Dynamics Test 2

8/12/2019 Dynamics Test 2

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READING QUIZ

1. What is the work done by the force F?A) F s B) F s

C) Zero D) None of the above.

s

s1 s2

F

2. If a particle is moved from 1 to 2, the work done on the

particle by the force, FRwill be

A) B)

C) D)

2

1

s

t

s

F ds

2

1

s

ts

F ds

2

1

s

ns

F ds 2

1

s

ns

F ds


2/104

APPLICATIONS

A roller coaster makes use of gravitational forces to assist the

cars in reaching high speeds in the valleys of the track.

How can we design the track (e.g., the height, h, and the radius

of curvature, r) to control the forces experienced by thepassengers?


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APPLICATIONS

(continued)

Crash barrels are often usedalong roadways for crash

protection.

The barrels absorb the cars

kinetic energy by deforming.

If we know the velocity of

an oncoming car and the

amount of energy that canbe absorbed by each barrel,

how can we design a crash

cushion?


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WORK AND ENERGY

Another equation for working kinetics problems involving

particles can be derived by integratingthe equation of motion

(F= ma) with respect to displacement.

This principle is useful for solving problems that involve

force, velocity, and displacement. It can also be used to

explore the concept ofpower.

By substituting at= v (dv/ds) into Ft= mat, the result is

integrated to yield an equation known as theprinciple of workand energy.

To use this principle, we must first understand how to

calculate the work of a force.


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WORK OF A FORCE (Section 14.1)

A force does workon a particle when the particle undergoes a

displacement along the line of action of the force.

Work is defined as theproductof force

and displacement componentsacting in

the same direction. So, if the angle

between the force and displacement

vector is q, the increment of work dU

done by the force is

dU = F ds cos q

By using the definition of the dot product

and integrating, the total work can be

written asr2

r1

U1-2 = F dr


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WORK OF A FORCE

(continued)

Work ispositiveif the force and the movement are in thesame direction. If they are opposing, then the work is

negative. If the force and the displacement directions are

perpendicular, the work is zero.

If Fis a function of position (a commoncase) this becomes

=s2

s1

F cos qdsU1-2

If both F and qare constant (F = Fc), this equation furthersimplifies to

U1-2= Fc cos q(s2- s1)


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WORK OF A WEIGHT

The work done by the gravitational force acting on a particle(or weight of an object) can be calculated by using

The work of a weight is the product of the magnitude of

the particles weight and its vertical displacement. If

Dy is upward, the work is negativesince the weightforce always acts downward.

- W (y2

y1

) = - W Dy- W dy =U1-2 =

y2

y1


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WORK OF A SPRING FORCE

When stretched, a linear elastic spring

develops a force of magnitude Fs= ks, where

k is the spring stiffnessand s is the

displacement from the unstretched position.

If a particle is attached to the spring, the force Fsexerted on the

particleisoppositeto that exerted on the spring. Thus, the work

done on the particle by the spring force will be negative or

U1-2 = [ 0.5 k (s2)20.5 k (s1)

2 ] .

The work of the spring force moving from position s1to position

s2is

= 0.5 k (s2)20.5 k (s1)

2k s dsFs dsU1-2

s2

s1

s2

s1==


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SPRING FORCES

1. The equations above are for linearsprings only! Recall

that a linear spring develops a force according to

F = ks (essentially the equation of a line).

3. Always double checkthe sign of the spring work after

calculating it. It is positive work if the force put on the object

by the spring and the movement are in the same direction.

2. The work of a spring is notjust spring force times distance

at some point, i.e., (ksi)(si). Beware, this is a trap that

students often fall into!

It is important to note the following about spring forces.


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PRINCIPLE OF WORK AND ENERGY

(Section 14.2 & Section 14.3)

U1-2is the work done by all the forcesacting on the particle as it

moves from point 1 to point 2. Work can be either apositive ornegativescalar.

By integrating the equation of motion, Ft= mat= mv(dv/ds), theprinciple of work and energycan be written as

U1-2= 0.5 m (v2)20.5 m (v1)2 or T1+ U1-2= T2

T1and T2are the kinetic energiesof the particle at the initial and final

position, respectively. Thus, T1= 0.5 m (v1)2 and T2= 0.5 m (v2)

2.

The kinetic energy is always apositive scalar (velocity is squared!).

So, the particles initial kinetic energy plus the work done by all the

forces acting on the particle as it moves from its initial to final position

is equal to the particles final kinetic energy.


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PRINCIPLE OF WORK AND ENERGY

(continued)

The principle of work and energy cannotbe used, in general, to

determine forces directed normalto the path, since these forces

do no work.

Note that the principle of work and energy (T1+ U1-2= T2) isnot a vector equation! Each term results in a scalar value.

Both kinetic energy and work have the same units, that of

energy! In the SI system, the unit for energy is called ajoule(J),

where 1 J = 1 Nm. In the FPS system, units are ftlb.

The principle of work and energy can also be applied to a system

of particlesby summing the kinetic energies of all particles in the

system and the work due to all forces acting on the system.


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The case of a body sliding over a rough surfacemerits special

consideration.

This equation is satisfied if P = kN. However, we know fromexperience that friction generates heat, a form of energy that doesnot seem to be accounted for in this equation. It can be shown thatthe work term (kN)s representsboththe external workof thefriction force and the internal workthat is converted into heat.

WORK OF FRICTION CAUSED BY SLIDING

The principle of work and energy would be

applied as

0.5m (v)2+ P s(kN) s = 0.5m (v)2

Consider a block which is moving over a

rough surface. If the applied force Pjust

balances the resultant frictional forcekN,a constant velocity v would be maintained.


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Given:When s = 0.6 m, the spring is

not stretched or compressed,

and the 10 kg block, which is

subjected to a force of F=

100 N, has a speed of 5 m/s

down the smooth plane.

Find: The distance swhen the block stops.

Plan: Since this problem involves forces, velocity and displacement,apply the principle of work and energy to determine s.

EXAMPLE


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Apply the principle of work and energybetween position 1

(s1= 0.6 m) and position 2 (s2). Note that the normal force (N)

does no work since it is always perpendicular to the

displacement.

EXAMPLE

(continued)Solution:

T1+ U1-2= T2

There is work done by three different forces;

1) work of a the force F =100 N;

UF= 100 (s2s1) = 100 (s2 0.6)

2) work of the block weight;

UW= 10 (9.81) (s2s1) sin 30= 49.05 (s2 0.6)

3) and, work of the spring force.

US= - 0.5 (200) (s20.6)2= -100 (s2 0.6)

2


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The work and energy equation will be

T1+ U1-2= T2

0.5 (10) 52 + 100(s20.6) + 49.05(s2 0.6) 100(s2 0.6)2= 0

125

+ 149.05(s20.6) 100(s2 0.6)2

= 0

EXAMPLE

(continued)

Solving for (s20.6),(s20.6) = {-149.05 (149.0524(-100)125)0.5} / 2(-100)

Selecting the positive root, indicating a positive spring deflection,(s20.6) = 2.09 m

Therefore, s2= 2.69 m


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CONCEPT QUIZ

1. If a spring stiffness is k = 5 s3 N/m and the spring is

compressed by s = 0.5 m, the work done on a particle

attached to the spring will be

A) 0.625 N m B) 0.625 N m

C) 0.0781 N m D) 0.0781 N m


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GROUP PROBLEM SOLVING

Given:Block A has a weight of 60 lb

and block B has a weight of 40lb. The coefficient of kinetic

friction between the blocks and

the incline is k= 0.1. Neglect

the mass of the cord and pulleys.

Find: The speed of block A after block B moves 2 ft up the

plane, starting from rest.

Plan: 1) Define the kinematic relationships between the blocks.

2) Draw the FBD of each block.

3) Apply the principle of work and energy to the system

of blocks. Why choose this method?


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GROUP PROBLEM SOLVING (continued)

Since the cable length is constant:

2sA

+ sB

= l

2DsA+ DsB= 0

When DsB= -2 ft => DsA= 1 ft

and 2vA+ vB= 0

=> vB= -2vA

Note that, by this definition of sAand sB, positive motion

for each block is defined as downwards.

Solution:

1) The kinematic relationshipscan be determined by definingposition coordinates sAand sB, and then differentiating.

sA sB


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Similarly, for block B:

NB= WBcos 30

2) Draw the FBD of each block.


(continued)

yx

NA

NA

2T

WA

A

6030

B

NB

NB

T WB

Sum forces in the y-direction for block A

(note that there is no motion in y-direction):

Fy= 0: NAWAcos 60= 0

NA= WAcos 60


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[0.5mA(vA1)2+ .5mB(vB1)

2] + [WAsin 602TNA]DsA

+ [WBsin 30T + N

B]Ds

B= [0.5m

A(v

A2)2+ 0.5m

B(v

B2)2]

3) Apply the principle of work and energyto the system (the

blocks start from rest).T1+ U1-2= T2

where vA1= vB1= 0, DsA= 1ft, DsB= -2 ft, vB= -2vA,NA= WAcos 60, NB= WBcos 30


(continued)

=> [0 + 0] + [60 sin 602T0.1(60 cos 60)](1)

+ [40sin 30T + 0.1(40 cos 30)](2)

= [0.5(60/32.2)(vA2)2+ 0.5(40/32.2)(-2vA2)

2]


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(continued)

Again, the Work and Energy equation is:

=> [0 + 0] + [60 sin 602T0.1(60 cos 60)](1)

+ [40sin 30T + 0.1(40 cos 30)](2)

= [0.5(60/32.2)(vA2)2+ 0.5(40/32.2)(-2vA2)

2]

Note that the work due to the cable tension force on each block

cancels out.

Solving for the unknown velocity yeilds

=> vA2= 0.771 ft/s


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ATTENTION QUIZ

1. What is the work done by the normal

force N if a 10 lb box is moved from Ato B ?

A) - 1.24 lb ft B) 0 lb ft

C) 1.24 lb ft D) 2.48 lb ft

2. Two blocks are initially at rest. How many equations would

be needed to determine the velocity of block A after block B

moves 4 m horizontally on the smooth surface?

A) One B) Two

C) Three D) Four

2 kg

2 kg


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READING QUIZ

1. The formula definition of power is ___________.

A) dU / dt B) Fv

C) Fdr/dt D)All of the above.

2. Kinetic energy results from _______.

A) displacement B) velocity

C) gravity D) friction


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APPLICATIONS

Engines and motors are often

rated in terms of their power

output. The power output of the

motor lifting this elevator is

related to the vertical force F

acting on the elevator, causing itto move upwards.

Given a desired lift velocity for the

elevator (with a known maximum

load), how can we determine the

power requirement of the motor?


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APPLICATIONS(continued)

The speed at which a truck

can climb a hill depends in

part on the power output of

the engine and the angle of

inclination of the hill.

For a given angle, how can we determine the speed of this

truck, knowing the power transmitted by the engine to the

wheels? Can we find the speed, if we know the power?

If we know the engine power output and speed of the truck, can

we determine the maximum angle of climb of this truck ?


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POWER AND EFFICIENCY

(Section 14.4)

Thus, power is a scalardefined as the product of the force

and velocitycomponents acting in the same direction.

Since the work can be expressed as dU = F dr, the power

can be written

P = dU/dt = (F dr)/dt = F (dr/dt) = F v

If a machine or engine performs a certain amount of work,

dU, within a given time interval, dt, the power generated can

be calculated asP = dU/dt

Poweris defined as the amount of workperformedper unitof time.


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POWER

Using scalar notation, power can be written

P = F v= F v cos qwhere qis the angle between the force and velocity vectors.

In the FPS system, power is usually expressed in units of

horsepower(hp) where

1 hp = 550 (ft lb)/s = 746 W .

So if the velocity of a body acted on by a force Fis known,

the power can be determined by calculating the dot productor by multiplying force and velocity components.

The unit of power in the SI system is the Watt(W) where

1 W = 1 J/s = 1 (N m)/s .


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EFFICIENCY

If energy input and removal occur at the same time, efficiencymay also be expressed in terms of the ratio of output energy

to input energy or

e= (energy output) /(energy input)

Machines will always have frictional forces. Since frictional

forces dissipateenergy, additional power will be required to

overcome these forces. Consequently, the efficiency of a

machine is always less than 1.

The mechanical efficiencyof a machine is the ratio of the

useful power produced (output power) to the power supplied

to the machine (input power) or

e= (power output) /(power input)


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PROCEDURE FOR ANALYSIS

Multiply the force magnitudeby the component of velocity

acting in the directionof Fto determine the power supplied

to the body (P = F v cos q ).

If the mechanical efficiencyof a machine is known, either

the power input or output can be determined.

Determine the velocityof thepointon the body at which the

force is applied. Energy methods or the equation of motion

and appropriate kinematic relations, may be necessary.

In some cases,powermay be found by calculating the workdone per unit of time(P = dU/dt).

Find the resultant externalforceacting on the body causing

its motion. It may be necessary to draw a free-body diagram.


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EXAMPLE

Given:A 50 kg block (A) is hoisted by the pulley

system and motor M. The motor has anefficiency of 0.8. At this instant, point P

on the cable has a velocity of 12 m/s

which is increasing at a rate of 6 m/s2.

Neglect the mass of the pulleys and

cable.Find: The power supplied to the motor at this

instant.

Plan:

1) Relate the cable and block velocities by defining positioncoordinates. Draw a FBD of the block.

2) Use the equation of motion to determine the cable tension.

3) Calculate the power supplied by the motor and then to the

motor.


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EXAMPLE (continued)

Solution:

Here sPis defined to a point on the cable. Also

sAis defined only to the lower pulley, since the

block moves with the pulley. From kinematics,

sP+ 2 sA= l

aP+ 2 aA= 0

aA= aP/ 2 = 3 m/s2 ()

sB

sm

1) Define position coordinatesto relate velocities.

Datum

SA

SP

Draw the FBD and kinetic diagramof the block:2T

WA

A

mAaA

A=


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2) The tension of the cablecan be obtained by applying theequation of motion to the block.

+ Fy= mAaA

2T 490.5 = 50 (3) T = 320.3 N

EXAMPLE

(continued)

3) Thepower supplied by the motoris the product of the force

applied to the cable and the velocity of the cable.

Po= F v= (320.3)(12) = 3844 W

Pi= Po/e= 3844/0.8 = 4804 W = 4.8 kW

Thepower supplied to the motoris determined using themotors efficiency and the basic efficiency equation.


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READING QUIZ

1. The linear impulse and momentum equation is obtained byintegrating the ______ with respect to time.

A) friction force B) equation of motion

C) kinetic energy D) potential energy

2. Which parameter is not involved in the linear impulse and

momentum equation?

A) Velocity B) Displacement

C) Time D) Force


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APPLICATIONS

A dent in an automotive fender

can be removed using an impulsetool, which delivers a force over a

very short time interval. To do so

the weight is gripped and jerked

upwards, striking the stop ring.How can we determine the

magnitude of the linear impulse

applied to the fender?

Could you analyze a carpentershammer striking a nail in the

same fashion?Sure!


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APPLICATIONS

(continued)

When a stake is struck by asledgehammer, a large impulse

force is delivered to the stake and

drives it into the ground.

If we know the initial speed of the

sledgehammer and the duration of

impact, how can we determine the

magnitude of the impulsive force

delivered to the stake?


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PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM

(Section 15.1)

This principle is useful for solving problems that involveforce, velocity, and time. It can also be used to analyze the

mechanics of impact (taken up in a later section).

The result is referred to as theprinciple of impulse andmomentum. It can be applied to problems involving both

linear and angular motion.

The next method we will consider for solving particle

kinetics problems is obtained by integrating the equation of

motion with respect to time.


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This equation represents the principle of linear impulse and

momentum. It relates the particles final velocity (v2) andinitial velocity (v1) and the forces acting on the particle as a

function of time.

Theprinciple of linear impulse and momentumis obtainedby integrating the equation of motion with respect to time.

The equation of motion can be written

F= m a= m (dv/dt)

Separating variables and integrating between the limits v= v1

at t = t1 and v= v2at t = t2results in

mv2mv1dvmFdt

v2

v1

t2

t1

==


(continued)


38/104


(continued)

Linear impulse: The integral Fdt is the linear impulse, denoted

I. It is a vector quantitymeasuring the effect of a force during itstime interval of action. Iacts in the samedirectionas Fand has

units of Ns or lbs.

Linear momentum: The vector mvis called the linear momentum,

denoted as L . This vectorhas the same directionas v. The linear

momentum vector has units of (kgm)/s or (slugft)/s.

The impulse may be determined by

direct integration. Graphically, it

can be represented by the area under

the force versus timecurve. If Fis

constant, then

I= F (t2t1) .


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The two momentum diagramsindicate direction

and magnitude of the particles initial and finalmomentum, mv1and mv2. The impulse diagramis

similar to a free body diagram, but includes the

time duration of the forces acting on the particle.


(continued)

The particles initial momentum plus the sum of all theimpulses applied from t1to t2is equal to the particlesfinal momentum.

The principle of linear impulse and momentum in

vectorform is written as

Fdtt2

t1 mv1+ = mv2


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IMPULSE AND MOMENTUM: SCALAR EQUATIONS

The scalar equations provide a convenient means for applying

the principle of linear impulse and momentum once the velocity

and force vectors have been resolved into x, y, z components.

Since the principle of linear impulse and momentum is a

vector equation, it can be resolved into its x, y, z component

scalar equations:

m(vx)1+ Fxdt = m(vx)2

m(vy)1+ Fydt = m(vy)2

m(vz)1+ Fzdt = m(vz)2

t2

t1

t2

t1t2

t1


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PROBLEM SOLVING

Establish the x, y, z coordinate system.

Forcesas functions of timemust be integratedto obtainimpulses. If a force is constant, its impulse is the product of

the forces magnitude and time interval over which it acts.

Resolve the force and velocity (or impulse and momentum)vectors into their x, y, z components, and apply theprinciple

of linear impulse and momentumusing its scalar form.

Draw the particles free body diagramand establish thedirectionof the particles initial and final velocities, drawingthe impulse and momentum diagramsfor the particle. Show

the linear momenta and force impulse vectors.


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EXAMPLE

Given:A 0.5 kg ball strikes the rough

ground and rebounds with thevelocities shown. Neglect the

balls weight during the time itimpacts the ground.

Find: The magnitude of impulsive force exerted on the ball.

Plan: 1) Draw the momentum and impulse diagramsof the

ball as it hits the surface.

2) Apply the principle of impulse and momentum todetermine the impulsive force.


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EXAMPLE


1) The impulse and momentum diagramscan be drawn as:

The impulse caused by the balls weight and the normalforce Ncan be neglected because their magnitudes are

very small as compared to the impulse from the ground.

+ =

Wdt 0

Ndt 0Fdt

mv2

30

mv1

45


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2) The principle of impulse and momentumcan be applied along

the direction of motion:

mv1+ Fdt = mv2

t2

t1

EXAMPLE

(continued)

0.5 (25 cos 45i 25 sin 45j) + Fdt

= 0.5 (10 cos 30i + 10 sin 30j)

t2

t1

The impulsive force vector is

I = Fdt = (4.509 i+ 11.34j ) Ns

Magnitude: I = 4.5092+ 11.342 = 12.2 Nst1

t2


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Given:The 20 kg crate is resting

on the floor. The motor Mpulls on the cable with a

force of F, which has a

magnitude that varies as

shown on the graph.

Find: The speed of the crate

when t = 6 s.

Plan:

1) Determine the force needed to begin lifting the crate, andthen the time needed for the motor to generate this force.

2) After the crate starts moving, apply the principle of

impulse and momentum to determine the speed of the

crate at t = 6 s.


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Solution:

F = mg = (20) (9.81) = 196.2 N

F = 196.2 N = 50 t

t = 3.924 s

2) Apply the principle of impulse and momentumfrom the time

the crate starts lifting at t1= 3.924s to t2= 6 s.

Note that there are two external forces (cable force and

weight) we need to consider.

1) The crate begins movingwhen the cable force F exceeds the

crate weight. Solve for the force, then the time.

A. The impulse due to cable force:

F dt = [0.5(250) 5 + (250) 1]0.5(196.2)3.924= 490.1 Ns3.924

6+


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B. The impulse due to weight:

+ ( mg) dt = 196.2 (6 3.924) = 407.3 Ns3.924

6

Now, apply the principle of impulse and momentum

mv1+ F dt = mv2 where v1= 0t2

t1

0 + 490.1 407.3 = (20) v2

=> v2= 4.14 m/s

+


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READING QUIZ

1. The internal impulses acting on a system of particles

always __________

A) equal the external impulses. B)sum to zero.

C) equal the impulse of weight. D) None of the above.

2. If an impulse-momentum analysis is considered during the

very short time of interaction, as shown in the picture, weight

is a/an __________

A) impulsive force.

B) explosive force.

C) non-impulsive force.

D) internal force.


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APPLICATIONS

Does the release velocity of the ball

depend on the mass of the ball?

As the wheels of this pitching machine

rotate, they apply frictional impulses to

the ball, thereby giving it linear

momentum in the direction of Fdt and

F dt.

The weight impulse, Wt is very smallsince the time the ball is in contact

with the wheels is very small.


50/104


This large crane-mounted hammer is

used to drive piles into the ground.

Conservation of momentum can be

used to find the velocity of the pile

just after impact, assuming the

hammer does not rebound off the pile.

If the hammer rebounds, does the pile velocity change from

the case when the hammer doesnt rebound ? Why ?

In the impulse-momentum analysis, do we have to consider

the impulses of the weights of the hammer and pile and the

resistance force ? Why or why not ?


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FOR A SYSTEM OF PARTICLES

(Section 15.2)

The linear impulse and momentum equation for this system

only includes the impulse of externalforces.

mi(v

i)2dtFimi(vi)1

t2

t1=+

For the system of particles shown,

the internal forces fibetween

particles always occur in pairs with

equal magnitude and oppositedirections. Thus the internal

impulses sum to zero.


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For a system of particles, we can define a fictitious centerof mass of an aggregate particle of mass mtot, where mtotis

the sum (mi) of all the particles. This system of particles

then has an aggregate velocity of vG= (mivi) / mtot.

MOTION OF THE CENTER OF MASS

The motion of this fictitious mass is based on motion of the

center of mass for the system.

The position vector rG= (miri) / mtot describes the motionof the center of mass.


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CONSERVATION OF LINEAR MOMENTUM FOR A

SYSTEM OF PARTICLES (Section 15.3)

When the sum of external impulsesacting on a systemof objects is zero, the linear impulse-momentum

equation simplifies to

mi(v

i)1 = mi(vi)2

This equation is referred to as the conservation of

linear momentum. Conservation of linear momentum

is often applied when particles collide or interact.

When particles impact, only impulsive forcescause a

change of linear momentum.

The sledgehammer applies an impulsive force to the stake. The weightof the stake is considered negligible, or non-impulsive, as compared to

the force of the sledgehammer. Also, provided the stake is driven into

soft ground with little resistance, the impulse of the ground acting on the

stake is considered non-impulsive.


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mvi= mAvA+ mBvB+ mCvC

100(20j) = 20(50i+ 50j) + 30(-30i-50k) + 50(vcx i+ vcyj+ vcz k)

Equating the components on the left and right side yields:0 = 1000900 + 50(vcx) vcx= -2 m/s

2000 = 1000 + 50 (vcy) vcy= 20 m/s

0 = -1500 + 50 (vcz) vcz= 30 m/s

So vc= (-2i+ 20j+ 30k) m/s immediately after the explosion.

EXAMPLE I



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EXAMPLE II

Find: The speed of the car A after collision if the cars

collide and rebound such that B moves to the right

with a speed of 2 m/s. Also find the average

impulsive force between the cars if the collision

place in 0.5 s.

Plan: Use conservation of linear momentumto find the

velocity of the car A after collision (all internal

impulses cancel). Then use theprinciple of impulse

and momentumto find the impulsive force by looking

at only one car.

Given: Two rail cars with masses

of mA= 20 Mg and mB=15 Mg and velocities as

shown.


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EXAMPLE II

(continued)

Conservation of linear momentum (x-dir):

mA(vA1) + mB(vB1) = mA(vA2)+ mB(vB2)

20,000 (3) + 15,000 (-1.5)

= (20,000) vA2+ 15,000 (2)

vA2= 0.375 m/s

The average force is

F dt = 52,500 Ns = Favg(0.5 sec); Favg= 105 kN

Solution:

Impulse and momentum on car A (x-dir):

mA (vA1)+ F dt = mA (vA2)

20,000 (3) - F dt = 20,000 (0.375)

F dt = 52,500 Ns


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READING QUIZ

1. When the motion of one or both of the particles is at an angle

to the line of impact, the impact is said to be ________

A) central impact. B) oblique impact.

C) major impact. D) None of the above.

2. The ratio of the restitution impulse to the deformation

impulse is called _________

A) impulse ratio. B) restitution coefficient.

C) energy ratio. D) mechanical efficiency.


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APPLICATIONS

The quality of a tennis ball is measured by the height of its

bounce. This can be quantified by the coefficient of

restitutionof the ball.

If the height from which the ball is dropped and the height of

its resulting bounce are known, how can we determine the

coefficient of restitution of the ball?


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APPLICATIONS

(continued)

In the game of billiards, it is important to be able to predict

the trajectory and speed of a ball after it is struck by

another ball.

If we know the velocity of ball A before the impact, howcan we determine the magnitude and direction of the

velocity of ball B after the impact?

What parameters do we need to know for this?


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IMPACT (Section 15.4)

Impactoccurs when two bodies collide during a very shorttime

period, causing large impulsive forces to be exerted between thebodies. Common examples of impact are a hammer striking a

nail or a bat striking a ball. Theline of impact is a line through

themass centers of the colliding particles. In general, there are

twotypes of impact:

Central impactoccurs when the

directions of motion of the two colliding

particles are along the line of impact.

Oblique impactoccurs when the direction

of motion of one or both of the particles is

at an angle to the line of impact.


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CENTRAL IMPACT

There are two primary equations used when solving impact

problems. The textbook provides extensive detail on their

derivation.

Central impacthappens when the velocities of the two objects

are along the line of impact (recall that the line of impact is a

line through the particles mass centers).vA vB

Line of impact

Once the particles contact, they may

deform if they are non-rigid. In any

case, energy is transferred between the

two particles.


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CENTRAL IMPACT

(continued)

In most problems, the initial velocities of the particles, (vA)1and

(vB)1, are known, and it is necessary to determine the final

velocities, (vA)2and (vB)2. So the firstequation used is the

conservation of linear momentum, applied along the line of impact.

(mAvA)1 + (mBvB)1 = (mAvA)2 + (mBvB)2

This provides one equation, but there are usually two unknowns,

(vA)2and (vB)2. So another equation is needed. Theprinciple ofimpulse and momentumis used to develop this equation, which

involves the coefficient of restitution, or e.


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The coefficient of restitution, e, is the ratio of the particlesrelative separation velocityafter impact, (vB)2(vA)2, to the

particles relative approach velocitybefore impact, (vA)1(vB)1.

The coefficient of restitution is also an indicator of the energy

lost during the impact.

The equation defining the coefficient of restitution, e, is

(vA)1- (vB)1

(vB)2(vA)2e =

If a value for eis specified, this relation provides the second

equation necessary to solve for (vA)2and (vB)2.

CENTRAL IMPACT

(continued)


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COEFFICIENT OF RESTITUTION

Plastic impact (e= 0): In a plastic impact, the relative

separation velocity is zero. The particles stick together and

move with a common velocity after the impact.

Elastic impact (e= 1): In a perfectly elastic collision, no

energy is lost and the relative separation velocity equals the

relative approach velocity of the particles. In practical

situations, this condition cannot be achieved.

In general, ehas a value between zero and one.

The two limiting conditions can be considered:

Some typical valuesof eare:

Steel on steel: 0.50.8 Wood on wood: 0.40.6

Lead on lead: 0.120.18 Glass on glass: 0.930.95


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IMPACT: ENERGY LOSSES

In aplastic collision(e= 0), the energy lost is a maximum,

although it does not necessarily go to zero. Why?

During a collision, some of the particles initial kinetic

energy will be lost in the form of heat, sound, or due to

localized deformation.

Once the particles velocities before and after the collision

have been determined, the energy lossduring the collisioncan be calculated on the basis of the difference in the

particles kinetic energy. The energy loss is

U1-2= T2T1 where Ti= 0.5mi(vi)2


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OBLIQUE IMPACT

Momentum of each particle is conserved in the directionperpendicularto

the line of impact (y-axis):

mA(vAy)1= mA(vAy)2 and mB(vBy)1= mB(vBy)2

In an oblique impact, one or both of the

particles motion is at an angle to the line of

impact. Typically, there will be four

unknowns: the magnitudesand directionsof

the final velocities.

Conservation of momentum and the coefficient

of restitution equation are applied alongthe line

of impact (x-axis):

mA(vAx)1+ mB(vBx)1= mA(vAx)2+ mB(vBx)2

e= [(vBx)2(vAx)2]/[(vAx)1(vBx)1]

The four equations required to solve for the unknowns are:



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In most impact problems, the initial velocities of the particles

and the coefficient of restitution, e, are known, with the final

velocities to be determined.

For obliqueimpact problems, the following equations are also

required, appliedperpendicularto the line of impact (y-dir.):

mA(vAy)1= mA(vAy)2 and mB(vBy)1= mB(vBy)2

For both central and obliqueimpact problems, the following

equations apply alongthe line of impact (x-dir.):

m(vx)1= m(vx)2 and e= [(vBx)2(vAx)2]/[(vAx)1(vBx)1]

Define the x-y axes. Typically, the x-axisis defined alongthe

line of impact and the y-axisis in the plane of contact

perpendicularto the x-axis.


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EXAMPLE


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Solve the impact problem by using x-y axes defined along and

perpendicular to the line of impact, respectively:

The momentum of the ball is conservedin

the y-dir:

m(vb)1 sin 30= m(vb)2 sin

(vb)2 sin = 10 m/s (1)

The coefficient of restitution applies in the x-dir:

e= [ 0(vbx)2 ] / [ (vbx)10 ]

0.75 = [ 0(-vb)2 cos ] / [ 20cos 300]

(vb)2 cos = 12.99 m/s (2)

EXAMPLE


Using Eqs. (1) and (2) and solving for the velocity and yields:

(vb)2 = (12.992+102)0.5 = 16.4 m/s

= tan-1(10/12.99)=37.6


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CONCEPT QUIZ

2. Under what condition is the energy lost during a collision

maximum?

A) e= 1.0 B) e =0.0

C) e= -1.0 D) Collision is non-elastic.

1. Two balls impact with a coefficient of restitution of 0.79.

Can one of the balls leave the impact with a kinetic energygreater than before the impact?

A) Yes B) No

C) Impossible to tell D) Dont pick this one!

RIGID BODY MOTION: ROTATION ABOUT A FIXED AXIS


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RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS

(Section 16.3)

The change in angular position, d, is called the

angular displacement, with units of either

radians or revolutions. They are related by1 revolution = (2) radians

When a body rotates about a fixed axis, any

point P in the body travels along a circular path.The angular position of P is defined by .

Angular velocity, , is obtained by taking thetime derivative of angular displacement:

= d/dt (rad/s) +Similarly, angular accelerationis

= d2/dt2 = d/dt or = (d/d) + rad/s2

RIGID BODY MOTION: ROTATION ABOUT A FIXED AXIS


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If the angular acceleration of the body is

constant, = C, the equations for angularvelocity and acceleration can be integrated

to yield the set of algebraicequations

below.

= 0+ C t= 0+ 0 t + 0.5 C t2

2= (0)2+ 2C(0)

0and 0are the initial values of the bodys

angular position and angular velocity. Notethese equations are very similar to the

constant acceleration relations developed for

the rectilinearmotion of a particle.

RIGID-BODY MOTION: ROTATION ABOUT A FIXED AXIS

(continued)

RIGID BODY ROTATION: VELOCITY OF POINT P


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The magnitude of the velocity of P is

equal to r (the text provides the

derivation). The velocitys direction is

tangent to the circular path of P.

In the vectorformulation, the magnitude

and direction of vcan be determined

from the cross productof and rp.Here rpis a vector from any point on the

axis of rotation to P.v= rp= r

The direction of vis determined by the

right-hand rule.

RIGID-BODY ROTATION: VELOCITY OF POINT P

RIGID BODY ROTATION: ACCELERATION OF POINT P


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The acceleration of P is expressed in terms of

its normal(an) and tangential(at) components.In scalar form, these are at= r and an=

2 r.

The tangential component, at, represents the

time rate of change in the velocity'smagnitude. It is directed tangentto the path of

motion.

The normal component, an, represents the time

rate of change in the velocitys direction. It is

directed towardthe centerof the circular path.

RIGID-BODY ROTATION: ACCELERATION OF POINT P

RIGID BODY ROTATION: ACCELERATION OF POINT P


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Using the vectorformulation, the acceleration

of P can also be defined by differentiating thevelocity.

a= dv/dt = d/dt rP+ drP/dt

= rP+ ( rP)

It can be shown that this equation reduces to

a= r2

r= at+ an

RIGID-BODY ROTATION: ACCELERATION OF POINT P

(continued)

The magnitudeof the acceleration vector is a = (at)2+ (an)

2


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EXAMPLE


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EXAMPLE

Given:The motor gives the blade an angular

acceleration

= 20 e-0.6t

rad/s2

, where t isin seconds. The initial conditions are that

when t = 0, the blade is at rest.

Find:The velocity and acceleration of the tip P of one of the

blades when t =3 s. How many revolutions has the bladeturned in 3 s ?

Plan: 1) Determine the angular velocity and displacement of the

blade using kinematics of angular motion.2) The magnitudes of the velocity and acceleration of

point P can be determined from the scalar equations of

motion for a point on a rotating body. Why scalar?

EXAMPLE (continued)


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EXAMPLE (continued)

Solution:

1) Since the angular accelerationis given as a function of time,

= 20 e-0.6trad/s2 , the angular velocity and displacementcan be found by integration.

= dt = 20 e-0.6tdt

= e-0.6t20(-0.6)

= dt

= e-0.6tdt = e-0.6t20

(-0.6)

20

(-0.6)2

Angular displacement

Also , when t = 3 s, =20 e-0.6(3) b= 3.306 rad/s2

when t = 3 s,= -5.510 rad/s

when t = 3 s,= 9.183 rad

= 1.46 rev.

A ( i d)


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2) The velocity of point Pon the the fan, at a radius of 1.75 ft,

is determined as

vP= r = (5.510)(1.75) = 9.64 ft/s

The magnitudeof the acceleration of P is determined by

aP= (an)2+ (at)

2 = (53.13)2+ (5.786)2 = 53.4 ft/s2

EXAMPLE (continued)

The normal and tangential components of accelerationof

point P are calculated asan= ()

2 r = (5.510)2 (1.75) = 53.13 ft/s2

at= r = (3.306)(1.75) = 5.786 ft/s2

READING QUIZ


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READING QUIZ

1. A body subjected to general plane motion undergoes a/an

A) translation.B) rotation.

C) simultaneous translation and rotation.

D) out-of-plane movement.

2. In general plane motion, if the rigid body is represented by a

slab, the slab rotates

A) about an axis perpendicular to the plane.

B) about an axis parallel to the plane.

C) about an axis lying in the plane.

D) None of the above.

APPLICATIONS


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APPLICATIONS

As a part of the design process for the truck, an engineer had to

relate the velocity at which the hydraulic cylinder extends and

the resulting angular velocity of the bin.

The dumping bin on the truck rotates

about a fixed axis passing through the

pin at A. It is operated by the extension

of the hydraulic cylinder BC.

The angular position of the bin can be

specified using the angular positioncoordinate and the position of point C

on the bin is specified using the

coordinate s.

APPLICATIONS (continued)


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The large window is opened using a

hydraulic cylinder AB.

The position B of the hydraulic

cylinder rod is related to the angular

position, , of the window.

A designer has to relate the translational

velocity at B of the hydraulic cylinder

and the angular velocity and accelerationof the window? How would you go

about the task?



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The position of the piston, x, can be defined as a function of

the angular position of the crank, . By differentiating x with

respect to time, the velocity of the piston can be related to the

angular velocity, w, of the crank. This is necessary when

designing an engine.

The stroke of the piston is defined as the total distance moved

by the piston as the crank angle varies from 0 to 180. How

does the length of crank AB affect the stroke?

ABSOLUTE MOTION ANALYSIS


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(Section 16.4)

The absolute motion analysis methodrelates

the position of a point, B, on a rigid body

undergoing rectilinear motion to the angular

position, , of a line contained in the body.

Usually the chain rulemust be used when taking the derivatives

of the position coordinate equation.

Once a relationship in the form of sB= f () is

established, the velocity and acceleration of

point B are obtained in terms of the angular

velocity and angular acceleration of the rigid

body by taking the first and second timederivativesof the position function.

The figure below shows the window using a hydraulic cylinder AB.

EXAMPLE I


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Given:The platform is constrained

to move vertically by the

smooth vertical guides. The

cam C rotates with a

constant angular velocity, w.

Find: The velocity and accelerationof platform P as a function of

the angle of cam C.

Use the fixed reference point O and define the position of

the platform, P, in terms of the parameter .

Take successive time derivatives of the position equation

to find the velocity and acceleration.

EXAMPLE I

Plan:

EXAMPLE I


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aP= d(rw cos ) / dt = rw(-sin ) (w) = rw2sin

Differentiating vPto find the acceleration,

O

By geometry, y = r + r sin By differentiating with respect to time,

vP= r cos () = rwcos

Note that the cam rotates with a constantangular velocity.

EXAMPLE II


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Given: Crank AB rotates at a constant velocity of w= 150

rad/s .

Find: The velocity of point P when = 30.

Plan: Define x as a function of and differentiate withrespect to time.

EXAMPLE II (continued)


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( )

vP= -0.2wsin + (0.5)[(0.75)2

(0.2sin )2]-0.5(-2)(0.2sin )(0.2cos ) w

vP= -0.2wsin [0.5(0.2)2 sin2w] / (0.75)2(0.2 sin )2

At = 30, w= 150 rad/s and vP= -18.5 ft/s = 18.5 ft/s

xP= 0.2 cos + (0.75)2(0.2 sin )2Solution:

ATTENTION QUIZ


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ATTENTION QUIZ

2. If vA=10 m/s and aA=10 m/s2, determine

the angular acceleration, a,when = 30.

A) 0 rad/s2 B) -50.2 rad/s2

C) -112 rad/s2 D) -173 rad/s2

1. The sliders shown below are confined to move in the

horizontal and vertical slots. If vA=10 m/s, determine theconnecting bars angular velocity when = 30.

A) 10 rad/s B) 10 rad/s

C) 8.7 rad/s D) 8.7 rad/s

READING QUIZ


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READING QUIZ

1. When a relative-motion analysis involving two sets of

coordinate axes is used, the x - y coordinate system willA) be attached to the selected point for analysis.

B) rotate with the body.

C) not be allowed to translate with respect to the fixed frame.

D) None of the above.

2. In the relative velocity equation, vB/A is

A) the relative velocity of B with respect to A.

B) due to the rotational motion.

C) rB/A.

D) All of the above.

APPLICATIONS


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APPLICATIONS

As the slider block A moves horizontally to the left with vA, it

causes the link CB to rotate counterclockwise. Thus vBis directed

tangent to its circular path.

Which link is undergoing general plane motion? Link AB or

link BC?

How can the angular velocity, of link AB be found?


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RELATIVE MOTION ANALYSIS


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(Section 16.5)

When a body is subjected to general plane motion, it undergoes a

combination of translationand rotation.

drB = drA + drB/A

Disp. due to translation and rotation

Disp. due to translation

Disp. due to rotation

Point A is called thebase pointin this analysis. It generally has a

knownmotion. The x- y frame translates with the body, but does not

rotate. The displacement of point B can be written:

RELATIVE MOTION ANALYSIS: VELOCITY


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The velocity at B is given as :(drB/dt) = (drA/dt) + (drB/A/dt)or

vB=vA+vB/A

Since the body is taken as rotating about A,

vB/A = drB/A/dt = rB/A

Here will only have a kcomponent since the axis of rotation

isperpendicularto the plane of translation.


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RELATIVE MOTION ANALYSIS: VELOCITY


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Furthermore, point B at the center of the wheel moves along a

horizontal path. Thus, vBhas a known direction, e.g., parallel

to the surface.

vB= vA + rB/A

When a wheel rolls without slipping, point A is often selected

to be at the point of contact with the ground.

Since there is no slipping, point A has zero velocity.

(continued)



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3. Write the scalar equations from the x and y components of

these graphical representations of the vectors. Solve for

the unknowns.

1. Establish the fixed x-y coordinate directions and draw a

kinematic diagramfor the body. Then establish the

magnitude and direction of the relative velocity vector vB/A.

Scalar Analysis:

2. Write the equation vB= vA + vB/A. In the kinematic diagram,

represent the vectors graphically by showing their

magnitudes and directions underneath each term.

The relative velocity equationcan be applied using either a

Cartesian vector analysis or by writing scalar x and y component

equations directly.



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Vector Analysis:

3. If the solution yields a negativeanswer, the sense of

direction of the vector is oppositeto that assumed.

2. Express the vectors in Cartesian vector form(CVN) and

substitute them into vB= vA + rB/A. Evaluate the cross

product and equate respective iandjcomponents to obtain

twoscalar equations.

1. Establish the fixed x - y coordinate directions and draw the

kinematic diagramof the body, showing the vectors vA, vB,

rB/Aand . If the magnitudes are unknown, the sense of

direction may be assumed.

(continued)

EXAMPLE I


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Given:Roller A is moving to the

right at 3 m/s.

Find: The velocity of B at the

instant = 30.

Plan:

1. Establish the fixed x - y directions and draw a kinematic

diagram of the bar and rollers.

2. Express each of the velocity vectors for A and B in terms

of their i,j, kcomponents and solve vB= vA + rB/A.

EXAMPLE I (continued)


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( )

Equating the iandjcomponents gives:

0 = 30.75

-vB=1.299

Express the velocity vectors in CVNvB= vA + rB/A

-vBj= 3 i + [ k

(-1.5cos30i+1.5sin 30j)]

-vBj= 3 i1.299 j0.75 i

Solution:

Solving: = 4 rad/s or = 4 rad/s k

vB= 5.2 m/s or vB= -5.2 m/sj

yKinematic diagram:

y

EXAMPLE II


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Given:Crank rotates OA with an

angular velocity of 12 rad/s.

Find: The velocity of piston B

and the angular velocity of

rod AB.

Plan:

Notice that point A moves on a circular path. The

directions of vA

is tangent to its path of motion.

Draw a kinematic diagram of rod AB and use

vB= vA+ AB rB/A.

EXAMPLE II (continued)S l ti


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By comparing the i,jcomponents:

i: 0 = -3.6 + 0.3 AB AB = 12 rad/s

j: vB = 0.5196AB vB = 6.24 m/s

Rod AB. Write the relative-velocity

equation:vB= vA+ AB rB/A

Solution:

Since crack OA rotates with an angularvelocity of 12 rad/s, the velocity at A

will be: vA= -0.3(12) i= -3.6 i m/s

Kinematic diagram of AB:

vBj = -3.6 i + AB k (0.6cos30 i 0.6sin30j )

vBj = -3.6 i + 0.5196 ABj+ 0.3 AB i

CHECK YOUR UNDERSTANDING QUIZ


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CHECK YOUR UNDERSTANDING QUIZ

1. If the disk is moving with a velocity at point

O of 15 ft/s and

= 2 rad/s, determine thevelocity at A.

A) 0 ft/s B) 4 ft/s

C) 15 ft/s D)11 ft/s

2. If the velocity at A is zero, then determine the angular

velocity, .

A) 30 rad/s B) 0 rad/s

C) 7.5 rad/s D) 15 rad/s

2 ftV=15 ft/s

A

O

Dynamics Test 2

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