Tuesday, 12 August 2014

Teaching Mechanical Energy Conservation

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Now, I move on to the conservation of mechanical energy. Here again, I will show you how I explain physics problem solutions in terms of fundamental principles. As always, the statement of the principle will be the first line of the problem solution. The problem I have chosen to illustrate the method requires the application of both Newton's second law and mechanical energy conservation.

The notation I have been using throughout this series is described in earlier articles, specifically "Teaching Kinematics", "Teaching Newton's Second Law", and "Solving Work-Energy Problems".

Problem. A tiny box of mass M starts from rest and slides down the frictionless surface of a cylinder of radius R. Show that the box leaves the surface when the angle between the radiial line to the box and the vertical axis is th = arccos(2/3).

Analysis. The box is touching just the frictionless cylindrical surface, which exerts an outward normal force N on it. The only other force on the box is its weight MG. The box is moving along a circular path, so we apply Newton's second law in the radial direction (outward positive). With the help of a free-body diagram, we have

... Newton's Second Law

... SUM(Fr) = MAr

... -MGcos(th) + N = M(-V**2)/R.

Since the cylindrical surface can only push (it can't pull), the box cannot stay on the surface unless the normal force N is greater than zero. Consequently, the box leaves the surface at the point where N = 0. From the last equation, this corresponds to

... cos(th) = (V**2)/RG.

But this doesn't tell us much if we don't know the speed V of the box, so let's see what we can learn by applying mechanical energy conservation. We use an inertial coordinate frame with the y axis vertical and the origin at the circular center of the cylinder. We equate the box's mechanical energies at the top of the cylinder and at the point where it leaves the cylinder. The initial position of the box is Yi = R, its initial speed is Vi = 0, its final position is Yu = Rcos(th) and its final speed is Vu = V, the speed when it leaves the surface. Now with the conservation of mechanical energy,

... Conservation of Mechanical Energy

... (MVi**2)/2 + MGYi = (MVu**2)/2 + MGYu

... (M0**2)/2 + MGR = (MV**2)/2 + MGRcos(th),

so... V**2 = 2GR(1 - cos(th)).

Finally, plugging this result into the equation for cos(th) we found previously, we have

... cos(th) = 2GR(1 - cos(th))/RG = 2 - 2cos(th),

and... th = arccos(2/3).

Dr William Moebs is a retired physics professor, who taught at two Universities: Indiana-Purdue Fort Wayne and Loyola Marymount University. You can see hundreds of examples illustrating how he emphasizes fundamental principles by consulting PHYSICS HELP

To know more about technical/applied-mechanics, simply just click here https://www.aticoexport.com/technical/applied-mechanics/

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