{"id":39,"date":"2017-06-30T18:40:20","date_gmt":"2017-06-30T18:40:20","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/atd-monroecc-physics\/?post_type=chapter&p=39"},"modified":"2017-07-27T21:13:02","modified_gmt":"2017-07-27T21:13:02","slug":"conservation-laws-3","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-monroecc-physics\/chapter\/conservation-laws-3\/","title":{"raw":"Conservation Laws","rendered":"Conservation Laws"},"content":{"raw":"

Concepts and Principles<\/h2>\r\n

The Impulse-Momentum Relation<\/h3>\r\nWe\u2019ve already used the impulse-momentum relation to analyze situations involving constant forces. The relation is typically applied in its component form:\r\n\r\n\"\"\r\n\r\nHopefully it\u2019s not too much of a stretch to argue that for forces that vary in magnitude or direction the simple summation over a time interval (Dt) must be replaced by an integral over an infinitesimal time (dt):\r\n\r\n\"\"\r\n\r\nRegardless of whether the forces acting on an object are constant or not, the impulse they exert on the object is precisely equal to the change in the object\u2019s momentum.\u00a0<\/strong>\r\n\r\n \r\n

The Work-Energy Relation<\/h2>\r\nOur previous encounter with the work-energy relation resulted in:\r\n\r\n\"\"\r\n\r\nwhere the forces acting on the object of interest were constant in both magnitude and direction. Again, for forces that vary, I will generalize this result to:\r\n\r\n\"\"\r\n\r\nwhere the angle f is the angle between the instantaneous force acting on the object and the instantaneous displacement of the object (dr). Thus, this angle can change as the object moves along its path. Technically, this integral is termed a line integral<\/em> and its evaluation can be rather complicated.\r\n\r\nAlso recall from our previous discussion of work-energy that this is not<\/em> a vector equation, meaning it is not applied independently in each of the coordinate directions.\r\n

Analysis Tools<\/h2>\r\n

Applying the Impulse-Momentum Relation<\/h3>\r\nLet\u2019s re-examine the same situation we examined at the beginning of the previous chapter, a rocket launched directly upward with a time-dependent thrust.\r\n\r\n \r\n
\r\n

A 2.0 kg toy rocket is fitted with an engine that provides a thrust roughly modeled by the function F(t) = (60 N\/s) t - (15 N\/s2) t2, for 0 < t < 4.0 s, and zero thereafter. The rocket is launched directly upward.<\/p>\r\n\r\n<\/div>\r\nFor analysis, we\u2019ll apply the impulse-momentum relation between:\r\n

Event 1: The instant the rocket leaves the launch-pad.<\/p>\r\n

Event 2: The instant the thrust drops to zero.<\/p>\r\n\"\"\r\n\r\nRemember from last chapter that the rocket does not leave the launch pad until 0.36 s after the engine is ignited.\r\n\r\n\"\"\r\n\r\nWhen the engine shuts off, the rocket is traveling at 42.5 m\/s upward.\r\n\r\n \r\n\r\nWe could also apply the impulse-momentum relation between:\r\n

Event 1: The instant the thrust drops to zero.<\/p>\r\n

Event 2: The instant the rocket reaches its maximum height.<\/p>\r\nDuring this interval, the only force acting on the rocket is the force of gravity, and the impulse-momentum relation is:\r\n\r\n\"\"\r\n\r\nThus, the rocket reaches its highest altitude 8.34 s after launch.\r\n\r\nIt\u2019s important to note that when a force is a function of time, it\u2019s relatively easy to integrate the function and determine the impulse. However, it should be clear that it would not<\/em> be easy to determine the work done by a force of this type. Since work is expressed as an integral of a force with respect to a displacement (dr), the force function has to be expressed in terms of position, r. In general, it\u2019s not an easy (or sometimes possible) task to \u201cconvert\u201d a function of time into a function of position, so work-energy is not a particularly useful way to analyze systems when the forces acting are time dependent. However, if the forces depend upon the position<\/em> of the object, work-energy is a powerful analysis tool.\r\n\r\n \r\n

Applying the Work-Energy Relation<\/h2>\r\n
\r\n\r\nA 0.15 kg ball is launched vertically upward by means of a spring-loaded plunger, pulled back 8.0 cm and released. It requires a force of about 10 N to push the plunger back 8.0 cm.\r\n\r\n<\/div>\r\n\"\"\u00a0<\/em>\r\n\r\nThe Force Exerted by a SpringThe force that the spring exerts on the ball depends on the amount by which the spring is compressed. The more the spring is compressed, the larger the force it exerts on the ball. A common model<\/em> is that the force exerted by a spring is directly proportional to the amount of deformation of the spring, in this case compression. Deformation (s) is defined to be the difference between the current length of the spring (L) and the equilibrium length (L0):\r\n\r\n\"\"\r\nIf we define the positive coordinate direction to point in the same direction as positive deformation (stretch), then\r\n\r\n\"\"\r\n\r\nwith the proportionality constant, k, referred to as the spring constant<\/em>.\r\n\r\n \r\n\r\nFor the plunger, since it takes 10 N to compress the spring by 8.0 cm,\r\n\r\n\"\"\r\n\r\nFor analysis, we\u2019ll apply the work-energy relation between:\r\n

Event 1: The instant the plunger is released.<\/p>\r\n

Event 2: The instant the ball reaches its maximum height.<\/p>\r\nFor these two events, work-energy looks like this:\r\n\r\n\"\"\r\n\r\nTo do the integral, we must express Fspring in terms of the variable of integration, y. For the coordinate system chosen, s and y are identical. Also note that the force of the spring and the direction of motion of the ball point in the same direction. Thus, f = 0.\r\n\r\n\"\"\r\n\r\nThe ball reaches a maximum height of 19 cm above the top of the plunger.\r\n\r\n \r\n

Elastic Potential Energy<\/h2>\r\nWhen an object interacts with a spring, or other elastic material, a common model is that the material reacts linearly, i.e., with a force directly proportional to the deformation of the material. It is possible to calculate the work done by the linear material in general, and to rewrite the work-energy relation in such a way as to incorporate the effects of this work from the start. This is referred to as constructing a potential energy function<\/em> for the work done by the elastic material. We did exactly the same thing earlier for the force of gravity.\r\n\r\n \r\n\r\nImagine a spring of spring constant k, initially deformed by a distance si. It changes its deformation, ultimately resulting in deformation sf. To calculate the work done by the spring on the object causing the deformation:\r\n\r\n\"\"\r\n\r\nChoosing a coordinate system in which s and r are interchangeable (the origin is located at the point where the spring is at its natural length and the direction of elongation is positive) and allowing the force and the displacement to be in the same direction (f = 0) results in,\r\n\r\n\"\"\r\n\r\nThe terms, 1\/2ks2, are referred to as elastic potential energy<\/em>.\r\n\r\n \r\n\r\nInserting this result into the work-energy relation results in\r\n\r\n\"\"\r\n\r\nwith the understanding that the forces remaining in the equation, which may do work on the system, do not include the force of gravity or the force of the spring. The work done by the force of gravity and the force of the spring are already included in the relation via the inclusion of the potential energy terms.\r\n\r\n \r\n

Applying the Work-Energy Relation with Elastic Potential Energy<\/h2>\r\n
\r\n

A 0.15 kg ball is launched vertically upward by means of a spring-loaded plunger, pulled back 8.0 cm and released. It requires a force of about 10 N to push the plunger back 8.0 cm.<\/p>\r\n\r\n<\/div>\r\nThe spring constant of the plunger is known to be 125 N\/m from above. Applying the work-energy relation with potential energy terms between the instant the plunger is released and the instant the ball reaches its maximum height results in,\r\n\r\n\"\"\r\n\r\nOf course, this results in the same answer as before.\r\n\r\n \r\n\r\nApplying Newton\u2019s Second Law to a Spring-Mass System\r\n\r\n\r\n\r\n
\u00a0\"\"<\/td>\r\nSince we didn\u2019t analyze systems involving springs in the previous dynamics section, we should make up for that omission now.\r\n\r\nThe system at left consists of a spring with spring constant k attached to a block of mass m resting on a frictionless surface. The origin of the coordinate system is located at the position in which the spring is unstretched.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow imagine the block is pulled to the right and let go. Hopefully you can convince yourself that the block will oscillate back and forth. Let\u2019s apply Newton\u2019s Second Law at the instant the mass is at an arbitrary position, x. The only force acting on the mass in the x-direction is the force of the spring.\r\n\r\n\"\"\r\n\r\nBecause of our choice of coordinate system, the stretch of the spring (s) is exactly equal to the location of the block (x). Therefore,\r\n\r\n\"\"\r\n\r\nNote that when the block is at a positive position, the force of the spring is in the negative direction and when the block is at a negative position, the force of the spring is in the positive direction. Thus, the force of the spring always acts to return the block to equilibrium.\r\n\r\nRearranging gives\r\n\r\n\"\"\r\n\r\nand defining a constant, w2, as\r\n\r\n\"\"\r\n\r\n(Granted, it seems pretty silly to define k\/m as the square of a constant, but just play along. You may also find it frustrating to learn that this \u201comega\u201d is not<\/em> an angular velocity. The block doesn\u2019t even have an angular velocity!)\r\n\r\nyields,\r\n\r\n\"\"\r\n\r\nTherefore, the position function for the block must have a second time derivative equal to the product of (- w2) and itself. The only functions whose second time derivative is equal to the product of a negative constant and itself are the sine and the cosine functions. Therefore, a solution to this differential equation[1]<\/a>\r\n\r\n\"\"\r\n\r\ncan be written:\r\n\r\n\"\"\r\n\r\nor equivalently with the sine function, where A and f are arbitrary constants.[2]<\/a>\r\n