Chapter ThreeKinetics of particles

Kinetics of particles• It is the study of the relations existing between

the forces acting on body, the mass of the body,and the motion of the body.

• It is the study of the relation betweenunbalanced forces and the resulting motion.

• Newton ’s first law and third law are sufficient

for studying bodies at rest (statics) or bodies in

motion with no acceleration.

• When a body accelerates ( change in velocity

magnitude or direction) Newton ’s second law is

required to relate the motion of the body to the

forces acting on it.

Force, mass and acceleration• Newton ’s Second Law: If the resultant force

acting on a particle is not zero the particle willhave an acceleration proportional to themagnitude of resultant and in the direction of theresultant.

• The basic relation between force and

acceleration is found in Newton's second law of

motion and its verification is entirely

experimental.

• Consider a particle subjected to constant forces

• We conclude that the constant is a measure of

some property of the particle that does not

change.

constaF

aF

aF

==== ...2

2

1

1

• This property is the inertia of the particle which is its resistance to rate of change of velocity.

• The mass m is used as a quantitative measure of inertia, and therefore the experimental relation becomes,

F=ma

• This relation provides a complete formulation of Newton's second law; it expresses not only that the magnitude F and a areproportional but also that the vector F and a have the same direction.

Equation of motion and solution of problems• When a particle of mass m acted upon by several forces. The

Newton’s second law can be expressed by the equation

• To determine the acceleration we must use the analysis used inkinematics, i.e

• Rectilinear motion

• Curvilinear motion

∑ = maF

Rectilinear Motion• If we choose the x-direction, as the direction

of the rectilinear motion of a particle of mass m, the acceleration in the y and z direction will be zero, i.e

∑∑∑

=

=

=

0

0

z

y

xx

F

FmaF

• Generally,

• Where the acceleration and resultant force are given by

ZZ

yy

xx

maFmaFmaF

=

=

=

∑∑∑

222zyx

zyx

aaaa

kajaiaa

++=

++=

( ) 222 )()( ∑∑∑∑

∑++=

++=

zyx

zyx

FFFF

kFjFiFF

Curvilinear motion• In applying Newton's second law, we shall make

use of the three coordinate descriptions of

acceleration in curvilinear motion.

Rectangular coordinates

Where and

∑∑

=

=

yy

xx

maFmaF

••

= xax

••

= yay

Normal and tangential coordinate

• Where∑∑

=

=

tt

nn

maF

maF

••

=== vava tn ,22

ρβρ

Polar coordinates

• Where and

∑∑

=

=

θθ maF

maF rr

2•••

−= θrrar

••••

+= θθ rran 2

Examples

Example 1• Block A has a mass of 30kg and block B has a mass of

15kg. The coefficient of friction between all plane

surfaces of contact are and .

Knowing that Ѳ=300 and that the magnitude of the

force P applied to block A is 250N, determine

a) The acceleration of block A ,and

b) The tension in the cord

15.0=sµ 10.0=kµ

Example 2• A small vehicle enters the top A of the circular path with a

horizontal velocity vo and gathers speed as it moves down the path.

• Determine an expression for the angle β to the position where thevehicle leaves the path and becomes a projectile. Evaluate yourexpression for vo=0. Neglect friction and treat the vehicle as aparticle

Example 3• The slotted arm revolves in the horizontal about the fixed vertical axis

through point O. the 2 Kg slider C is drawn towards O at the constant

rate of 50mm/s by pulling the cord S. At the instant for which

r=225mm, the arm has a counterclockwise angular velocity ω=6rad/s

and is slowing down at the rate of 2rad/s2 .for this instant determine

the tension T in the cord and the magnitude N of the force exerted on

the slider by the sides of the smooth radial slot. Indicate which side, A

or B, of the slot contacts the slider. (problem 3/69)

work and kinetic energy

• The method of work and energy directly relates force, mass,

velocity, and displacement.

• We apply this method:

When intervals of motion are involved where the change in

velocity or the corresponding displacement of the particle is

required.

• Integration of the forces with respect to the displacement

of the particle leads to the equation of work and energy.

Work of Force

( ) ( )dzFdyFdxF

kdjdidkFjFiFdU

FdsdUrdFdU

zyx

zyxzyx

++=

++•++=

=•=

αcos

( )

( )∫

∫∫

∫

++=

==

•=

→

→

→

2

1

2

1

2

1

2

1

21

21

21

cos

A

Azyx

s

st

s

s

A

A

dzFdyFdxFU

dsFdsFU

rdFU

α

( ) xFU ∆=→ αcos21

Work of a constant force in rectilinear motion

yWyyWU

WyWyWdyU

WdydUy

y

∆−=−−=

−=−=

−=

→

→ ∫)( 1221

2121

2

1

When ∆y is negative (moves down), the work is positive

Work of the Force of Gravity

Work of the Force Exerted by a Spring

( ) xFFU

kxkxkxdxU

kxdxFdxdUkxF

x

x

∆+−=

−=−=

−=−==

→

→ ∫

2121

22

2121

21

21

212

1

If the spring returning to the undeformed position, then positive energy

Work of a gravitational Force

12221

2

2

2

1r

GMmr

GMmdrr

GMmU

drr

MmGFdrdU

rMmGF

r

r

−=−=

−=−=

=

∫→

Kinetic Energy of a Particle

21

22 2

1212

1

2

1

mvmvvdvmdsF

mvdvdsFdsdvmv

dtds

dsdvm

dtdvmmaF

v

v

s

st

t

tt

−==

=

====

∫∫

2211

1221

TUTTTU

=+−=

→

→

Power and Efficiency

• Friction energy dissipated by heat and reduce kinetic energy

inputpoweroutputpower

vFdt

rdFdt

dUpower

=

•=•

==

η

Potential Energy

WyVVVU

WyWyWdyU

g

gg

y

y

=

−=

−=−=

→

→ ∫

2121

2121

)()(

2

1

When Vg2 >Vg1, potential energy increases, and U 1-2is negative

Potential Energy of the body with respect to the force of gravity

rWR

rMmGV

rGMm

rGMmdr

rGMmU

g

r

r

2

12221

2

1

==

−=−= ∫→

R is from the center of the earth

it should be noted that the expression just obtained for the potential energy of a

body with respect to gravity is valid only as long as the weight of body can be

assumed to remain constant, i.e, as long as the displacements of the body are small

compared with the radius of the earth.

2

2121

22

2121

21

)()(

21

212

1

kxV

VVU

kxkxkxdxU

e

ee

x

x

=

−=

−=−=

→

→ ∫

Potential Energy of the body with respect to the elastic force F

Only the initial and final deflection of the spring are neededDeflection of the spring is measured from its undeformed position

Equation of work and energy

ofsw UUUU

whereTU

21212121

21

→→→→

→

++=

∆=

Where =work done by other(non conservative/ pathdependant) forces

ofU 21→

EU

VVTU

TUVV

of

geof

ofeg

∆=

∆+∆+∆=

∆=+∆−∆−

→

→

→

21

21

21

Advantages and disadvantages of work and energy method

• Advantage; 1. No need to calculate values between A1 and

A2. Only final stages are counted 2. All scalars so can be

added easily 3. Forces that do no work are ignored

• Disadvantage; can not determine accelerations, can not

determine accelerations and forces that do no work

Example 1• For the slider collar shown in the figure, if m=0.5kg,

b=0.8m, and h=1.5m, and if the velocity of the collar as itstrikes the base B is 4.70m/s after the release of thecollar from rest, calculate the work Q of friction. Whathappens to the energy that is lost?

• The 0.8kg collar slides with negligible frictionon the fixed rod in the vertical plane. If thecollar starts from rest at A under the action of8N horizontal force, calculate its velocity v asit hits the stop at B.

Example 2

• The 6kg cylindrical collar is released from restin the position shown and drops onto the spring.Calculate the velocity v of the cylinder whenthe spring has been compressed 50mm.

Example 3

• The two springs, each of stiffness k=1.2KN/m, are if equallength and un-deformed when θ=0. If the mechanism isreleased from rest in the position θ=20, determine itsangular velocity θ’ when θ=0. the mass m of each sphere is3kg. Treat the sphere as particles and neglect the massesof the light rods and springs.

Example 4

Impulse and momentum

• Work and energy is obtained by integrating theequation of F=ma with respect to thedisplacement of the particle.

• Impulse and momentum can be generated byintegrating the equation of motion (F=ma) withrespect to time.

Linear impulse and momentum• Consider a particle of a mass m which

is subjected to several forces in space.

• The product of the mass & the velocityis defined as the linear momentum.

)1..(..........)( GForvmdtdvmF ∑∑ ===

vmG =

• Equation (1) states that the resultant of allforces acting on a particle equals its time rateof change of linear momentum.

• It is valid as long as the mass m of the particleis not changing with time.

• The scalar components of equation (1) are:∑∑∑ === zzyyxx

GFGFGF ,,

• The effect of the resultant force on the linearmomentum of the particle over a finite period of timesimply by integrating with respect to time t.

• The linear momentum at time t2 is G2=mV2 and the linearmomentum at time t1 is G1=mV1.

• The product of force and time is called linear impulse.

GGGFdt

dGFdtt

t

∆=−=

=

∫∑

∑2

1

12

∑ F

• The total linear impulse on a mass m equals

the corresponding change in linear momentum

of m.

( ) ( )

( ) ( )

( ) ( )

−=

−=

−=

=+

∫∑

∫∑

∫∑

∫∑

12

12

12

21

2

1

2

1

2

1

2

1

zz

t

tz

yy

t

ty

xx

t

tx

t

t

mVmVdtF

mVmVdtF

mVmVdtF

GFdtG The impulse integral is a vector

Scalar impulse momentum eqns.

Conservation of linear momentum• If the resultant force on a particle is zero

during an interval of time, its linear momentumG remains constant. In this case the linearmomentum of a particle is said to be conserved.

210 GGorG ==∆

Angular impulse and momentum• The moment of a linear momentum vector

about the origin O is defined as the angularmomentum of P & is given by the productrelation for moment of vector:-

• The angular momentum then is a vectorperpendicular to the plane A defined by

vm

oH

vmrH ×=0

vr &

• The scalar component of angular momentum

is:-

)()(),(

)()()(

yvxvmHxvzvmHzvyvmH

vvvzyxkji

mH

kyvxvmjxvzvmizvyvmvmrH

xyzzxyyzx

zyx

o

xyzxyzo

−=−=−=

=

−+−+−=×=

• If represents the resultant of all forces acting on the particlesP, the moment Mo about the origin O is the vector cross product.

• The moment about the fixed point O of all forces acting on M equalsthe time rate of change of angular momentum of M about O.

∑ F

....(*)..........∑

∑ ∑

=

×+×⇒×+×=

×=×=

oo

o

o

HMvmrvmvvmrvmrH

vmrFrM

• To obtain the effect of moment on the angular

momentum of the particle over a finite period of time;

• The product of moment & time is angular impulse the total

angular impulse on M about a fixed point O equals the

corresponding change in angular momentum of M about O.

112212

12

2

1

& vmrHvmrHwhere

HHHdtM

oo

ooo

t

to

×=×=

∆=−=∫∑

∑ oM

∫∑ =+2

1

21

t

tooo HdtMH

Conservation of angular momentum• If the resultant moment about a fixed point O of all

forces acting on a particle is zero during the interval

of time, equation (*) requires that its angular

momentum H0 about that point remains constant.

210 ooo HHorH ==∆

Impact

Impact• Refers to the collision b/n two bodies

and is characterized by the generation

of relatively large contact forces that

act over a very short interval of time.

Direct central impact• Consider the collinear motion of two spheres of masses m1

and m2 travelling with velocities V1 & V2. If V1 is greaterthan V2, collision occurs with the contact forces directedalong the line of centers.

• In as much as the contact forces are equal& opposite during impact; the linearmomentum of the system remainsunchanged.

'' 22112211 vmvmvmvm +=+

• For given masses & initial conditions, the momentumequation contains two unknowns, v1’ & v2’, an additionalrelationship is required.

• The relationship must reflect the capacity of thecontacting bodies to recover from the impact & can beexpressed by the ratio e of the magnitude of therestoration impulse to the magnitude of the deformationimpulse. This ratio is called the coefficient of restitution.

• Fr – contact force during restoration period

• Fd – contact force during deformation period

( )[ ]( )[ ]

( )[ ]( )[ ] 2.....................''

1...............''

2

2

22

22

1

1

11

11

particleforvvvv

vvmvvm

dtF

dtFe

particleforvvvv

vvmvvm

dtF

dtFe

o

o

o

ot

td

t

tr

o

o

o

ot

td

t

tr

o

o

o

o

−−

=−−

==

−−

=−−−−−−

==

∫

∫

∫

∫

• According to classical theory of impact, the value

e=1 means that the capacity of the two particles to

recover equal their tendency to deform.

• The value e=0, on the other hand describes

inelastic or plastic impact where the particles cling

together after collision & the loss of energy is a

maximum.

b) Oblique central impact• Here the initial and final velocities are not parallel.

• The spherical particles of mass m1 & m2 have initialvelocities v1 & v2 in the same plane & approach each otheron a collision course.

• The direction of the velocity vector are measured fromthe direction tangent to the contacting surfaces.

( ) ( )

( ) ( ) ,cos,sin

,cos,sin

222222

111111

θθ

θθ

vvvv

vvvv

tn

tn

=−=

=−=

There will be four unknown namely, (v1’)n, (v1’)t, (v2’)n, & (v2’)t1) Momentum of the system is conserved in the n-direction,

2) & 3) The momentum for each particle is conserved in the t-

direction since there is no impulse on either particle in the t-

direction

tt

tt

vmvmvmvm

)'()()'()(

2222

1111

==

3) the coefficient of restitution, the velocity component in the n- direction,

( ) ( )( ) ( )nn

nn

vvvv

e21

12 ''−−

=

Example 1• The 4kg cart, at rest at t=0, is acted on by a

horizontal force that varies with time t asshown. Neglect friction and determine thevelocity of the cart at t=1Sec and t=3Sec

Example 2• The loaded mine skip has a mass of 3Mg. The hoisting

drum produces a tension T in the cable according to thetime schedule shown. If the skip is at rest againest A awhen the drum is activated, determine the speed v of theskip when t=6Sec. Friction loss may be neglected.

Example 3• a particle with a mass of 4kg has a position vector in

meters given by ,where t is the time inseconds. For t=3sec, determine the magnitude of theangular momentum of the particle and the magnitude ofthe moment of all forces on the particle, both about theorigins of the coordinates.

ktjtitr 323 2 −−=

Example 4• A particle of mass m moves with negligible friction

on a horizontal surface and is connected to a lightspring fastened at O. At position A the particle hasthe velocity VA=4m/s. Determine the velocity VBof the particle as it passes position B.

Example 5• The assembly of two 5Kg spheres is rotating freely about the vertical

axis at 40rev/min with θ=900. if the force F necessary to maintain the

given position is increased to raise the base collar and reduce θ to 600,

determine the new angular velocity ω. Also determine the work U done

by F in changing the configuration of the system. Assume that the mass

of the arms and the collars is negligible.

Example 6• The magnitude and direction of the velocities of

two identical frictionless balls defore strike eachother are as shown in the figure. Assume e=0.90,

A) determine the magnitude and direction of the velocity ofeach ball after impact.

B) calculate the percentage loss of energy due to the impact.

Example 7• Sphere A has a mass of 23Kg and a radius of 75mm while sphere

B has a mass of 4Kg and a radius of 50mm. If the spheres aretravelling initially along the parallel paths with the speedsshown, determine the velocities of the spheres immediatelyafter impact. Specify the angles θA and θB with respect to thex-axis made by the rebound velocity vectors. The coefficient ofrestitution is 0.4 and friction is neglected.

Example 8• The 2Kg sphere is projected horizontally with velocity of 10m/s

against the 10kg carriage that is backed up by the spring with a

stiffness of 1600N/m. The carriage is initially at rest with the

spring uncompressed. If the coefficient of restitution is 0.6,

calculate the rebound velocity v’, the rebound angle, and the

maximum travel δ of the carriage after impact.