Kinetics of Particles: Newton’s second law

12.1 Introduction• 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 (changes in velocity magnitude or direction), Newton’s second law is required to relate the motion of the body to theNewton s second law is required to relate the motion of the body to the forces acting on it.

• Newton’s second law:• Newton s second law:

- A particle will have an acceleration proportional to the magnitude of the resultant force acting on it and in the direction of the resultant gforce.

- The resultant of the forces acting on a particle is equal to the rate of h f li f h i lchange of linear momentum of the particle.

- The sum of the moments about O of the forces acting on a particle is equal to the rate of change of angular momentum of the particle

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equal to the rate of change of angular momentum of the particle about O.

Kinetics of Particles: Newton’s second law

12.2 Newton’s Second Law of Motion• Newton’s Second Law: If the resultant force acting on a

particle is not zero, the particle will have an acceleration proportional to the magnitude of resultant and in theproportional to the magnitude of resultant and in the direction of the resultant.

• Consider a particle subjected to constant forces,

maF

aF

aF mass,constant

3

3

2

2

1

1 ===== L

• When a particle of mass m is acted upon by a force the acceleration of the particle must satisfy

,Fr

amF rr= amF =

• Acceleration must be evaluated with respect to a Newtonian frame of reference, i.e., one that is not f f f , ,accelerating or rotating.

• If force acting on particle is zero? Particle will not accelerate i e it will remain stationary or

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Particle will not accelerate, i.e., it will remain stationary or continue on a straight line at constant velocity.

Kinetics of Particles: Newton’s second law

12.3 Linear Momentum of a Particle• Replacing the acceleration by the derivative of the

velocity yieldsvelocity yields

=∑ dtvdmF

r

rr

( )

particletheofmomentumlinear=

==

LdtLdvm

dtd

r

rr

particletheofmomentumlinear =L

• Linear Momentum Conservation Principle: If the resultant force on a particle is zero, the linear momentum of the particle remains constant in both magnitude and direction.g

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Kinetics of Particles: Newton’s second law

Systems of Unitsy• Of the units for the four primary dimensions (force, mass,

length, and time), three may be chosen arbitrarily. The f th t b tibl ith N t ’ 2 d Lfourth must be compatible with Newton’s 2nd Law.

• International System of Units (SI Units): base units are• International System of Units (SI Units): base units are the units of length (m), mass (kg), and time (second). The unit of force is derived,

⎞⎛( ) 22 smkg1

sm1kg1N1 ⋅

=⎟⎠⎞

⎜⎝⎛=

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Kinetics of Particles: Newton’s second law

12.5 Equations of Motionq• Newton’s second law provides

amF rr=∑∑

• Solution for particle motion is facilitated by resolving vector equation into scalar component equations, e.g., f t l tfor rectangular components,

( ) ( )kajaiamkFjFiF zyxzyxrrrrrr

++=++∑

zmFymFxmFmaFmaFmaF

zyx

zzyyxx&&&&&& ===

===

∑∑∑∑∑∑

• For tangential and normal components,maFmaF nntt == ∑∑

ρ

2vmFdtdvmF nt == ∑∑

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Kinetics of Particles: Newton’s second law

12.6 Dynamic Equilibriumy q• Alternate expression of Newton’s second law,

amF 0=−∑ rr

ectorinertial vam ≡−∑

r

• With the inclusion of the inertial vector, the system f f i h i l i i lof forces acting on the particle is equivalent to zero.

The particle is in dynamic equilibrium.

• Methods developed for particles in staticMethods developed for particles in static equilibrium may be applied, e.g., coplanar forces may be represented with a closed vector polygon.

• Inertia vectors are often called inertial forces as they measure the resistance that particles offer to changes in motion, i.e., changes in speed or g , , g pdirection.

• Inertial forces may be conceptually useful but are lik h d i i l f f d

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not like the contact and gravitational forces found in statics.

Kinetics of Particles: Newton’s second law

Sample Problem 12.3pAPPROACH:

h i h ki i f h bl k d• What is the kinematics of the blocks and pulley?

• What is the equation of motion for the blocks and pulley?

• Combine the kinematic relationships The two blocks shown start from rest. The horizontal plane and the pulley are frictionless, and the pulley is assumed

pwith the equations of motion to solve for the accelerations and cord tension.

frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of each block and the tension in the cord

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tension in the cord.

Kinetics of Particles: Newton’s second law

Sample Problem 12.3p

O

ABAB aaxy 21

21 ==

x

y

O

:AAx amF =∑

y

( ) AaT kg1001 =

:BBy amF =∑

( )( ) ( ) B

BBB

aT

amTgm

kg300sm81.9kg300 22

2

=−

=−

( )( ) BaT kg300-N29402 =

:0==∑ CCy amF

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y

02 12 =− TT

Kinetics of Particles: Newton’s second law

Sample Problem 12.4pSOLUTION:

• The block is constrained to slide down the wedge. Therefore, their motions are dependent. Express the acceleration of block as the acceleration of wedge plusblock as the acceleration of wedge plus the acceleration of the block relative to the wedge.

The 12-N block B starts from rest and

• Write the equations of motion for the wedge and block.

slides on the 30-N wedge A, which is supported by a horizontal surface.

l i f i i d i ( ) h

• Solve for the accelerations.

Neglecting friction, determine (a) the acceleration of the wedge, and (b) the acceleration of the block relative to the

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wedge.

Kinetics of Particles: Newton’s second law

Sample Problem 12.4pSOLUTION:

• The block is constrained to slide down the wedge. Therefore, their motions are dependent. Express the acceleration of block as the acceleration of wedge plusblock as the acceleration of wedge plus the acceleration of the block relative to the wedge.

• Write the equations of motion for the wedge and block.

• Solve for the accelerations.

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Kinetics of Particles: Newton’s second law

Sample Problem 12.5pSOLUTION:

• Resolve the equation of motion for theResolve the equation of motion for the bob into tangential and normal components.

• Solve the component equations for the normal and tangential accelerations.

S l f th l it i t f th

The bob of a 2-m pendulum describes an arc of a circle in a vertical plane If

• Solve for the velocity in terms of the normal acceleration.

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 h fi d th l it d lshown, find the velocity and accel-

eration of the bob in that position.

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Kinetics of Particles: Newton’s second law

Sample Problem 12.5pSOLUTION:

• Resolve the equation of motion for theResolve the equation of motion for the bob into tangential and normal components.

• Solve the component equations for the normal and tangential accelerations.

S l f th l it i t f th• Solve for the velocity in terms of the normal acceleration.

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Kinetics of Particles: Newton’s second law

Sample Problem 12.6pSOLUTION:

• The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path The forces acting on the carof the path.The forces acting on the car are its weight and a normal reaction from the road surface.

Determine the rated speed of a highway curve of radius ρ = 400 m

• Resolve the equation of motion for the car into vertical and normal componentshighway curve of radius ρ 400 m

banked through an angle θ = 18o. The rated speed of a banked highway curve is the speed at which a car should

components.

• Solve for the vehicle speed.is the speed at which a car should travel if no lateral friction force is to be exerted at its wheels.

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Kinetics of Particles: Newton’s second law

Sample Problem 12.6pSOLUTION:

• The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path The forces acting on the carof the path.The forces acting on the car are its weight and a normal reaction from the road surface.

• Resolve the equation of motion for the car into vertical and normal componentscomponents.

• Solve for the vehicle speed.

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Kinetics of Particles: Newton’s second law

12.7 Angular Momentum of a Particleg• moment of momentum or the angular

momentum of the particle about O.=×= VmrHO

rrr

kjirrr

• is perpendicular to plane containingOHr

Vmrrr and

φsinrmVHO =

zyx

Omvmvmv

zyxkji

Hr

=θ

φ

θ

&2mr

vrmO

=

=

• Derivative of angular momentum with respect to time,

×+×=×+×=O amrVmVVmrVmrH rrrr&rrr&r&r

zyx

∑∑

=

×=

O

O

MFr

r

r

∑ O

• From Newton’s second law: the sum of the moments about O of the forces acting on the particle = the rate

f h f h l f h i l

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of change of the angular momentum of the particle about O.

Kinetics of Particles: Newton’s second law

12.8 Eqs of Motion in Radial & Transverse Componentsq p

( )θ&2∑

• Consider particle at r and θ, in polar coordinates,

( )( )θθ

θ

θθ &&&&

&&

rrmmaFrrmmaF rr

2

2

+==

−==

∑∑

• This result may also be derived from conservation of angular momentum,

( )θθ

θ &

&

mrdFr

mrHO

2

2

=

=

∑ ( )( )( )

θθ

θθ

&&&

&&&& rrrm

mrdt

Fr

22 +=

=∑

( )θθθ &&&& rrmF 2+=∑

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Kinetics of Particles: Newton’s second law

12.9 Conservation of Angular Momentumg• When only force acting on particle is directed

toward or away from a fixed point O, the particle i id t b i d l fis said to be moving under a central force.

• Since the line of action of the central force passes through O and0∑ HM &rrthrough O, and 0∑ == OO HM

constant==× OHVmrrrr

• Position vector and motion of particle are in a plane perpendicular to .OH

r

M it d f l t• Magnitude of angular momentum,

000 sinconstantsin

φφ

VmrVrmHO

===

000 sinφVmr

momentumangularconstant

2

2 ==H

mrH

O

O θ&or

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massunit momentumangular 2 === hr

mHO θ&

Kinetics of Particles: Newton’s second law

12.9 Conservation of Angular Momentumg• Radius vector OP sweeps infinitesimal area

21 θdrdA 221=

D fi θθ &2121 ddA l l• Define === θθ 2212

21 r

dtdr

dtdA areal velocity

• Recall for a bod mo ing nder a central force• Recall, for a body moving under a central force,

constant2 == θ&rh

• When a particle moves under a central force, its areal velocity is constant.

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Kinetics of Particles: Newton’s second law

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

the earth on a satellite is an important example ofthe earth on a satellite is an important example of gravitational force.

• Newton’s law of universal gravitation - two particles of mass M and m attract each other with equal and opposite force directed along the line connecting the particles,

Mm2

ngravitatio ofconstant =

=

Gr

MmGF

4

49

2

312

slbft104.34

skgm1073.66

⋅×=

⋅×= −−

• For particle of mass m on the earth’s surface,ft232m819 ==== gmgMGmW

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222 s2.32

s81.9gmg

RmW

Kinetics of Particles: Newton’s second law

Sample Problem 12.7pSOLUTION:

• Write the radial and transverseWrite the radial and transverse equations of motion for the block.

• Integrate the radial equation to find an

A block B of mass m can slide freely on

expression for the radial velocity.

• Substitute known information into the transverse equation to find anA block B of mass m can slide freely on

a frictionless arm OA which rotates in a horizontal plane at a constant rate .0θ&

the transverse equation to find an expression for the force on the block.

a) the component v of the velocity of B

Knowing that B is released at a distance r0 from O, express as a function of ra) the component vr of the velocity of B

along OA, and b) the magnitude of the horizontal force

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exerted on B by the arm OA.

Kinetics of Particles: Newton’s second law

Sample Problem 12.7p• Integrate the radial equation to find an

expression for the radial velocity.

dvdrdvdvdrdvv

dtdr

drdv

dtdvvr r

rrr

r ==== &&&

==

====

rr

rr

rrr

drrdrrdvvdrdvv

dtdr

drdv

dtdvvr

20

2 θθ &&

&&&

SOLUTION:• Write the radial and transverse

equations of motion for the block∫∫ =r

r

v

rr

rr

drrdvv

drrdrrdvvr 2

00

0

θ

θθ

&

equations of motion for the block.

::

θθ amFamF rr

==

∑∑ ( )

( )θθθ

&&&&

&&&

rrmFrrm

20 2

+=

−=

r00( )2

022

02 rrvr −=θ

S b tit t k i f ti i t th:θθ amF =∑ ( )θθ rrmF 2+= • Substitute known information into the transverse equation to find an expression for the force on the block.

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( ) 2120

2202 rrmF −= θ

Kinetics of Particles: Newton’s second law

Sample Problem 12.8pSOLUTION:

Si h lli i i d• Since the satellite is moving under a central force, its angular momentum is constant. Equate the angular momentum at A and B and solve for the velocity at B.

A satellite is launched in a direction parallel to the surface of the earth with a velocity of 18820 km/h from yan altitude of 240 km. Determine the velocity of the satellite as it reaches it maximum altitude of 2340 km. Themaximum altitude of 2340 km. The radius of the earth is 6345 km.

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Kinetics of Particles: Newton’s second law

Sample Problem 12.8pSOLUTION:

Si h lli i i d• Since the satellite is moving under a central force, its angular momentum is constant. Equate the angular momentum at A and B and solve for the velocity at B.

constantsin == OHvrm φ

=

=

AAB

BBAA

rrvv

vmrvmr

( ) ( )( ) km23406345

km2406345km/h18820++

=

Br

( )

km/h4.14269=Bv

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