Chapter 12 - site.iugaza.edu.pssite.iugaza.edu.ps/mhaiba/files/2012/01/ch12-Kinetics-of-Particles... · Introduction 2/25/2014 6:39 PM Dr. Mohammad Suliman Abuhaiba, PE •Newton’s

Chapter 12 Kinetics of Particles: Newton’s Second Law

2/25/2014 6:39 PM

Dr. Mohammad Suliman Abuhaiba, PE

Introduction

2/25/2014 6:39 PM


• Newton’s first and third laws are sufficient for the

study of bodies at rest (statics) or bodies in

motion with no acceleration.

• When a body accelerates, Newton’s 2nd law is

required to relate the motion of the body to the

forces acting on it.

2

Introduction

2/25/2014 6:39 PM


Newton’s second law:

• A particle will have an acceleration proportional to

magnitude of resultant force acting on it and in the

direction of the resultant force.

• Resultant of forces acting on a particle = rate of

change of linear momentum of the particle.

• Sum of moments about O of forces acting on a

particle = rate of change of angular momentum of

the particle about O.

3

Newton’s 2nd of Motion

2/25/2014 6:39 PM


• Newton’s 2nd: If resultant force acting

on a particle is not zero, particle will

have an acceleration proportional to

magnitude of resultant & in the

direction of resultant.

• Consider a particle subjected to

constant forces,

ma

F

a

F

a

F mass,constant

3

3

2

2

1

1

4

• If force acting on particle is zero, particle will not

accelerate, i.e., it will remain stationary or continue

on a straight line at constant velocity.

Newton’s 2nd of Motion

2/25/2014 6:39 PM


• When a particle of mass m is acted upon by a

force acceleration of the particle must satisfy ,F

amF

• Acceleration must be evaluated wrt a

Newtonian frame of reference (one

that is not accelerating or rotating).

5

Linear Momentum of a Particle

dt

Ldvm

dt

d

dt

vdmF

• Linear Momentum Conservation Principle:

If resultant force on a particle is zero, the linear

momentum of particle remains constant in both

magnitude & direction.

2/25/2014 6:39 PM


6

Systems of Units

• International System of Units (SI

Units): length (m), mass (kg), and

time (second)

22 s

mkg1

s

m1kg1N1

ft

slb1

sft1

lb1slug1

sft32.2

lb1lbm1

2

22

• U.S. Customary Units: force (lb),

length (ft), and time (second)

2/25/2014 6:39 PM


7

Equations of Motion • Newton’s 2nd law: amF

zmFymFxmF

maFmaFmaF

kajaiamkFjFiF

zyx

zzyyxx

zyxzyx

• For tangential & normal

components,

2vmF

dt

dvmF

maFmaF

nt

nntt

2/25/2014 6:39 PM


8

Dynamic

Equilibrium

2/25/2014 6:39 PM


• Alternate expression of Newton’s

2nd law,

ectorinertial vam

amF

0

• Particle is in dynamic equilibrium.

• Methods developed for particles in

static equilibrium may be applied

• Inertial forces = measure the

resistance that particles offer to

changes in motion, i.e., changes in

speed or direction.

9

Sample Problem 12.1

A 200-lb block rests on a

horizontal plane. Find the

magnitude of the force P

required to give the block

an acceleration of 10 ft/s2

to the right. The

coefficient of kinetic

friction between the block

and plane is mk 0.25.

2/25/2014 6:39 PM


10

Sample

Problem 12.1

N

NF

g

Wm

k

25.0

ft

slb21.6

sft2.32

lb200

2

2

m

x

y

O

Solution:

• Resolve equation of motion for the

block into two rectangular

component equations.

:maFx

lb1.62

sft10ftslb21.625.030cos 22

NP

0yF 0lb20030sin PN

• The two equations may be solved

for the applied force P and the

normal reaction N.

lb1.62lb20030sin25.030cos

lb20030sin

PP

PN

lb151P

2/25/2014 6:39 PM


11

Sample Problem 12.3

The two blocks shown start

from rest. The horizontal

plane and the pulley are

frictionless, and the pulley

is assumed to be of

negligible mass.

Determine the acceleration

of each block and the

tension in the cord.

2/25/2014 6:39 PM


12

Solution:

• Write the kinematic relationships

for the dependent motions and

accelerations of the blocks.

ABAB aaxy21

21

x

y

Sample Problem 12.3

• Write equations of motion for

blocks and pulley.

:AAx amF AaT kg1001

O

2/25/2014 6:39 PM


13

Sample Problem 12.3

:BBy amF

B

B

BBB

aT

aT

amTgm

kg300-N2940

kg300sm81.9kg300

2

22

2

:0 CCy amF

02 12 TT

2/25/2014 6:39 PM


14

Sample Problem 12.3

N16802

N840kg100

sm20.4

sm40.8

12

1

221

2

TT

aT

aa

a

A

AB

A

• Combine kinematic relationships with

equations of motion to solve for

accelerations and cord tension.

ABAB aaxy21

21 AaT kg1001

A

B

a

aT

21

2

kg300-N2940

kg300-N2940

0kg1002kg150N2940

02 12

AA aa

TT

x

y

O

2/25/2014 6:39 PM


15

Sample Problem 12.4

The 12-lb block B starts from

rest and slides on the 30-lb

wedge A, which is supported

by a horizontal surface.

Neglecting friction, determine:

a. Acceleration of the wedge

b. Acceleration of the block

relative to the wedge.

2/25/2014 6:39 PM


16

Sample Problem 12.4

Solution:

• The block is constrained to slide down the

wedge. Therefore, their motions are

dependent. ABAB aaa

:30cos ABABxBx aamamF

30sin30cos

30cos30sin

gaa

aagWW

AAB

ABABB

:30sin AByBy amamF

30sin30cos1 ABB agWWN

2/25/2014 6:39 PM


• Write equations of motion for wedge and

block.

x

y

:AAx amF

AA

AA

agWN

amN

1

1

5.0

30sin

17

Sample

Problem 12.4 AA agWN 15.0

• Solve for accelerations:

30sinlb12lb302

30coslb12sft2.32

30sin2

30cos

30sin30cos2

30sin30cos

2

1

A

BA

BA

ABBAA

ABB

a

WW

gWa

agWWagW

agWWN

2sft07.5Aa

30sinsft2.3230cossft07.5

30sin30cos

22AB

AAB

a

gaa

2sft5.20ABa

2/25/2014 6:39 PM


18

Sample Problem 12.5

The bob of a 2-m pendulum

describes an arc of a circle

in a vertical plane. If the

tension in the cord is 2.5

times the weight of the bob

for the position shown, find

the velocity and

acceleration of the bob in

that position.

2/25/2014 6:39 PM


19

Sample Problem 12.5

Solution:

• Resolve equation of motion for

the bob into tangential &normal

components.

• Solve the component equations

for normal & tangential

accelerations.

:tt maF

30sin

30sin

ga

mamg

t

t

2sm9.4ta

2/25/2014 6:39 PM


20

Sample Problem 12.5

:nn maF

30cos5.2

30cos5.2

ga

mamgmg

n

n

2sm03.16na

• Solve for velocity in terms of

normal acceleration.

22

sm03.16m2 nn avv

a

sm66.5v

2/25/2014 6:39 PM


21

Sample Problem 12.6 Determine the rated speed

of a highway curve of

radius = 400 ft banked

through an angle q = 18o.

The rated speed of a

banked highway curve is

the speed at which a car

should travel if no lateral

friction force is to be

exerted at its wheels.

2/25/2014 6:39 PM


22

Sample Problem

12.6

Solution:

• The car travels in a horizontal

circular path with a normal

component of acceleration

directed toward center of the

path.

• Resolve equation of motion

for the car into vertical &

normal components.

:0 yF

q

q

cos

0cos

WR

WR

:nn maF

q

q

q

2

sincos

sin

v

g

WW

ag

WR n

• Solve for vehicle speed.

18tanft400sft2.32

tan

2

2 qgv

hmi1.44sft7.64 v

2/25/2014 6:39 PM


23

Howe Work Assignment #12-1

10, 17, 23, 30, 37, 44, 51

Due Saturday 1/3/2014

2/25/2014 6:39 PM


24

Angular Momentum of a Particle

• moment of momentum or

angular momentum of particle about O.

VmrHO

zyx

O

mvmvmv

zyx

kji

H

• is perpendicular to plane containing OH

Vmr

and

q

q

2

sin

mr

vrm

rmVHO

2/25/2014 6:39 PM


25

• Derivative of angular momentum

wrt time,

O

O

M

Fr

amrVmVVmrVmrH

• Newton’s 2nd law: sum of moments

about O of forces acting on the

particle is equal to rate of change of

angular momentum of the particle

about O.

2/25/2014 6:39 PM


Angular Momentum of a Particle

26

Equations of Motion in Radial &

Transverse Components

qqq qq rrmmaFrrmmaF rr 2 ,2

Consider particle at r and q, in polar coordinates,

qq

qqq

q

q

q

rrmF

rrrmmrdt

dFr

mrHO

2

222

2

This result may also be derived from

conservation of angular momentum,

2/25/2014 6:39 PM


27

Conservation of Angular Momentum

• When only force acting on

particle is directed toward or

away from a fixed point O, the

particle is said to be moving

under a central force.

• Since the line of action of the

central force passes through O,

and 0 OO HM

constant OHVmr

2/25/2014 6:39 PM


28


• Position vector and motion of

particle are in a plane

perpendicular to .OH

• Magnitude of angular

momentum,

000 sin

constantsin

Vmr

VrmHO

massunit

momentumangular

constant

2

2

hrm

H

mrH

O

O

q

q

2/25/2014 6:39 PM


29


• Radius vector OP sweeps

infinitesimal area qdrdA 2

21

• Define qq 2

212

21 r

dt

dr

dt

dA areal

velocity

• Recall, for a body moving under a

central force, constant2 qrh

• When a particle moves under a

central force, its areal velocity is

constant.

2/25/2014 6:39 PM


30

Newton’s Law of Gravitation • Gravitational force exerted by the sun on a planet or by earth

on a satellite is an important example of gravitational force.

• Newton’s law of universal gravitation - two particles of

mass M & m attract each other with equal and opposite force

directed along the line connecting the particles,

4

49

2

312

2

slb

ft104.34

skg

m1073.66

ngravitatio ofconstant

G

r

MmGF

• For particle of mass m on the earth’s surface,

222 s

ft2.32

s

m81.9 gmg

R

MGmW

2/25/2014 6:39 PM


31

Sample Problem 12.7 A block B of mass m can slide

freely on a frictionless arm OA

which rotates in a horizontal

plane at a constant rate

Knowing that B is released at a

distance r0 from O, express as a

function of r

a. the component vr of velocity

of B along OA

b. magnitude of horizontal force

exerted on B by the arm OA.

.0q

2/25/2014 6:39 PM


32

Sample Problem

12.7

Solution:

• Write the radial and

transverse equations of

motion for the block.

:

:

qq amF

amF rr

qq

q

rrmF

rrm

2

0 2

• Integrate the radial equation to

find an expression for the

radial velocity.

r

r

v

rr

rr

rr

rrr

drrdvv

drrdrrdvv

dr

dvv

dt

dr

dr

dv

dt

dvvr

r

0

2

0

0

2

0

2

q

qq

20

220

2 rrvr q

• Substitute known information

into the transverse equation to

find an expression for the force

on the block. 2120

2202 rrmF q

2/25/2014 6:39 PM


33

Sample Problem 12.8 A satellite is launched in a

direction parallel to the

surface of the earth with a

velocity of 18820 mi/h

from an altitude of 240 mi.

Determine the velocity of

the satellite as it reaches

its maximum altitude of

2340 mi. The radius of

the earth is 3960 mi.

2/25/2014 6:39 PM


34

Sample Problem 12.8 Solution:

• Satellite is moving under a central force, its angular

momentum is constant.

mi23403960

mi2403960hmi18820

constantsin

B

AAB

BBAA

O

r

rvv

vmrvmr

Hvrm

hmi12550Bv

2/25/2014 6:39 PM


35

Home Work Assignment #12-2

66, 74, 81

Due Monday 3/3/2014

2/25/2014 6:39 PM


36

Computer Problem

12.C1

Due Monday

3/3/2014

2/25/2014 6:39 PM


37

Documents

Chapter 12 - site.iugaza.edu.pssite.iugaza.edu.ps/mhaiba/files/2012/01/ch12-Kinetics-of-Particles... · Introduction 2/25/2014 6:39 PM Dr. Mohammad Suliman Abuhaiba, PE •Newton’s