FE Review

Dynamics

G. MauerUNLV

Mechanical Engineering

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

horiz. distance d = 20 m

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

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 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

horiz. distance d = 20 m

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:

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:

Polar coordinates

Polar coordinates

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

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

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

Example cont’d: Problem 2.198 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

2.9 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

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!

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!

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

Problem 3.27 in Book:Find accel of Mass AStart with:

(A)Newton’s Law for A.(B)Newton’s Law for A

and B(C) Free-Body analysis of

A and B(D) Free-Body analysis of

A

Problem 3.27 in Book:Find accel of Mass AStart with:

(A)Newton’s Law for A.(B)Newton’s Law for A

and B(C) Free-Body analysis of

A and B(D) Free-Body analysis of

A

Problem 3.27 in Book cont’dNewton applied to mass B gives:

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

(D)Fu = 2T- mB*g-2T = 0

Problem 3.27 in Book cont’dNewton applied to mass B gives:

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

(D)Fu = 2T- mB*g-2T = 0

Problem 3.27 in Book cont’dNewton applied to mass A gives:

(A)Fx = 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

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

Problem 3.27 in Book cont’dNewton applied to mass A gives:

(A)Fx = 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

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

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(B)20 m(C) 80 m(D) 10 m(E) none of the above

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

Conservative Forces

A conservative force is one for which the work done is independent of the path taken

Another way to state it:The work depends only on the initial

and final positions,not on the route taken.

Conservative Forces

T1 + V1 = T2 + V2

Potential Energy

Potential energy is energy which results from position or configuration. An object may have the capacity for doing work as a result of its position in a gravitational field. It may have elastic potential energy as a result of a stretched spring or other elastic deformation.

Potential Energy

elastic potential energy as a result of a stretched spring or other elastic deformation.

Potential Energy

Potential Energy

y

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 work done by friction is

(A) -150 Nm(B) 150 Nm(C) 250 Nm(D) -250 Nm(E) 500 Nm

(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

vmrHo

Angular Momentum

Linear Momentum

vmG

Rot. about Fixed Axis Memorize!

rωr

dt

dv

Page 336:

at = x r

an = x ( x r)

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

16.4 Motion Analysis

http://gtrebaol.free.fr/doc/flash/four_bar/doc/

Approach

1. Geometry: DefinitionsConstants

Variables

Make a sketch

2. Analysis: Derivatives (velocity and acceleration)

3. Equations of Motion

4. Solve the Set of Equations. Use Computer Tools.

Example

Bar BC rotates at constant BC. Find the angular Veloc. of arm BC.

Step 1: Define the Geometry

Example

Bar BC rotates at constant BC. Find the ang. Veloc. of arm BC.Step 1: Define the Geometry

A

i

JB

C

(t) (t)

vA(t)

O

Geometry: Compute all lengths and angles as f((t))

All angles and distance AC(t) are time-variant

A

i

JB

C

(t) (t)

vA(t)

O

Velocities: = -dot is given.

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

Analysis: Solve the rel. Veloc. Vector equation conceptually

Seen from O: vA = x OA

A

i

JB

C

(t)

vA(t)

O (t)

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

Analysis: Solve the rel. Veloc. Vector equation

Seen from O: vA = x OA

A

i

JB

C

(t)

vA(t)

O (t)

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

Analysis: Solve the rel. Veloc. Vector equation

Seen from C: vCollar + BC x AC(t) A

i

JB

C

(t)

O (t)

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

BC x AC(t)

vA,rel

Analysis: Solve the rel. Veloc. Vector equation numerically

A

i

JB

C

(t)

vA(t)

O (t)

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:

OA

0

0

wOA

rA

l cos ( )

l sin ( )

0

BC

0

0

wBC

rAC

l cos ( )

l sin ( )

0

Analysis: Solve the rel. Veloc. Vector equation numerically

A

i

JB

C

(t)

vA(t)

O (t)

Vector Analysis: OA rA vCOLL BC rAC

LEFT_i l wOA sin ( ) RIGHT_i l wBC sin ( ) vcoll cos ( )

LEFT_j l wOA cos ( ) RIGHT_j l wBC sin ( ) vcoll sin ( )

Here: BC is given as -2 rad/s (clockwise). Find OA

Analysis: Solve the rel. Veloc. Vector equation numericallyA

i

JB

C

(t)

vA(t)

O (t)

Solve the two vector (i and j) equations :

Given

l wOA sin ( ) l wBC sin ( ) vcoll cos ( )

l wOA cos ( ) l wBC sin ( ) vcoll sin ( )

vec Find wOA vcoll( ) vec4.732

0.568

A

i

JB

C

(t)

vA(t)

O (t)

Recap: The analysis is becoming more complex.

•To succeed: TryClear Organization from the start

•Mathcad

•Vector Equation = 2 simultaneous equations, solve simultaneously!

fig_05_011

16.6 Relative Velocity

vA = vB + vA/B

Relative Velocity

vA = vB + vA/B

= VB (transl)+ vRot

vRot = x r

Seen from O:vB = x rSeen from A:

vB = vA + AB x rB/A

Seen from O:vB = x rSeen from A:

vB = vA + AB x rB/A

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

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

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)

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

i

J

B

AvA = const

Centrip. r* AB 2

r*

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

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

r* AB2 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)

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*

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

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

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

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

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

Find: a B and AB

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

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

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:

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

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