Vector Mechanics for Engineers: Dynamics Chapter 13 Notes

Chapter

13

Vector Mechanics for Engineers: Dynamics

RVCC

Kinetics of Particles:

Energy and Momentum Methods

Chapter

14

Introduction

• Previously, problems dealing with the motion of particles were solved through the fundamental equation of motion,

Current chapter introduces two additional methods of analysis. .

a m F  

• Method of work and energy: directly relates force, mass, velocity and displacement (F, m, v, d).

• Method of impulse and momentum: directly relates force, mass, velocity, and time (F, m, v, t).

Work of a Force

• Differential vector dr is the particle displacement. • Work of the force is

... ... ... cos    



• Work is a scalar quantity, i.e., it has magnitude and sign but not direction. It can be >0, <0 (F opposite to displacement), zero (F perpendicular to

displacement).

force. length 

• Dimensions of work are Units are



   

Work of a Force

• Object in a curvilinear motion

• Work of a force during a finite displacement,



















• Work is represented by the area under the curve of F_t plotted against s.

The work done by the external total force F to move the particle from A1 to A2, is equal to the integral of the tangential component of that force with respect to the distance along the path.

Work of a Force

• Work of a constant force in rectilinear motion,



F



x

U

dx

F

U













cos

cos

• Work of the force of gravity, from previous slide:

• Work of the weight is equal to product of weight W and vertical displacement y.

• Work of the weight is positive when the body moves down.

W = -W j

The work is independent of the path and is equal to weight times vertical displacement

Wdy

dz

F

dy

F

dx

F

dU















Work of a Force

• Magnitude of the force exerted by a spring is proportional to deflection,





• Work of the force exerted by spring,

.... ... ... 2 1 2 1       



• Work of the force exerted by spring is positive when x₂ < x₁, i.e., when the spring is returning to its undeformed position.

• Work of the force exerted by the spring is equal to negative of area under curve of F plotted against x,





F = kx is actually a static relationship which is true only when

Particle Kinetic Energy: Principle of Work & Energy

• Consider a particle of mass m acted upon by force F

• Integrating from A₁ to A₂ , 2 1 2 1 2 2 2 1 2 1 2 1 mv mv dv v m ds F v v s s t 





The work of the force is equal to the change in kinetic energy of the particle.

F



J m N m s m kg s m kg 2 2 2 2 1 _                  mv T

energy

kinetic

mv

T

T

T

U









Work of a Force

Work of a gravitational force (assume particle M

occupies fixed position O while particle m follows path shown), 1 2 2 2 1 2 2 1 r Mm G r Mm G dr r Mm G U dr r Mm G Fdr dU r r        



The work of F while m moves from A to A’

2 3 12 2

s

kg

m

10

73

.

66

where

since









G

r

Mm

G

F

r

mgR

r

WR

r

Mm

G

F







Applications of the Principle of Work and Energy

at A₂. In A₁ no speed. Consider work & kinetic energy.

• Velocity found without determining

expression for acceleration and integrating. • All quantities are scalars and can be added

directly.

• Forces which do no work are eliminated from

Draw FBD in a general point A

Consider the two forces and the work they do from A₁ to A₂.

What is the work done by P? None! Because………

Applications of the Principle of Work and Energy

energy cannot be applied to directly determine the acceleration of the pendulum bob.

• Moreover, calculating the tension in the cord (that does no work) requires supplementing the method of work and energy with an application of Newton’s second law.

• Suppose you want to know P when it passes in A₂:…………. gl v₂  2 gl v where ₂  2

Sample Problem 13.1

An automobile weighing 4000 lb is driven down a 5o _{incline at a speed of} 60 mi/h when the brakes are applied causing a constant total breaking force of 1500 lb.

Determine the distance traveled by the automobile as it comes to a stop.

FBD

Problem 1

Hints:

1. Draw FBD 2. Evaluate work.

3. Apply work-energy principle.

Calculate the velocity v of the 50-kg crate when it reaches the bottom of the chute at B if it is given an initial velocity of 4 m/s down the chute at A. The coefficient of kinetic friction is 0.30.

Sample Problem 13.2

Two blocks are joined by an inextensible cable as shown. If the system is released from rest, determine the velocity of block

A after it has moved 2 m. Assume that the

coefficient of friction between block A and the plane is m_k = 0.25 and that the pulley is weightless and frictionless.

SOLUTION:

• Apply the principle of work and

Problem 4

Hints:

1. Apply work-energy principle to the two crates independently

Sample Problem 13.3

A spring is used to stop a 60 kg package which is sliding on a horizontal surface. The spring has a constant k = 20 kN/m and is held by cables so that it is initially compressed 120 mm. The package has a velocity of 2.5 m/s in the position shown and the maximum deflection of the spring is 40 mm.

Determine (a) the coefficient of kinetic friction between the package and surface and (b) the velocity of the package as it passes again through the position shown.

Problem 5

P is a force that is pushing the crate upward.

It is not the force of the spring.

Homework:

Solving Problems on Your Own

Sample Problem 13.4

A 2000 lb car starts from rest at point 1 and moves without friction down the track shown.

Determine:

a) the force exerted by the track on the car at point 2, and

b) the minimum safe value of the radius of curvature at point 3.

Problem 6

FBD

Hints:

1. Draw FBD.

2. Work of weight is…

3. Use work-energy principle and

Newton’s second law.

Packages having a mass of 2 kg are delivered from a conveyor to a smooth circular ramp with a velocity of v_O = 1m/s. If the

radius of the ramp is 0.5 m, determine the angle θ = θ_max at

Problem 7

Hints:

1. Draw FBD.

2. Apply motion for tangential and normal coordinates.

3. You then need another equation

Let’s solve the same problem using instead only the equations of motion

Packages having a mass of 2 kg are delivered from a conveyor to a smooth circular ramp with a velocity of v_O = 1m/s. If the

radius of the ramp is 0.5 m, determine the angle θ = θ_max at

Power and Efficiency

v

F

dt

r

d

F

dt

dU

Power

Power





















• Dimensions of power are work/time or force*velocity. Units for power are

W 746 s lb ft 550 hp 1 or s m N 1 s J 1 (watt) W 1       input power output power input work k output wor U) of rate the is (constant machine a of efficiency      forces friction of because . ... 

Problem 8

Hints: 1. FBD

2. Use power equation to find T 3. Apply Newton

The power winch A hoists the 800 lb log up the 30o _{incline at a constant speed of 4} ft/sec. If the power output of the winch is 6 hp, compute the coefficient of kinetic friction μ_k between the log and the incline.

Problem 8.b

Continuing on the previous problem, if the power is suddenly increased to 8 hp,

what is the corresponding instantaneous acceleration a of the log? [still consider the same temporary v = 4 ft/s; need to use Newton]

Sample Problem 13.5

The dumbwaiter D and its load have a combined weight of 600 lb, while the counterweight C weighs 800 lb.

Determine the power delivered by the electric motor M when the dumbwaiter

(a) is moving up at a constant speed of

8 ft/s and (b) has an instantaneous

velocity of 8 ft/s and an acceleration of 2.5 ft/s2_{, both directed upwards.}

If a problem states that the power is constant (or you need to find the average power) and time and v are your variables (no info on x), you can solve as:

Separate the variables

Equation where x does not appear

If you need x instead of t:

Equation where t does not appear

If velocity is not constant!





mav

P

If velocity is not constant but acceleration is

Problem 9

1. Apply kinematics to find acceleration (constant) 2. Consider previous slide reasoning to find P

Homework:

Solving Problems on Your Own

Conservative Forces

particle from point 1 to point 2 is independent of the path followed

Force is conservative

Weight force

Conservative, since it depends only on the vertical

displacement, not on the path Spring

force

Conservative, since it depends only on the compression or

extension of the spring Friction

force

Non-Conservative, since it depends on the path (the longer,

Work-Energy Principle Highlighting

Conservative Forces

• We have seen:

T

T

U





where represents the work done by all forces

acting on the particle

2 1

U

U

T

T

Gravitational Potential Energy

Weight force

A rock that falls on a pile of brick from a point A at +y high,

will brake the pile. In A the brick has a gravitational

potential energy that is associated to its weight, since capable of doing (positive) work Potential energy is a measure of the amount of

work a conservative force will do A

At –y the gravitational potential energy is negative since the weight does a negative work when the particle is moved back at y=0 (datum) where V = 0

Potential Energy

• If gravitational force is the only force acting on the body, the work of the force of gravity can be written as :

• Work is independent of path followed; it

depends only on the initial and final values of the function Wy.

 Wy V_g

potential energy of the body with respect to force of gravity.

   





• Units of work and potential energy are the same: J m N   Wy V_g

• The choice of datum from which the elevation y is measured is arbitrary.

Elastic Potential Energy

Spring force

The spring force has always the capacity or potential of doing (positive) work on a particle when it is under a deformed position. The

work is done when returning to the unstretched position

Potential energy is a measure of the amount of work a conservative force will do

Potential Energy

• Work of the force exerted by a spring depends only on the initial and final deflections of the spring,

2 2 2 1 2 1 2 1 2 1 kx kx U _  

• The potential energy of the body with respect to the elastic force,

   

• Note that the preceding expression for V_e is valid only if the deflection of the spring is measured from its undeformed position.

Conservation of Energy

• Work of a conservative force,

2 1

2

1 V V

U _{ conservative}  

• Concept of work and energy, 1

2 2

1

T

T

U





• If only conservative forces are acting on the body, it follows that

constant













V

T

E

V

T

V

T

• When a particle moves under the action of conservative forces, the total mechanical energy is constant.   W V T W V T     1 1 1 1 0 gl v₂  2 Recall

In A₁ the energy is entirely potential In A₂ the energy is entirely kinetic

(1)

Summarizing

2 1

2

1

V

V

U





• Concept of work and energy,

1 2

2

1

T

T

U





(1)

(2)

•Notice that: we started this chapter on equation (2).

•The work was inclusive of all forces we called “external and internal”.

•We notice that some of these forces are conservative (gravity and elastic); we now indicate them with V₁ and V₂

•This allow us to separate conservative forces and non-conservative forces (an external pushing force P)

•If we only have conservative forces, it means equation (2) becomes:

If also non-conservative forces are used, then we should write a general equation

2 1 2 1 V V U _   2 2 1 1

V

T

V

T







V

U

T

V

T









Conservation of Energy

• Friction forces are not conservative. Total

mechanical energy of a system involving

friction decreases.

• Mechanical energy is dissipated by friction into

thermal energy.

• Total energy

is always constant, but other forms

of energy could take place: a generator converts

mechanical energy into electrical, a gasoline

engine converts chemical energy into mechanical,

a nuclear reactor converts mass into thermal

Conservation of Energy

T₁ + V₁ = T₂ + V₂

If we consider only conservative forces, the equation:

is referred as the conservation of (mechanical) energy

It states that during motion the kinetic energy must be transformed into potential energy and vice-versa.

That is, if the body increases its kinetic energy, its potential energy will decrease

Total energy of the ball at the initial position At distance h/2 2 2 ) ( 2 ₀ 2 0 2 h g y y a v v   _g  

The total energy E at point 2 is:

gh

v



1

2

3

The total energy E at point 3 is:

Wh

Wh

V

T

E







0





When and how to use the

Principle of Conservation of Energy

T

+ V

+ U

= T

+ V

Summarizing:

we started with the principle of

work & energy

:

T

+ U

= T

If the forces acting on the body are conservative forces (spring

and/or gravitational) then the equation becomes the

Conservation

of energy

:

If also forces non-conservative act on the body then the equation

becomes:

Sample Problem 13.6

A 20 lb collar slides without friction along a vertical rod as shown. The spring attached to the collar has an undeflected length of 4 in. and a constant of 3 lb/in.

If the collar is released from rest at position 1, determine its velocity after it has moved 6 in. to position 2.

Problem 10

Hints:

1. In the energy principle, consider that you also have forces that may do work (i.e.T and friction, but friction is not considered here).

2. T is constant but is its path constant….?

3. What about the reactions of the guides on the slider, do they do work?

4. Choose datum at A to compare with slides.

The 10 kg slider moves with negligible friction up the inclined guide. The attached spring has a stiffness of 60 N/m and is stretched 0.6 m in position A, where the slider is released from rest. The 250 N force is constant and the pulley offers negligible resistance to the motion of the cord. Calculate the velocity v_c of the slider as it passes point C.

Sample Problem 13.7

The 0.5 lb pellet is pushed against the spring and released from rest at A. Neglecting friction, determine the smallest deflection of the spring for which the pellet will travel around the loop and remain in contact with the loop at all times.

Problem 11

Hints:

1. Apply the conservation of energy. 2. What is the work of the force of

the cable?

3. How are you going to evaluate it? 4. Choose datum at the top of the

gantry.

The gantry structure in the photo is used to test the response of an airplane during a crash. The plane, having a mass of 8 Mg, is hoisted back until θ = 60o _{and then the pull-back cable}

AC is released when the plane is at rest. Determine the speed of the plane just before it crashes into the ground at θ = 15o_{. Also, what is the maximum tension developed in the}

supporting cable during the motion? Neglect the effect of lift caused by the wings during the motion and the size of the airplane.

Linear Momentum of a Particle

• Replacing the acceleration by the derivative of the velocity yields

 

particle

the

of

momentum

linear













L

L

dt

L

d

v

m

dt

d

dt

v

d

m

F











Linear Momentum Conservation Principle

:

If the resultant force on a particle is zero, the

linear momentum of the particle remains

Useful in many cases involving velocity and needed for

using the principle of impulse and momentum.

Problem 12

1 mi = 5280 ft

When evaluating velocity, you need to calculate it as distance covered divided by time.

Sample Problem 14.2

(linear momentum for a system of

particles)

A 20-lb projectile is moving with a

velocity of 100 ft/s when it explodes into 5 and 15-lb fragments. Immediately after the explosion, the fragments travel in the directions q_A = 45o and q_B = 30o.

Problem 13

Hints:

1. All initial velocities are zero

2. Are there any horizontal external forces?

Car A (4000lb) and car B (3700 lb) are at rest on a flatcar also at rest. Car A and B then accelerate and quickly reach constant speeds relative to the flatcar of 7.65 ft/s and 7.50 ft/s, respectively, before decelerating to a stop at the opposite end of the flatcar. Knowing that the speed of the flatcar is 1.02 ft/s to the right when the cars are moving at constant speeds, determine the weight of the flatcar. Neglect friction and rolling resistance.

What happens if there is an external force acting on the system?

Linear momentum will not be conserved!

That is, the final momentum of the system will be different from

the initial one because of the action of the external force

Principle of Impulse and Momentum

• From Newton’s second law,

 

1 impulse of the force

2 1 v m v m F dt F t t          



 



This method is typically used in problems

involving

Force

,

Time

and

Velocity

.

Principle of Impulse and Momentum

m

v

v

m





Imp





The final momentum of the particle can be obtained by adding vectorially its initial momentum and the impulse of the force during the time interval.

• Dimensions of the (linear) impulse of a force are

force*time.

• Units for the impulse of a force are





N   2   

Notice that, since Δt will always be small, F has to be high in order for the Impulse to be comparable to the linear momentum.

Principle of Impulse and Momentum

If more forces act on a particle

2 1 mv

v

m  

If no external forces act on a particle

Total momentum is conserved This occurs either if the resultant of the external forces is zero, or

(as seen in next slides) when Δt is very short and all the external forces are non-impulsive (and will be neglected).

Impulsive Motion

• Force acting on a particle during a very short time interval that is large enough to cause a significant change in momentum is called an

impulsive force.

• When impulsive forces act on a particle,

2 1 F t mv

v

m 



• When a baseball is struck by a bat, contact occurs over a short time interval but force is large enough to change sense of ball motion.

• Nonimpulsive forces are forces for which is small and therefore, may be

neglected (weight of a body, force of a spring only if the time interval is small!).

t F

Sample Problem 13.10

An automobile weighing 4000 lb is driven down a 5o _{incline at a speed of} 60 mi/h when the brakes are applied, causing a constant total braking force of 1500 lb.

Determine the time required for the automobile to come to a stop.

Sample Problem 13.11

A 4 oz baseball is pitched with a

velocity of 80 ft/s. After the ball is hit by the bat, it has a velocity of 120 ft/s in the direction shown. If the bat and ball are in contact for 0.015 s,

determine the average impulsive force exerted on the ball during the impact.

Problem 12

Not the typical impulse/momentum problem!

The 100 kg stone is originally at rest on the smooth horizontal surface. If a towing force of 200 N, acting at an angle of 45o _{is applied to the stone for 10 seconds,} determine the final velocity and the normal force which the surface exerts on the stone during the time interval.

Hints:

• the problem involves F, v, t: what principle do we use to solve it….?

• ……….! • Draw FBD.

Problem 13

Hints:

•Apply principle of Impulse and Momentum to A and B independently

•Draw FBD before starting any calculation •Apply equation of dependent motion

Blocks A and B have a mass of 3 kg and 5kg, respectively. If the system is released from rest, determine the velocity of block B in

Conservation of Linear Momentum for a System

of Particles

F

t

mv

mv













F

dt

mv

mv











For a system of particles:

or

Where the integral (or the sum) involves only impulsive external* forces (the impulsive internal forces cancel each other (action-reaction)).

If the sum of the external impulses is zero, then:

2

1

mv

mv







The total momentum of the particles is conserved

This principle is often applied when particles moving freely collide or interact Note that the momentum is conserved but in general their total energy is not conserved

In problems with system of particles, often Equation 2 is applied and allows the evaluation of a velocity; then Equation 1 is applied to only one particle to find the impulsive force(s).

Equation 2

Sample Problem 13.12

A 10 kg package drops from a chute into a 24 kg cart with a velocity of 3 m/s. Knowing that the cart is initially at rest and can roll freely, determine (a) the final velocity of the cart, (b) the impulse exerted by the cart on the package, and (c) the fraction of the initial energy lost in the impact.

Problem 15

Hints:

• To apply the impulse –momentum, you need velocities before and after impact. Note that the hammer is in free fall, therefore conservation of energy… • Note that weight of hammer, of pile and reaction of sand are all non-impulsive… • Since there is no rebound, the two objects have the same final speed.

•Consider three (3) different moments: 0 before the hammer falls; 1 when the

An 800 kg rigid pile P is driven into the ground using a 300 kg hammer H. The hammer falls from rest at a height y_o = 0.5 m and

strikes the top of the pile. Determine the impulse which the hammer imparts on the pile if the pile is surrounded entirely by loose sand so that after striking, the hammer does not rebound off the pile.

x y

Recalling the example on the textbook

+

=

t = 0

The hammer is not released yet.

Right at the impact before they move: W_H, W_p and R_s are t = 1 R_s W_H W_P -R_I R_I t = 2

They move together v₂

Homework:

Solving Problems on Your Own

Problem 16

Hints:

1. Are there any horizontal external forces…?

Problem 17

Hints:

1. At the beginning the boat is at rest

Homework:

Solving Problems on Your Own

Impact

• Impact: Collision between two bodies which

occurs during a small time interval and during which the bodies exert large forces on each other.

• Line of Impact: Common normal to the surfaces

in contact during impact.

• Central Impact: Impact for which the mass

centers of the two bodies lie on the line of impact; otherwise, it is an eccentric impact..

Direct Central Impact

• Direct Impact: Impact for which the velocities of

the two bodies are directed along the line of impact.

Oblique Central Impact

• Oblique Impact: Impact for which one or both of

the bodies move along a line other than the line of impact.

Direct Central Impact

• Bodies moving in the same straight line,

v_A >v_B .

• Upon impact the bodies undergo a

period of deformation, at the end of which,

they are in contact and moving at a common velocity u.

• A period of restitution follows during

which the bodies either regain their original shape or remain permanently deformed (the restitution impulse pushes the particles apart).

• We wish to determine the final velocities of the two bodies. Considering a system with both particles, no external forces act. Then the total momentum of the two body system is preserved:

B B A A B B A Av m v m v m v m     

• A second relation between the final velocities is required. Apply the principle to each particle.

Direct Central Impact









Particle A deforms under the action of B

During the period of restitution (when A regain its shape) consider again the action of B on A:

It is possible to define a coefficient of restitution:

Direct Central Impact

B B v u u v e     • Combining the relations leads to the desired

second relation between the final velocities.





A

B v e v v v    

• Perfectly plastic impact, e = 0: v_B  v_A  v

_?

_...

_...

_...

_...

• Perfectly elastic impact, e = 1:

Total energy and total momentum conserved.

...

...

...

?

Note that only in the case of perfectly elastic impact the total energy of the two particles as well as their total momentum is conserved.

The principle of work and energy cannot be used for impact problems since it is not known how the internal forces of deformation and restitution vary.

However, knowing the particle’s velocities before and after the collision, the energy loss during collision can be calculated as

1 2 2 1 T T U    _

If perfectly elastic impact If plastic impact

Problem 16

Hints:

•Consider four (4) different moments: 0 before the ball is released after stretching the cord; 1 when the ball hits the ceiling, 2 when the ball bounce back after the collision and finally 3 when is will be all stretched.

• To apply the impulse –momentum, you need velocity before the impact, therefore conservation of energy…

• Apply impulse-momentum (impact ball-ceiling)…

• Apply equation of conservation of energy to the ball just after the collision to determine stretch. •Weight of ball is non-impulsive

Oblique Central Impact

• Final velocities are

unknown in magnitude and direction. Four equations are required (choosing n and t axes).

• Normal component of total

momentum of the two particles is conserved.

 

 

 

 

A v m v m v m v m     

Consider particles perfectly smooth and frictionless (i.e. no external impulse) Impulse is only due to internal forces along n axis

• Normal components of relative velocities before and after impact are related by the coefficient of

   

v





v





e



   

v



v



• No tangential impulse component; tangential component of momentum for each particle is conserved.

Sample Problem 13.15

The magnitude and direction of the velocities of two identical

frictionless balls before they strike each other are as shown. Assuming

e = 0.9, determine the magnitude

and direction of the velocity of each ball after the impact.

Problem 17

Hints:

• What kind of impact?

• Assume the four unknown components of v after impact all in the positive directions (if come out negative means opposite directions)

Components of initial velocities (along x and y or n and t):

Two smooth disks A and B, having a mass of 1 kg and 2 kg

respectively, collide with the velocities shown. If the coefficient of restitution for the disks is e = 0.75, determine the velocity of each disk after the collision.

Sample Problem 13.14

A ball is thrown against a frictionless, vertical wall. Immediately before the ball strikes the wall, its velocity has a magnitude v and forms angle of 30o with the horizontal. Knowing that

e = 0.90, determine the magnitude and

direction of the velocity of the ball as it rebounds from the wall.

Sample Problem 13.17

A 30 kg block is dropped from a height of 2 m onto the the 10 kg pan of a

spring scale. Assuming the impact to be perfectly plastic, determine the

maximum deflection of the pan. The constant of the spring is k = 20 kN/m.

Homework:

Solving Problems on Your Own