Final Exam Review

Final Exam Review

Please Return Loan Clickers to the MEG office after Class!

Today!

Final ExamWed. Wed. May 14

8 – 10 a.m.

Always work from first Principles!

Review

Always work from first Principles!Kinetics:

Free-Body AnalysisNewton’s Law

Constraints

Review

gA

i

JUnit vectors

B

L

G

1. Free-Body

Review

gA

i

JUnit vectors

B

L

G

1. Free-Body

B_x

B_ymg

gA

i

JUnit vectors

B

L

G

2. Newton

B_x

B_ymg

Moments about B: -mg*L/2 = IB*a with IB = m*L1/3

gA

i

JUnit vectors

B

L

G

3. Constraint

B_x

B_ymg

aG = a*L/2 = -3g/(2L) * L/2 = -3g/4

1. Free-Body

aCart,x = const

A

i

J

g

R= 0.8mh= 0.05m

b

R R-hmg

A_yA_x

N

aCart,x = const

A

i

J

g

R= 0.8mh= 0.05m

b

R R-hmg

A_yA_x

N

2. Newton

Moments about Center of Cylinder:A_xFrom triangle at left:

Ax*(R-h) –b*mg = 0acart*(R-h) –b*g = 0

aCart,x = const

A

i

J

g

R= 0.8mh= 0.05m

b

R R-hmg

A_yA_x

N

2. Newton

N = 0 at impending rolling, thus Ay = mg

Ax = m*acart

Kinematics (P. 16-126)

CTR

4r

-2r*i + 2r*j

A (x0,y0)

B (d,h)v0

g

horiz.

distance = dx

yh

X-Y Coordinates

Point Mass Dynamics

Normal and Tangential CoordinatesVelocity Page 53 tusv *

Normal and Tangential Coordinates

Polar coordinates

Polar coordinates

Polar coordinates

ABAB VVV /

We Solve Graphically (Vector Addition)

12.10 Relative (Constrained) Motion

vB

vA

vB/A

Example : Sailboat tacking against Northern Wind

BoatWindBoatWind VVV /

2. Vector equation (1 scalar eqn. each in i- and j-direction)

500

150

i

Constrained Motion

LB

A

i

JvA = const

vA is given as shown.Find vB

Approach: Use rel. Velocity:vB = vA +vB/A

(transl. + rot.)

NEWTON'S LAW OF INERTIA A body, not acted on by any force, remains in

uniform motion. NEWTON'S LAW OF MOTION

Moving an object with twice the mass will require twice the force.

Force is proportional to the mass of an object and to the acceleration (the change in velocity).

F=ma.

Rules1. Free-Body Analysis, one for each mass

3. Algebra:Solve system of equations for all unknowns

2. Constraint equation(s): Define connections.You should have as many equations as Unknowns.COUNT!

0 = 30 0

g

i

J

m

M*g

M*g*sin*i

-M*g*cos*j

Mass m rests on the 30 deg. Incline as shown. Step 1: Free-Body Analysis. Best approach: use coordinates tangential and normal to the path of motion as shown.

Mass m rests on the 30 deg. Incline as shown. Step 1: Free-Body Analysis.

Step 2: Apply Newton’s Law in each Direction:

0 = 30 0

g

i

J

m

M*g

M*g*sin*i

-M*g*cos*j

xmxForces *i*sin*g*m)_(

)_(0j*cos*g*m-N )_( onlystaticyForces

N

Friction F = mk*N:Another horizontal

reaction is added in negative x-direction.

0 = 30 0

g

i

J

m

M*g

M*g*sin*i

-M*g*cos*j

xmNkxForces *i*)*sin*g*m()_( m

)_(0j*cos*g*m-N )_( onlystaticyForces

N mk*N

Energy Methods

Only Force components in direction of motion do WORK

oductScalarrdFdW

Pr_

The potential energy V is defined as:

The work is defined as

dr*F- W - V

𝑾= 𝑭 ∗𝒅𝒓 𝒐𝒓𝑾=𝑻∗𝒅𝝑

The work-energy relation: The relation between the work done on a particle by the forces which are applied on it and how its kinetic energy changes follows from Newton’s second law.

Work of

Gravity

Conservative Forces:

Gravity is a conservative force:

• Gravity near the Earth’s surface:

• A spring produces a conservative force:U s

12

kx 2

U g mgy

U g GMm

R

Rot. about Fixed Axis Memorize!

rωr

dtdv

Page 336:

at = a x r

an = w x ( w x r)

fig_05_012Meriam Problem 5.71Given are: wBC wBC 2 (clockwise), Geometry: equilateral trianglewith l 0.12 meters. Angle 60

180 Collar slides rel. to bar AB.

GuessValues:(outwardmotion ofcollar ispositive)

wOA 1

vcoll 1

Vector Analysis: wOA rA vCOLL wBC rAC Mathcad does not evaluate cross products symbolically, so the LEFT andRIGHT sides of the above equation are listed below. Equaling the i- and j-terms yields two equations for the unknowns wOA and vCOLL

Enter vectors:

Mathcad EXAMPLE

fig_05_012Mathcad Example

part 2:Solving the vector equations

fig_05_012Mathcad Examples

part 3Graphical Solution

wBC

wBC X rAC wOA X rOA

ARM BC: VA= wBC X rAC

Right ARM OA:VA =wOA X rOA

Collar slides rel. to Arm BCat velocity vColl. The angle

of vector vColl = 60o

Find: vB and wAB

Graphical Solution Veloc. of Bi

J

B

AvA = const

w Counterclockw.

vB

Given: Geometry andVA

vB = vA + vB/A

wAB rxvA +vA = const

vA isgiven

vB = ?

wAB rx

wAB rx

Find: vB and wAB

i

J

B

AvA = const

w Counterclockw.

vB

Given: Geometry andVA

vB = vA + vB/A

wAB rxvA +vA = const

vA isgiven

Solution:vB = vA + wAB X r

wAB rx

wAB rx

vB = 3 ft/s down, = 60o

and vA = vB/tan. The relative velocity vA/B is found from the vector eq.

(A)vA = vB+ vA/B ,vA/B points

() vA = vB+ vA/B ,vA/B points

(C) vB = vA+ vA/B ,vA/B points

(D) VB = vB+ vA/B ,vA/B points

vB vA

x

y

vB

vA

vA/B

The instantaneous center of Arm BD is located at Point:

(A) B(B) D(C) F(D) G(E) H

AAB

B

BD

D (t)

a (t)

vD(t)i

J

E

O G

F

H

Rigid Body Acceleration

Stresses and Flow Patterns in a Steam TurbineFEA Visualization (U of Stuttgart)

Find: a B and aAB

Look at the Accel. o f B re lative to A :i

J

B

AvA = const

wC o u n te rc lo ck w

.

vB

G iven: G eom etry andV A ,aA , vB , wAB

r

aB = aA + aB/A ,centr+ aB/A ,angular

r* wAB2 r* a+

r

Find: a B and aAB

Look at the Accel. o f B re la tive to A :W e know:1. Centripetal: m agnitude rw2 anddirection (inward). If in doubt, com putethe vector product wx(w*r)

i

J

B

AvA = const

C entrip .r* wAB

2



r* wAB2 r* a+

r

Find: a B and aAB


2. T he DIRECTION o f the angular accel(norm al to bar AB )

i

J

B

AvA = const

Centrip . r* wAB 2

r* a



r* wAB2 r* a+

Find: a B and aAB


2. The DIRECTION o f the angular accel(norm al to bar AB)

3. The DIRECTION o f the accel of point B(horizonta l a long the constra int)

i

J

B

AvA = const

Centrip. r* wAB 2

Angular r* a



r* wAB2 r* a+

aB

r

B

A

vA = constw

C entrip . r* wAB 2

aB

Angular r* a

r is the vector from referencepoint A to po int B

r

i

J

W e can add graphically:S tart w ith C entipetal



Find: a B and aAB

W e can add graphically:S tart w ith C entipetal

aB = aA + a B/A ,centr+ aB/A ,angular

aB

r* a r* wAB2

Result:a is < 0 (c lockwise)

aB is negative (to theleft)

B

AvA = const

w

Centrip. r* wAB2

r is the vector fromreference point A to po int B

r

i

J

N owC om plete the

Triangle:


Find: a B and aAB

At =90o, arad = v2/r or v2 = 4.8*0.3 thus v = 1.2

The accelerating Flywheel has R=300 mm. At =90o, the accel of point P is -1.8i -4.8j m/s2. The velocity of point P is

(A) 0.6 m/s(B) 1.2 m/s(C) 2.4 m/s(D) 12 m/s(E) 24 m/s

At =90o, v = 1.2 w = v/r = 1.2/0.3 = 4 rad/s

The accelerating Flywheel has R=300 mm. At =90o, the velocity of point P is 1.2 m/s. The wheel’s angular velocity w is

(A) 8 rad/s(B) 2 rad/s(C) 4 rad/s(D) 12 rad/s(E) 36 rad/s

Accelerating Flywheel has R=300 mm. At =90o, the accel is -1.8i -4.8j m/s2. The magnitude of the angular acceleration a is

(A) 3 rad/s2

(B) 6 rad/s2

(C) 8 rad/s2

(D) 12 rad/s2

(E) 24 rad/s2

At =90o, atangential = a*R or a = -1.8/0.3 = -6 rad/s2

fig_06_002

fig_06_002Plane Motion3 equations:S Forces_xS Forces_yS Moments about G

a*.....:

*.....................

*:

GG

y

x

IMRotation

ymF

xmFnTranslatio

fig_06_005

fig_06_005Parallel Axes TheoremPure rotation about fixed point P

2*dmII GP

Rigid Body Energy MethodsChapter 18 in

Hibbeler, Dynamics

Stresses and Flow Patterns in a Steam TurbineFEA Visualization (U of Stuttgart)

PROCEDURE FOR ANALYSIS

Problems involving velocity, displacement and conservative force systems can be solved using the conservation of energy equation.

• Potential energy: Draw two diagrams: one with the body located at its initial position and one at the final position. Compute the potential energy at each position using

V = Vg + Ve, where Vg= W yG and Ve = 1/2 k s2.

• Apply the conservation of energy equation.

• Kinetic energy: Compute the kinetic energy of the rigid body at each location. Kinetic energy has two components: translational kinetic energy, 1/2m(vG)2, and rotational kinetic energy,1/2 IGw2.

Impulse and Momentum

This equation represents the principle of linear impulse and momentum. It relates the particle’s final velocity (v2) and initial velocity (v1) and the forces acting on the particle as a function of time.

The principle of linear impulse and momentum is obtained by 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 = v2 at t = t2 results in

mv2 – mv1dvmF dtv2

v1

t2

t1

==

PRINCIPLE OF LINEAR IMPULSE AND MOMENTUM(continued)

IMPACT (Section 15.4)

Impact occurs when two bodies collide during a very short time period, causing large impulsive forces to be exerted between the bodies. Common examples of impact are a hammer striking a nail or a bat striking a ball. The line of impact is a line through the mass centers of the colliding particles. In general, there are two types of impact:

Central impact occurs when the directions of motion of the two colliding particles are along the line of impact.Oblique impact occurs when the direction of motion of one or both of the particles is at an angle to the line of impact.

CENTRAL IMPACT

There are two primary equations used when solving impact problems. The textbook provides extensive detail on their derivation.

Central impact happens 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.

CENTRAL IMPACT (continued)

In most problems, the initial velocities of the particles, (vA)1 and (vB)1, are known, and it is necessary to determine the final velocities, (vA)2 and (vB)2. So the first equation used is the conservation of linear momentum, applied along the line of impact.

(mA vA)1 + (mB vB)1 = (mA vA)2 + (mB vB)2

This provides one equation, but there are usually two unknowns, (vA)2 and (vB)2. So another equation is needed. The principle of impulse and momentum is used to develop this equation, which involves the coefficient of restitution, or e.

The coefficient of restitution, e, is the ratio of the particles’ relative separation velocity after impact, (vB)2 – (vA)2, to the particles’ relative approach velocity before 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 e is specified, this relation provides the second equation necessary to solve for (vA)2 and (vB)2.

CENTRAL IMPACT (continued)

OBLIQUE IMPACT

Momentum of each particle is conserved in the direction perpendicular to 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 magnitudes and directions of the final velocities.

Conservation of momentum and the coefficient of restitution equation are applied along the 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:

End of Review

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Final Exam Review