EGR 232 Engineering Dynamics Fall 2012 Lecture 16: Conservation of Energy Today: Homework Due Conservation of Energy Read Chap 13, Sections 6 to 9 Homework: Chap. 13, Problems 57, 61, 67, 76

Lecture Outcomes:Following today's lecture you should be able to -- explain what potential energy -- explain conservative and nonconservative forces -- explain the conservation law of energy -- Use law of conservation of energy to determine particle motion.

Work: The product of a force acting through a parallel distance. Doing work on a system changes the energy of the system.

Kinetic Energy: Energy of motion. Bodies which have mass and velocity also have energy.

Potential Energy: Energy of position. When work is done on a body by a conservative force, there is an energy (or potential to do work) which the body has because of its position.Gravitational Potential Energy (near the earth's surface): where m is the mass, g the acceleration of gravity, and y is the height with respect to the reference height at which the energy is defined as zero.

Gravitational Potential Energy (general formula):

where G is the Universal constant of Gravitation equal to 66.73x10-12 m3/(kg-s2) or 3.4.4x10-9 ft4/(lb-s4), M and m are the masses of the bodies, and r is the distance between the two bodies. Notice that the reference position where potential energy is zero is when the two bodies are an infinite distance away from each other.

Spring (Elastic) Potential Energy:

where k is the elastic or spring constant of the material and x is the deformation (compression or extension) of the material.

Conservative vs. Nonconservative Forces:A force is conservative if its work U1-2 is independent of the path followed by the particle as it moves from one position to another.

A distinction is made between conservative and nonconservative forces because the work done by a conservative force may be reclaimed completely as mechanical energy. The work done by a nonconservative force usually results in an increase of a combination of thermal energy and mechanical energy. It is not possible to reclaim all the energy transformed into thermal energy back to useful mechanical work.

The significance of this is that only conservative forces yield changes of energy that may be described as reversible changes of potential energy.

Example of Conservative force. Example of Nonconservative force:

For a conservative force:

Conservation of Energy: If only conservative forces act on a body, then the total energy state of the body remains constant.

F path A

x1 x2

Ff

g

path B

path A

y2

y1

ref y = 0

W

F

Work-Energy Principle:The most general form of the work-energy equation includes kinetic energy terms, potential energy terms, and a work term to deal with any nonconservative forces.

This can be written as

Example 1: A weight of 50 N is dropped from a height of 20 cm on to a spring with an elastic constant of 1000 N/m. a) what is the speed of the weight as it first contacts the spring,b) what is the maximum compression of the spring as it slows down the mass?

Conservation of Energy: = constant energy

W = mg so if W = 50 N then m = 50/9.81 = 5.09 kg

State 1: T1 = 0 since v1 = 0 Vg1=mgy1=(50N)(0.2m)= 10 J

State 2: T2 = (1/2)mv2 = (0.5)(5.09)v2 = (2.59kg)v2 Vg2 = 0 y = 0 at ref height

State 3: T3 = 0 since v3 = 0 Vg3=mgy3=(50N)(-y) Ve3 = (1/2)k y2=(500N/m)y2

Comparing states 1 and 2:

y

20 cm

v

ref

= 1.96 m/s

Comparing states 1 and 3:

Solving for y3

The answer is the positive value: 0.2 m below the original unstretched position.

Example 2: Problem 13:67

Blocks A and B have masses of 4 kg and 1.5 kg respectively, and are connected by a cord and pulley system and released from rest in the position shown with the spring undeformed. Ignore any friction. Knowing that the the constant of the spring is 300 N/m, determine a) the velocity of the block B after it has moved 150 mmb) the maximum velocity of the block Bc) the maximum displacement of block B

Solution:

Conservation of Motion: V + T = constantState 1: As shown and at rest and undeformed spring. Ref height for B is 0.State 2: Block B is 150 mm below starting state.State 3: Max. velocity stateState 4: Max displacement of B

Kinetic Energy:

Potential Energy:

Constraint Equation:

State 1: no velocity, both blocks at ref. height, spring undeformed:

State 2: block B has moved 150 mm….therefore block A and spring have both moved 75 mm. Both blocks now have velocity

A4 kg

300 N/m

1.5 kg

B

but recall that

so

State 3: maximum velocity of Block B (which occurs at same max for velocity of A.).from

rewrite this in terms of xB and vB.

solving for vA

To find the position of maximum velocity, take the derivative and set it equal to zero.

This can be reduced to

Therefore

vA = 0.537 m/stherefore: xB = 2 xA = 0.196 m

vB = 2 vA = 1.075 m/s

Case 4: maximum displacement of b will occur when the kinetic energy again drops to zero.

so either

or

Workout ProblemA force P is slowly applied to a plat that is attached to two spring and causes a deflection x0. In each of the two cases shown derive an expression for the constant ke, in terms of k1 and k2 of he single spring equivalent to the given system, that is, of the single spring which will undergo the same deflection x0 when subjected to the same force P.

EGR232 Dynamics: Homework Set 16 Fall 2012Problem 13:57A 1.2 kg collar C may slide without friction along a horizontal rod. It is attached to three springs, each of constant k = 400 N/m and 150 mm undeformed length. Knowing that the collar is released from rest in the position shown, determine the maximum speed it will reach in the ensuing motion.

EGR232 Dynamics: Homework Set 16 Fall 2012Problem 13:61A 500 g collar can slide without friction on the curved rod BC in a horizontal plane. Knowing that the undeformed length of the spring is 80 mm and that k = 400 kN/m, determine a) the velocity that the collar should be given at A to reach B with zero velocity. b) the velocity of the collar when it eventually reaches C.

EGR232 Dynamics: Homework Set 16 Fall 2012Problem 13:67The system shown is in equilibrium when ϕ = 0 degrees. Knowing that initially ϕ = 90 degrees and that block C is given a slight nudge when the system is in that position, determine the speed of the block as it passes through the equilibrium position ϕ = 0. Neglect the weight of the rod.

EGR232 Dynamics: Homework Set 16 Fall 2012Problem 13:76The 2-lb ball at A is suspended by an inextensible cord and given an initial horizontal velocity of 16 ft/s. If l = 2 ft and xB = 0, determine yB so that the ball will enter the basket.