dynamics 8-1 - valparaiso university slides/dynamicsslides.pdfprofessional publications, inc. ferc...

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8-1DynamicsOverview

Dynamics—the study of moving objects.

Kinematics—the study of a body’s motion independent of the forces on the body.

Kinetics—the study of motion and the forces that cause motion.


8-2DynamicsKinematics—Rectangular Coordinates


8-3a1DynamicsKinematics—Polar Coordinates


8-3a2DynamicsKinematics—Polar Coordinates


8-4aDynamicsKinematics—Circular Motion


8-4b1DynamicsKinematics—Circular Motion

Angular velocity =

Angular acceleration =

Tangential acceleration =

Normal acceleration =


8-4b2DynamicsKinematics—Circular Motion

Example (FEIM):A turntable starts from rest and accelerates uniformly at 1.5 rad/s2. Howmany revolutions does it take for the rotational frequency to reach 33.33rpm?

!

" = 2#f = 2#rad

rev

$

% &

'

( ) 33.33

rev

min

$

% &

'

( )

1 min

60 s

$

% &

'

( ) = 3.49

rad

s

!

" =f

0

t

# $dt =f

0

t

# %tdt =%t

f

2

2

!

t ="

#

!

" =#

2

2$=

3.49 rad

s( )2

2( ) 1.5 rad

s2( )

= 4.06 rad

!

n =4.06 rad

2"rad

rev

= 0.65 revolution


8-5a1DynamicsKinematics—Projectile Motion

!

s = s0

+ v0t + 1

2at

2

v = v0

+ a0t

v2 = v

0

2 + 2a0

s " s0( )

Constant acceleration formulas:



Example 1 (FEIM):A projectile is launched at 180 m/s at a 30° incline. The launch pointis 150 m above the impact plane. Find the maximum height, flighttime, and range.



!

" h max

= v0sin# + at = 0

tmax

= $v0

sin#

a= $

180m

s

%

& '

(

) * 0.5( )

$9.8m

s

%

& '

(

) *

= 9.18 s

h = h0+ v

0t sin# + 1

2at

2

hmax

= 150 m + 180m

s2

%

& '

(

) * (9.18 s)(0.5)$ 1

2( ) 9.8m

s2

%

& '

(

) * (9.18 s)2 = 563 m

himpact

= 0 = 150 m + 90t $ 4.9t2

By the quadratic equation, timpact

= 19.9 s

cos30° = 0.866

Rimpact

= R0+ v

0t

impactcos#

= 0 + 180m

s

%

& '

(

) * (19.9 s)(0.866) = 3100 m


8-5b1DynamicsKinematics—Projectile Motion

Example 2 (FEIM):A bomber flies horizontally at 275 km/h and an altitude of 3000 m. Atwhat viewing angle from the bomber to the target should the bomb bedropped?


8-5b2DynamicsKinematics—Projectile Motion

!

himpact

= 1

2at2 = 1

2( ) "9.8m

s2

#

$ %

&

' ( t

2 = "3000 m

t =2h

impact

g= 24.7 s

275km

hr

#

$ %

&

' ( 1000

m

km

#

$ %

&

' (

1 hr

3600 s

#

$ %

&

' ( = 76.39 m/s

Rimpact

= R0

+ v0t = 0 + 76.39

m

s

#

$ %

&

' ( 24.7 s( ) = 1887 m

) = arctanR

impact

himpact

= arctan1887 m

3000 m= 32.2°


8-6aDynamicsKinetics—Newton’s 2nd Law of Motion

For a constant mass,

One-dimension motion


8-6b1DynamicsKinetics—Newton’s 2nd Law of Motion

Example (FERM prob. 6, p. 15-5):


8-6b2DynamicsKinetics—Newton’s 2nd Law of Motion

Example (FERM prob. 6, p. 15-5):


8-7aDynamicsKinetics—Impulse and Momentum

Impulse (constant mass in one dimension)

Momentum

Impulse-Momentum Principle


8-7b1DynamicsKinetics—Impulse and Momentum

Example 1 (FEIM):

A 0.046 kg marble attains a velocity of 76 m/s in a slingshot. Contactwith the slingshot is 1/25 of a second. What is the average force onthe marble during the launch?

!

Fave

=m"v

"t=

(0.046 kg) 76m

s

#

$ %

&

' (

0.04 s= 87.4 N


8-7b2DynamicsKinetics—Impulse and Momentum

Example 2 (FEIM):

A 2000 kg cannon fires a 10 kg projectile horizontally at 600 m/s. Ittakes 0.007 s for the projectile to pass through the barrel. What is therecoil velocity if the cannon is not restrained? What average forcemust be exerted on the cannon to keep it from moving?

!

F =m"v

"t=

(10 kg)(600m

s)

0.007 s= 8.57#105 N

!

mprojectile

"vprojectile

= mcannon

"vcannon

(10 kg)(600m

s) = (2000 kg)(v

cannon)

vcannon

= 3m

s= initial recoil velocity


8-8aDynamicsWork & Energy

Work

Kinetic Energy of a Mass

Kinetic Energy of a Rotating Body

Potential Energy• Gravity

• Spring (linear)Fs = kx where the spring iscompressed a distance x

!

W = KE2"KE

1= 1

2m(v

2

2" v

1

2)

!

W = KE2"KE

1= 1

2I(#

2

2" #

1

2)

!

W = PE2"PE

1= mg(h

2" h

1)

!

W = PE2"PE

1= 1

2k(x

2

2" x

1

2)


8-8bDynamicsWork & Energy

Conservation of Energy• For a closed system (no external work), the change in potential

energy equals the change in kinetic energy.

!

PE1"PE

2= KE

2"KE

1

PE1+KE

1= PE

2+KE

2

• For a system with external work, W equals ΔPE + ΔKE.

!

W1"2

= (PE1#PE

2) + (KE

2#KE

1)

PE1+KE

1+W

1"2= PE

2+KE

2


8-8cDynamicsWork & Energy

Impacts: Momentum is alwaysconserved.Elastic Impacts: Kinetic energy isconserved.

!

m1v

1+ m

2v

2= m

1" v 1+ m

2" v 2

m1

= m2sov

1+ v

2= " v

1+ " v

2= 0.85# 0.53 = 0.32

1

2m v

1

2+ 1

2m v

2

2= 1

2m " v

1

2

+ 1

2m " v

2

2

v1

2+ v

2

2= " v

1

2

+ " v 2

2

Solving two equations and two unkowns:

Therefore, (A) is correct.

!

" v 1

= #0.53 m/s

" v 2

= 0.85 m/s

Example 1 (FEIM):Two identical balls collide alongtheir centerlines in an elasticcollision. The initial velocity ofball 1 is 0.85 m/s. The initialvelocity of ball 2 is –0.53 m/s.What is the relative velocity ofeach ball after the collision?

(A) –0.53 m/s and 0.85 m/s (B) –0.72 m/s and 1.2 m/s (C) –5.1 m/s and 1.2 m/s (D) 0.98 m/s and 1.8 m/s


8-8dDynamicsWork & Energy

Example 2 (FEIM):Ball A of 200 kg is traveling at 16.7 m/s. It strikes stationary ball B of200 kg along the centerline. What is the velocity of ball A after the collision? Assume the collision is elastic.(A) –16.7 m/s(B) –8.35 m/s(C) 0(D) 8.35 m/s

There are two possible solutions for these equations.

Since there must be a change in the collision, ball A’s velocity must be 0.Therefore, (C) is correct.

!

mA

= mBsov

A+ v

B= " v

A+ " v

B= 16.7 m/s

vA

2+ v

B

2= " v

A

2

+ " v B

2

!

" v A

= 0, " v B

= 16.7 m/s

or

" v A

= 16.7 m/s, " v B

= 0


8-8eDynamicsWork & Energy

Inelastic Impacts:Kinetic energy does not have to beconserved if some energy is convertedto another form.

Example 1 (FEIM):A ball is dropped from an initial heightho. If the coefficient of restitution is0.90, how high will the ball rebound?(A) 0.45ho(B) 0.81ho(C) 0.85ho(D) 0.9ho

Since there must be a change in thecollision, ball A’s velocity must be 0.Therefore, (B) is correct.

!

" v 1# " v

2= #e(v

1# v

2)

where e = coefficient of restitution

" v 1=

m2v

2(1+ e) + (m

1# em

2)v

1

m1+ m

2

" v 2

=m

1v

1(1+ e) + (em

1#m

2)v

2

m1+ m

2

!

mgh = 1

2mv2

v = 2gho

" v = #ev = #e 2gho

= 2g " h

" h = e2ho

= (0.9)2ho

= 0.81ho


8-8fDynamicsWork & Energy

Example 2 (FEIM):Two masses collide in a perfectly inelastic collision. What is the velocity of the combined mass after collision?

(A) 0(B) 4 m/s(C) –5m/s(D) 10 m/s

Therefore, (B) is correct.

!

m1

= 4m2 v

1= 10 m/s v

2= "20 m/s

!

m1v

1+ m

2v

2= m

3v

3

m3

= m1+ m

2= 5m

2

4m2

10m

s

"

# $

%

& ' + m

2(20

m

s

"

# $

%

& ' = 5m

2v

3

5m2v

3= 40

m

s

"

# $

%

& ' m2

( 20m

s

"

# $

%

& ' m2

v3

= 4 m/s

“Perfectly inelastic” means the masses collide and stick together.


8-9aDynamicsKinetics

Friction

Example (FEIM):A snowmobile tows a sled with a weight of 3000 N. It accelerates up a15° slope at 0.9 m/s2. The coefficient of friction between the sled and thesnow is 0.1. What is the tension in the tow rope?

!

Fslope

= Frope

" (Ffriction

+ Fgravity

) = maslope

Frope

= Ffriction

+ Fgravity

+ maslope

!

= mg sin15° + mgµ cos15° + maslope

= (3000)(0.2588) + (3000 N)(0.1)(0.9659)

+3000 N

9.8m

s2

"

#

$ $ $

%

&

' ' '

0.9m

s2

"

# $

%

& '

= 1342 N


8-9b1DynamicsKinetics

Plane Motion of a Rigid BodySimilar equations can be written for the y-direction or any othercoordinate direction.

Example (FEIM):A 2500 kg truck skids with a deceleration of 5 m/s2. What is thecoefficient of sliding friction? What are the frictional forces and normalreactions (per axle) at the tires?


8-9b2DynamicsKinetics

The force of deceleration is equal to the friction force.

The moment about the center of gravity, MA, must be equal to zero.

!

Fdeceleration

= (2500 kg) 5m

s2

"

# $

%

& ' = F

friction= µmg

!

MA

=" (4.5 m)NB# (2 m)mg # (1 m)(F

deceleration) = 0

!

= (4.5 m)NB" (2 m)(2500 kg) 9.8

m

s2

#

$ %

&

' ( " (2500 kg) 5

m

s2

#

$ %

&

' ( = 0

!

NB

= 13,667 N

!

NA

= mg "NB

= (2500 kg) 9.8m

s2

#

$ %

&

' ( "13667 N = 10833 N

!

µ = 5m

s2

"

# $

%

& ' 9.8

m

s2

"

# $

%

& ' = 0.51

!

= µ(2500 kg) 9.8m

s2

"

# $

%

& '


8-10a1DynamicsRotation

Rotation about Fixed Axis



There are three simultaneous equations for the movement of the mass,the cylinder, and the relationship between the two:

Example (FEIM):A mass m is attached to a rope wound around a cylinder of mass mCand radius r. What is the acceleration of the falling mass? What is therope tension?

!

mg "F = ma

Fr = I#

#r = a



Rearranging,

!

Fr = Ia

r

!

mg "m

C

2a = ma

!

F =m

Cr

2a

2r2

=m

C

2a

!

a = gm

m +m

C

2

Solving for F,

!

F =m

C

2a = g

mCm

2 m +m

C

2

"

# $

%

& '

=mm

C

2m + mC

g


8-10bDynamicsRotation

Centripetal Force• The force required to keep a body rotating about an axis.

(r is the distance from the center of mass to the center of rotation.)

Centrifugal Force• The “reaction” to centripetal force.• The centrifugal force, like any inertia force, should not be used in

free-body diagrams.

Example (FEIM):A 2000 kg car travels 65 km/hr around a curve of radius 60 m.

What is the centripetal force?

!

65 km/hr = 18.06 m/s

Fc

=m v2

r=

(2000 kg) 18.06m

s

"

# $

%

& '

2

60 m= 10900 N


8-10cDynamicsRotation

Banking Curves

!

64km

hr

"

# $

%

& ' 1000

m

km

"

# $

%

& '

1hr

3600 s

"

# $

%

& ' = 17.8

m

s

( = arctanv2

gr

"

# $

%

& ' = arctan

17.8m

s

"

# $

%

& '

9.8m

s

"

# $

%

& ' (150 m)

"

#

$ $ $ $

%

&

' ' ' '

= 12.1°

Example (FEIM):A 2000 kg car travels at 64 km/hr around a banked curve with a radiusof 150 m. What should the angle between the roadway and thehorizontal be so tire friction is not needed to prevent sliding?


8-10dDynamicsRotation

Free Vibration Spring and Mass:

Natural Frequency:

Solution:

For initial conditions x(0) = x0 and x'(0) = v0,

For initial conditions x(0) = x0 and x'(0) = 0,

Torsional Free Vibration:

Natural frequency:

Solution:

!

"n

=k

t

I


8-10e1DynamicsRotation

Example (FEIM):A 54 kg mass is supported by three springs, as shown. The startingposition is 5.0 cm down from the equilibrium position. No externalforces act on the mass after it is released. What are the maximumvelocity and acceleration?



From the solution to the differential equation,

!

x(t) = x0cos"

nt +

v0

"n

#

$ %

&

' ( sin"

nt

v(t) = ) x (t) = *x0"

nsin"

nt + v

0cos"

nt

!

But, v0

= 0; so v(t) = "x0#

nsin#

nt

!

x0

= 0.05 m

k = k1+ k

2+ k

3= 1750

N

m+1750

N

m+ 4375

N

m= 7875

N

m

!

= "0.604 m/s sin12.08t

!

"n

=k

m=

7875 N

m54 kg

= 12.08 rad/s

v = # x = ($0.05 m) 12.08 rad

s

%

& '

(

) * sin12.08t



The maximum velocity is when sin 12.08t = 1. Because the motion isoscillatory, the maximum velocity occurs in both directions.

!

vmax

= ±0.604 m/s

a = " v = (#0.05 m) 12.08 rad

s

$

% &

'

( )

2

cos12.08t

amax

= ±7.30 m/s2


8-11a1DynamicsReview

A yoyo (mass m, inertia I about center of gravity) is placed on ahorizontal surface with a coefficient of friction µ. The string, wrappedunderneath, is pulled with a force F. Determine if the yoyo will slip on thefloor and which direction it will rotate as a function of F.

The two equations for the movement of the center of gravity and therotation are

!

F + R = ma

Fri+ Rr

o= I"

Example:



!

No slip: R < µmg " a = #ro$

There are two possibilities for thethird equation: slip or no slip.

Solving these 3 equations with 3unknowns (α, a, and R) leads to

!

" = #Fr

o# r

i

I + mro

2$" < 0

!

a =Fr

or

o" r

i( )I + mr

o

2#a > 0

!

R = "FI + mr

i

2

I + mro

2

The yoyo is moving to the right whilewrapping itself on the string. Thesolution for R gives the condition forno slip.

!

F < µmgI + mr

o

2

I + mri

2



The yoyo always moves to the right, but can rotate forward or backarddepending on the value for F.

Slip: R = –µmgSolving these 3 equations lead to

!

a =F "µmg

m

!

=Fr

i"µmgr

o

I

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