev 5) (are rad/rev) = 31.4 rad.

Now, we can apply the basic principles:

....Angular velocity Versus the angular Position...Rotation equation of motion

..........W**2 = Wo**2 + 2A(Q - Qo)...........................................Sum (Text) = IA

..........0**2 = Wo**2 + 2A(Q - Qo).............................................fR + B(0) + N(0) = IA

..........0 * 2 = (8.38 rad/s) * 2 + 2 A (31.4 rad)... UkNR = IA

therefore...A = - 1.12 rad/s * 2.

With this a Uk = 0.70 with I = 15 kgm value inserted in the rotation of the motion equation, we have

N = - IA / UkR (kgm 15 * 2) = (-1.12 rad/s**2)/(0.70)(0.80 m) = 30 n.

Problem. A pulley radius r and moment of inertia is free to rotate without friction around a horizontal axis through the Centre of the pulley. A string is wrapped around the pulley and the free end of the chain is linked to a mass M. cylinder when the bottle was released, falls and turns the pulley. What are the cylinder acceleration and angular acceleration of the pulley? What is the chain tension?

Analysis. We can apply Newton's second law to the bottle and the equation of rotation of the movement to the pulley. The angular acceleration of the pulley and linear acceleration of the cylinder are bound by a = AR.

The cylinder is affecting only the string (strain Tupward). the weight of the cylinder is MG. The pulley is affecting string (strain Tdownward) and support its central axis (B). Pulley weight acts to the axis of rotation of the pulley. Only the voltage (point R arm) exerts a couple around the central axis of the pulley.

Using free body for two-body diagrams, we apply the second law of Newton (with the positive direction downwards) to the bottle and the equation of rotation movement pulley (with positive clockwise):

...............................The second law of Newton...Rotation equation of motion

..................................SUM(Fy) = MAy....................................Sum (Text) = IA

...................................MG - T = McAy....................................TR = IA

After combining A = AR, and these two equations find us, after a bit of arithmetic, which

..................A = MGR * 2 / (M. * 2 + I)...T = GIG / (M. * 2 + I)...A = MGR / (M. * 2 + I)

Again, we see physical problems solved by consulting basic principles. In short, this is my message. Teach students to resolve Physics problems starting with the basics.

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 fundamental principles consulting

USING PHYSICS.

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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.

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