Final Exam Review. Please Return Loan Clickers to the MEG office after Class! Today!

Final Exam Review

Please Return Loan Clickers to the MEG office after Class!

Today!

Always work from first Principles!

Review

Always work from first Principles!Kinetics:

Free-Body AnalysisNewton’s Law

Constraints

Review

g

A

i

JUnit vectors

B

L

G

1. Free-Body

Review

g

A

i

JUnit vectors

B

L

G

1. Free-Body

B_x

B_y

mg

g

A

i

JUnit vectors

B

L

G

2. Newton

B_x

B_y

mg

Moments about B: -mg*L/2 = IB*a with IB = m*L2/3

g

A

i

JUnit vectors

B

L

G

3. Constraint

B_x

B_y

mg

aG = *L/2 = -g*3/4

1. Free-Body

aCart,x = const

A

i

J

g

R= 0.8m

h= 0.05m

b

R R-hmg

A_yA_x

N

aCart,x = const

A

i

J

g

R= 0.8m

h= 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= 0.8m

h= 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

Kinematics (P. 16-126)

CTR

4r

-2r*i + 2r*j

Feedback

Overall, when comparing traditional Homework formats with Mastering, I

prefer

(A) Paper submission of Homework (B) Electronic Submission

Feedback

For me, the most useful benefit of Mastering is

(A) Hints while developing the solution to a problem

(B) Instant grading of results(C) Practice Exams

A (x0,y0)

B (d,h)v

0g

horiz.

distance = dx

yh

X-Y Coordinates

Point Mass Dynamics

A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters.

The travel time t to Point B is

(A) t = 4 s(B) t = 1 s(C) t = 0.5 s(D) t = 2 sUse g = 10 m/s2

A (x0,y0)

B

v0 g

horiz. distance d = 20 m

h

x

y

Use g = 10 m/s2

A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters.

The travel time t to Point B is

(A) t = 4 s(B) t = 1 s(C) t = 0.5 s(D) t = 2 sUse g = 10 m/s2

A (x0,y0)

B

v0 g


h

x

y

Use g = 10 m/s2

st

tsmm

tgtvty

2

*/10*5.020

**5.0*)sin(*0)(22

2

A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters at time t = 2s.

The start velocity v0 is

(A) v0 = 40 m/s(B) v0 = 20 m/s(C) v0 = 10 m/s(D) v0 = 5 m/s

A (x0,y0)

B

v0 g


h

x

y

Use g = 10 m/s2

A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters at time t = 2s.

The start velocity v0 is

(A) v0 = 40 m/s(B) v0 = 20 m/s(C) v0 = 10 m/s(D) v0 = 5 m/s

A (x0,y0)

B

v0 g


h

x

y

Use g = 10 m/s2

smv

svm

tvtx

/100

2*1*020

*)cos(*0)(

12.7 Normal and Tangential Coordinatesut : unit tangent to the pathun : unit normal to the path

Normal and Tangential CoordinatesVelocity Page 53 tusv *

Normal and Tangential Coordinates

Fundamental Problem 12.27 ttn uau

va **

2

(A) constant•(B) 1 m/s2

•(C) 2 m/s2

•(D) not enough information•(E) 4 m/s2

The boat is traveling along the circular path with = 40m and a speed of v = 0.5*t2 , where t is in seconds. At t = 4s, the normal acceleration is:

http://session.masteringengineering.com/myct/problemView?preview=item&problemID=1137833&frame=banner&openedFrom=itemLibrary

Fundamental Problem 12.27 ttn uau

va **

2

(A) constant•(B) 1 m/s2

•(C) 2 m/s2

•(D) not enough information•(E) 4 m/s2

2/1*2_:2__

*5.0*2/

smastAt

tdtdva

t

t

The boat is traveling along the circular path with = 40m and a speed of v = 0.5*t2 , where t is in seconds. At t = 4s, the normal acceleration is:

http://session.masteringengineering.com/myct/problemView?preview=item&problemID=1137833&frame=banner&openedFrom=itemLibrary

Polar coordinates

Polar Coordinates

Point P moves on a counterclockwise circular path, with r =1m, dot(t) = 2 rad/s. The radial and tangential accelerations are:•(A) ar = 4m/s2 a = 2 m/s2

•(B) ar = -4m/s2 a = -2 m/s2

•(C) ar = -4m/s2 a = 0 m/s2

•(D) ar = 0 m/s2 a = 0 m/s2

e

Unit vectors

Cr

disk = 10 rad/s

er

B

Point B moves radially outward from center C, with r-dot =1m/s, dot(t) = 10 rad/s. At r=1m, the radial acceleration is:•(A) ar = 20 m/s2

•(B) ar = -20 m/s2

•(C) ar = 100 m/s2

•(D) ar = -100 m/s2

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

Given:r(t) = 2+2*sin((t)), dot=

constantThe radial velocity is

(A) 2+2*cos((t ))*-dot,(B) -2*cos((t))*-dot(C) 2*cos((t))*-dot(D) 2*cos((t))(E) 2* +2*cos((t ))*-dot

Given:r(t) = 2+2*sin((t)), dot= constantThe radial velocity is

(A) 2+2*cos((t ))*-dot,(B) -2*cos((t))*-dot(C) 2*cos((t))*-dot(D) 2*cos((t))(E) 2* +2*cos((t ))*-dot

Constrained Motion

L

B

A

i

J

vA = const

vA is given as shown.Find vB

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

(transl. + rot.)

Vr = 150 mm

The conveyor belt is moving to the left at v = 6 m/s. The angular velocity of the drum (Radius = 150 mm) is

(A) 6 m/s(B) 40 rad/s(C) -40 rad/s(D) 4 rad/s(E) none of the above

yE

Xc

c

XBxA

A B

The rope length between points A and B is:

•(A) xA – xB + xc

•(B) xB – xA + 4xc

•(C) xA – xB + 4xc

•(D) xA + xB + 4xc

Omit all constants!

Given: v0 = const.The vertical velocity

component of point A (in y-direction) is

(A)vA,y = v0*tan()(B) vA,y = v0*cot()(C) vA,y = v0*cos((D)vA,y = 2*v0

Given: v0 = const.The velocity of point A

in vertical y-direction is

(A)vA,y = v0*tan()(B) vA,y = v0*cot()(C) vA,y = v0*cos((D)vA,y = 2*v0

(E) vA,y = v0/cos()

Vy/vx =cot

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.

Dynamics

M1: up as positive:Fnet = T - m1*g = m1 a1

M2: down as positive.Fnet = F = m2*g - T = m2 a2

3. Constraint equation:a1 = a2 = a

Equations

From previous:T - m1*g = m1 a

T = m1 g + m1 a Previous for Mass 2:m2*g - T = m2 a

Insert above expr. for Tm2 g - ( m1 g + m1 a ) = m2 a

( m2 - m1 ) g = ( m1 + m2 ) a( m1 + m2 ) a = ( m2 - m1 ) g

a = ( m2 - m1 ) g / ( m1 + m2 )

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

-M*g*cosj

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

-M*g*cosj

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

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

N

Friction F = k*N:Another horizontal

reaction is added in negative x-direction.

0 = 30 0

g

i

J

m

M*g

M*g*sin

-M*g*cosj

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

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

N k*N

Mass m rests on the 30 deg. Incline as shown. The free-body reaction seen by the incline in j-direction is

(A) -mg*sin30o

(B) +mg*sin30o (C) -mg*cos30o

(D) +mg*cos30o

(E) None of the above

0 = 30 0

g

i

J

m

Mass m rests on the 30 deg. Incline as shown. The free-body reaction seen by the incline in j-direction is

(A) -mg*sin30o


(D) +mg*cos30o


0 = 30 0

g

i

J

m

mg

mg*cos()

Mass m rests on the 30 deg. Incline as shown. The static friction required to keep the mass from sliding in i-direction is

(A) -mg*sin30o


(D) +mg*cos30o


0 = 30 0

g

i

J

m

Mass m rests on the 30 deg. Incline as shown. The static friction required to keep the mass from sliding in i-direction is

(A) -mg*sin30o


(D) +mg*cos30o


0 = 30 0

g

i

J

m

mg*sin()

Newton applied to mass B gives:

(AFu = 2T = mB*aB (B) Fu = -2T + mB*g = 0(C) Fu = mB*g-2T = mB*aB

(DFu = 2T- mB*g-2T = 0

(AFu = 2T = mB*aB (B) Fu = -2T + mB*g = 0(C) Fu = mB*g-2T = mB*aB

(DFu = 2T- mB*g-2T = 0

Newton applied to mass B gives:

(AFx = T +F= mA*ax ; Fy = N - mA*g*cos(30o) = 0(B) Fx = T-F= mA*ax Fy = N- mA*g*cos(30o) = mA*ay

(C) Fx = T = mA*ax ; Fy = N - mA*g*cos(30o) =0

(DFx = T-F = mA*ax ; Fy = N-mA*g*cos(30o) =0

Newton applied to mass A gives:

Energy Methods

Only Force components in direction of motion do WORK

oductScalar

rdFdW

Pr_

Work of

Gravity

Work of a

Spring

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.

A car is traveling at 20 m/s on a level road, when the brakes are suddenly applied and all four wheels lock. k = 0.5. The total distance traveled to a full stop is (use Energy Method, g = 10 m/s2)

(A) 40 m(B20 m(C) 80 m(D) 10 m(E) none of the above

A car is traveling at 20 m/s on a level road, when the brakes are suddenly applied and all four wheels lock. k = 0.5. The total distance traveled to a full stop is (use Energy Method, g = 10 m/s2)

(A) 40 m(B20 m(C) 80 m(D) 10 m(E) none of the above

metersg

vs

sdistbrakeforSolve

sgmvmTT

k

k

10

400

**2

:____

*****2/1012

20

20

Collar A is compressing the spring after dropping vertically from A. Using the y-reference as shown, the work done by gravity (Wg) and the work done by the compression spring (Wspr) are

(A) Wg <0, Wspr <0(B) Wg >0, Wspr <0(C) Wg <0, Wspr >0(D) Wg >0, Wspr >0

y

Collar A is compressing the spring after dropping vertically from A. Using the y-reference as shown, the work done by gravity (Wg) and the work done by the compression spring (Wspr) are

(A) Wg <0, Wspr <0(B) Wg >0, Wspr <0(C) Wg <0, Wspr >0(D) Wg >0, Wspr >0

y

sdFW

*

Potential Energy

• For any conservative force F we can define a potential energy function U in the following way:

– The work done by a conservative force is equal and opposite to the change in the potential energy function.

• This can be written as: r1

r2 U2

U1

W Fgdr U

U U2 U1 W

rFgd

rr

r1

r2

Hooke’s Law• Force exerted to compress a spring is proportional to the

amount of compression.

Fs kx

PEs 1

2kx 2

Conservative Forces:

Gravity is a conservative force:

• Gravity near the Earth’s surface:

• A spring produces a conservative force: Us 1

2kx2

Ug mgy

Ug GMm

R

(Use Energy Conservation) A 1 kg block slides d=4 m down a frictionless plane inclined at =30 degrees to the horizontal. The speed of the block at the bottom of the inclined plane is

(A) 1.6 m/s(B) 2.2 m/s(C) 4.4 m/s(D) 6.3 m/s(E) none of the

above

dh

(Use Energy Conservation) A 1 kg block slides d=4 m down a frictionless plane inclined at =30 degrees to the horizontal. The speed of the block at the bottom of the inclined plane is

(A) 1.6 m/s(B) 2.2 m/s(C) 4.4 m/s(D) 6.3 m/s(E) none of the

above

dh

PE of block released = KE of block = PE gained by spring

Height dropped, h d sin 4sin 30m 2m

Potential energy released, mgh 19.8m / s2 2m 19.6J

Kinetic energy of the block = 1

2mv2 19.6J

v U2

m 19.6

2

16.3 m/s

A child of mass 30 kg is sliding downhill while the opposing friction force is 50 N along the 5m long incline (3m vertical drop). The change of potential energy is

(A) -882 Nm(B) 882 Nm(C) 1470 Nm(D) -1470 Nm(E) None of the

above

A child of mass 30 kg is sliding downhill while the opposing friction force is 50 N along the 5m long incline (3m vertical drop). The change of potential energy is

Change of PE = mg(hfinal-h0)

= 30*9.8*(0-3) = -882 Nm (A) -882 Nm(B) 882 Nm(C) 1470 Nm(D) -1470 Nm(E) None of the

above

Rot. about Fixed Axis Memorize!

rωr

dt

dv

Page 336:

at = x r

an = x ( x r)

Arm BD is rotating with constant dot= , while point D moves at vD*i. Seen from D, the velocity vector at B is:

(A) vB = vD*i - BD**cosiBD*sinj(B) vB = vD*i - BD**cosi BD*sinj(C) vB = vD*i + BD**cosiBD*sinj(D) vB = - BD**cosiBD*sinj(E) none of the above

i

J

BD

(t)

B

AB

B

D

(t),(t)

A vD(t)

Meriam Problem 5.71Given are: BC wBC 2 (clockwise), Geometry: equilateral triangle

with l 0.12 meters. Angle 60

180 Collar slides rel. to bar AB.

GuessValues:(outwardmotion ofcollar ispositive)

wOA 1

vcoll 1

Vector Analysis: OA rA vCOLL BC 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 OA and vCOLL

Enter vectors:

Mathcad EXAMPLE

Mathcad Example

part 2:Solving the vector equations

Mathcad Examples

part 3Graphical Solution

BC

BC X rAC

OA X rOA

ARM BC: VA= BC X rAC

Right ARM OA:VA =OA X rOA

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

of vector vColl = 60o

Find: vB and AB

Graphical Solution Veloc. of Bi

J

B

AvA = const

ABCounterclockw.

vB

Given: Geometry andVA

vB = vA + vB/A

AB rxvA +vA = const

vA isgiven

vB = ?

AB rx

AB rx

Find: vB and AB

i

J

B

AvA = const

ABCounterclockw.

vB

Given: Geometry andVA

vB = vA + vB/A

AB rxvA +vA = const

vA isgiven

Solution:vB = vA + AB X r

AB rx

AB rx

vB = 3 ft/s down, = 60o

and vA = vB/tanThe relative velocity vA/B is found from vector eq.

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

(BvA = 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 = 3 ft/s down, = 60o

and vA = vB/tanThe relative velocity vA/B is found from vector eq.

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

(BvA = 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

Rigid Body Acceleration

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

Find: a B and AB

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

J

B

AvA = const

ABC o u n te rc lo ck w

.

vB

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

r

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

r* AB2 r* +

r

Find: a B and AB

Look at the Accel. o f B re la tive to A :

W e know:

1. Centripetal: m agnitude r2 anddirection (inward). If in doubt, com putethe vector product x(*r)

i

J

B

AvA = const

C entrip .r* AB

2



r* AB2 r* +

r

Find: a B and AB


W e know:


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

i

J

B

AvA = const

Centrip. r* AB 2

r*



r* AB2 r* +

Find: a B and AB


W e know:


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

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

i

J

B

AvA = const

Centrip. r* AB 2

Angular r*



r* AB2 r* +

aB

r

B

A

vA = const

AB

C entrip . r* AB 2

aB

Angular r*

r is the vector from reference

point A to point B

r

i

J

W e can add graphically:S tart w ith C entipeta l



Find: a B and AB

W e can add graphically:S tart w ith C entipeta l


aB

r* r* AB

2

Result: is < 0 (c lockwise)

aB is negative (to theleft)

B

AvA = const

AB

C entrip . r* A B2

r is the vector from

reference point A to point B

r

i

J

N owC om plete the

Triangle:


Find: a B and AB

The instantaneous center of Arm BD is located at Point:

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

AAB

B

BD

D (t)

(t)

vD(t)i

J

E

O G

F

H

fig_06_002

Plane Motion3 equations: Forces_x Forces_y Moments about G

fig_06_002

Plane Motion3 equations: Forces_x Forces_y Moments about G

*.....:

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

*:

GG

y

x

IMRotation

ymF

xmFnTranslatio

fig_06_005

Parallel Axes TheoremPure rotation about fixed point P

2*dmII GP

Describe the constraint(s) with an Equation

Constrained Motion: The system no longer has all three

Degrees of freedom

6.78

Given: I_G=m*k2=300*1.5^2 = 675 kgm^2. The angular accel of the rocket is

Thrust T = 4 kN

F_y = m*a = 300*8.69N

(A) 0.102 rad/s2

(B) 0.31 rad/s2

(C) 3.1 rad/s2

(D) 5.9 rad/s2

6.78

Given: I_G= 675 kgm^2, m = 300 kg. The angular accel of the rocket is

Thrust T = 4 kN

F_y = m*a = 300*8.69N

(A) 0.102 rad/s2

(B) 0.31 rad/s2

(C) 3.1 rad/s2

(D) 5.9 rad/s2

Answer: (sum of moments about G = I_G*alpha)4000N * sin(1deg)*3m = 675*alpha

alpha = 0.31 rad/s^2

Final Exam Review. Please Return Loan Clickers to the MEG office after Class! Today!

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Transcript of Final Exam Review. Please Return Loan Clickers to the MEG office after Class! Today!