kinetics of particles study of the relations existing between the

R.Ganesh Narayanan, IITG

Kinetics of particles

• Newton’s first law and third law are sufficient for studying bodies at rest (- statics) or bodies in motion with no acceleration

study of the relations existing between the forces acting on a body, the mass of the body, and the motion of the body

• When a body accelerates (change in velocity magnitude or direction), Newton’s second law is required to relate the motion of the body to theforces acting on it

• 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 the direction of the resultant.

Ist law: A particle remains at rest or continues to move with uniform velocity (in a straight line with const. speed) if there is no unbalanced force acting on it


Consider a particle subjected to constant forces,

ma

F

aF

aF mass,constant

3

3

2

2

1

1 ====

Characteristic of particle considered

When a particle of mass m is acted upon by a force ‘F’, the acceleration of the particle must satisfy

amF =

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 Ist law

Magnitudes of F & a are proportional; F & a vectors are in same direction

Body subjected to several forces, ΣF = ma


Linear momentum of a particle

Replacing the acceleration by the derivative of the velocity yield

( )

particletheofmomentumlinear =

==

=Σ

L

dtLd

vmdt

ddtvd

mF

‘L’ - Same direction as that of velocity of particle – mv (vector)

Mass is constant

L = mv => ΣF = dL/dt

The rate of change of linear momentum (dL/dt) is zero when ΣF = 0 => If the resultant force acting on the particle is zero, linear momentum of the particle ‘L’ is constant, in magnitude and direction

This is called as Principle of conservation of momentum => another version of Newton’s 1st law

Unit: kg. (m/s)


Equations of motion

Newton’s second law provides amF =∑

( ) ( )

zmFymFxmF

maFmaFmaF

kajaiamkFjFiF

zyx

zzyyxx

zyxzyx

===

===

++=++

∑∑∑∑∑∑

∑

0 0- W For a projectile without air resistance

For tangential and normal components,

ρ

2vmF

dt

dvmF

maFmaF

nt

nntt

==

==

∑∑

∑∑

In rectangular components,


Dynamic equilibrium

• Alternate expression of Newton’s second law,

ectorinertial vam

amF 0

−

=−∑

=

Particle is in equilibrium under given forces and inertia vector– called Dynamic equilibrium

In coplanar force system, all forces can be made into closed vector polygon and can be solved (OR)

Sum of force components including inertia vector can be equated to zero

ΣFx = 0; ΣFy = 0 Rectangular components


Tangential components can also be used, -mat & -man

Tangential component: Measure of resistance of particle to change in speed

Normal component: Tendency of particle to leave its path

Inertia vectors or inertia forces are measure of resistance that particles offer for motion

These forces are not considered like g, contact forces etc. and hence dynamic equilibrium method is not used


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.

Beer/Johnston, 12.3

Write the kinematic relationships for the dependent motions and accelerations of the blocks.

ABAB aaxy21

21 ==

Apply Newton’s II law to blocks A, B, pulley C

Write equations of motion for blocks and pulley: :AAx amF =∑

( ) AaT kg1001 =:BBy amF =∑

( )( ) ( )( ) B

B

BBB

aT

aT

amTgm

kg300-N2940

kg300sm81.9kg300

2

22

2

=

=−

=−

x

y

T2 = 2940-150 aA


02 12 =− TT

Put T1, T2 in above equn., 2940-150aA-2 (100aA) = 0; aA = 8.4 m/s2

aB = 8.4/2 = 4.2 m/s2; T1 = 100 (8.4) = 840 N

T2 = 2 (840) 1680 N


Beer/Johnston, 12.14

A BA light train made up of two cars is traveling at 88 km/h when the brakes are applied to both cars. Knowing that car A has a weight of 24947.56 kg and carB has a weight of 19958 kg and that the braking force is 31137.5 N on each car, determine (a) the distance traveled by the train before it comes to a stop, (b) the force in the coupling between the cars while the train is slowing down.

88 km/h

Consider Fb as braking force

Fb

Find ax

(a) Relate ‘ax’ to ‘Xf’ by ax = v (dv/dx); Apply BCs and by integration of ‘x’and ‘v’; Find ‘Xf’

Fb Fb

Fb FcA

(b) Find Fc from this


A 2-kg ball revolves in a horizontal circle as shown at a constant speed of 1.5 m/s. Knowing that L = 600 mm, determine (a) the angle θ that the cord forms with the vertical, (b) the tension in the cord.

12.36

θT

W

ΣFy = m (ay) = 0 =>T cos θ-w = 0

ΣFx = m (ax) => T sinθ = mama

a is towards circle center = v2/ρ = v2/Lsinθ

Solve these equns. and get θ, t


A curve in a speed track has a radius of 200 m and a rated speed of 180 km/h. Knowing that a racing car starts skidding on the curve when traveling at a speed of 320 km/h, determine (a) the banking angle θ, (b) the coefficient of static friction between the tires and the track under the prevailing conditions

12.53

Rated speed is the speed at which a car should travel if no lateral friction force is to be exerted on its wheels

Put F=0 and find θ Find µ from µ = F/N

wN

F

θ

X

Y

ma


Angular momentum of particle

• moment of momentum or the angular momentum of the particle about O.

=×= VmrHO

• is perpendicular to plane containingOH Vmr and

zyx

O

mvmvmv

zyx

kji

H =We have Hx, Hy, Hz

• Derivative of angular momentum with respect to time,

• It follows from Newton’s second law that the sum of the moments about O of the forces acting on the particle is equal to the rate of change of the angular momentum of the particle about O

∑∑

=

×=

×+×=×+×=

O

O

M

Fr

amrVmVVmrVmrH

0


Eqs of Motion in Radial & Transverse Components

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

( )θθ

θ

θθ rrmmaF

rrmmaF rr

2==

==

∑∑ 2

+

Motion under central force

• 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, and0∑ == OO HM

O constant==× HVmr

rmv sinφ = r0mv0 sinφ0


The 2.7 kg collar B slides on the frictionless arm AA′. The arm is attached to drum D and rotates about O in a horizontal plane at the rate θ = 0.8t,where θ and t are expressed in rad/s and seconds, respectively. As the arm-drum assembly rotates, a mechanism within the drum releases cord so that the collar moves outward from O with a constant speed of 0.457 m/s. Knowing that at t = 0, r = 0, determine the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on B by arm AA′.

dr/dt = r = 0.457 m/s; ∫dr = ∫0.457 dt => r = 0.457t; r = 00

r

0

t

( )( )θθ

θ

θθ rrmmaF

rrmmaF rr

2==

==

∑∑ 2

+ar = -0.292 t

3 m/s2; aθ = 1.09 m/s2


TQ

=

M ar

M aθ

T = Q => -m ar = m aθ

0.292 t3 = 1.09 => t = 1.66 s


• Previously, problems dealing with the motion of particles were solved through the fundamental equation of motion, F = ma

• We introduce two additional methods of analysis

• Method of work and energy: directly relates force, mass, velocity and displacement

• Method of impulse and momentum: directly relates force, mass, velocity, and time

Energy and momentum methods


• Work of the force is

dzFdyFdxF

dsF

rdFdU

zyx ++=

=

•=

αcos

Work of a force

• Work is a scalar quantity, i.e., it has magnitude and sign but not direction.

force. length ×• Dimensions of work are Units are( ) ( )( )m 1N 1 J 1 =joule

•Work of finite displacement

•Work of force of gravity

•Work of the force exerted by a spring

Method of Work & energy


Kinetic energy of particle

dvmvdsF

ds

dvmv

dt

ds

ds

dvm

dt

dvmmaF

t

tt

=

==

==

• Consider a particle of mass m acted upon by force, F

• Integrating from A1 to A2 ,

energykineticmvTTTU

mvmvdvvmdsFv

v

s

st

==−=

−==

→

∫∫

221

1221

212

1222

12

1

2

1

• The work of the force F is equal to the change in kinetic energyof the particle => Principle of work & energy

KE:

Unit: Nm or Joule

Scalar quantity

T2 = T1 + U (1-2)


T2

T1

• To determine velocity of pendulum bob at A2. Consider work & kinetic energy.

• Force P acts normal to path and does no work.

glv

vg

WWl

TUT

2

21

0

2

22

2211

=

=+

=+ →

• Velocity found without determining expression for acceleration and integrating

• All quantities are scalars and can be added directly

• Forces which do no work are eliminated from the problem.

Advantages:


ΣFt = m at = 0

• Principle of work and energy cannot be applied to directly to determine the acceleration of the pendulum bob

• Calculating the tension in the cord requires supplementing the method of work and energy with an application of Newton’s second law

• As the bob passes through A2 ,

Wl

gl

g

WWP

l

v

g

WWP

amF nn

32

22

=+=

=−

=∑

Problem with more particles: KE can be obtained for each particle separately and KE can be added for all particles

T1 + U (1-2) = T2

T – arithmetic sum of the KEs of particles involved; U (1-2) = work of all the forces acting on the particles


Power: Rate at which work is done

Power = dU/dt = F.dr/dt = F.v

Unit: Nm/s

Efficiency = Power output / Power input < 1


Two blocks are joined by an inextensible cable as shown. If the system is released from rest, determine the velocity of block A after it has moved 2 m. Assume that the coefficient of friction between block A and the plane is µk = 0.25 and that the pulley is weightless and frictionless.

Apply the principle of work and energy separately to blocks A and B.

( )( )( )

( ) ( )

( ) ( )( ) ( ) 221

221

2211

2

kg200m2N490m2

m2m20

:

N490N196225.0

N1962sm81.9kg200

vF

vmFF

TUT

WNF

W

C

AAC

AkAkA

A

=−

=−+

=+

====

==

→

µµ

+ ve


( )( )

( ) ( )

( ) ( )( ) ( ) 221

221

2211

2

kg300m2N2940m2

m2m20

:

N2940sm81.9kg300

vF

vmWF

TUT

W

c

BBc

B

=+−

=+−

=+

==

→

• When the two relations are combined, the work of the cable forces cancel. Solve for the velocity

( )( ) ( )( ) ( )

( ) 221

221

kg500J 4900

kg300kg200m2N490m2N2940

v

v

=

+=−

sm 43.4=v


A 40.8 kg package is at rest on an incline when a constant force P is applied to it. The coefficient of kinetic friction between the package and the incline is 0.35. Knowing that the speed of the package is 0.6m/s after it has moved 0.9 m up the incline, determine the magnitude of the force P.


400

N

F = 0.35 N

20°

50°

P

T1 = 0

N = 400 Cos 20 + P sin 50 Put N in U (1-2)

U (1-2) = T2 – T1 => P = 737.9 N


A 2000-kg automobile starts from rest at point A on a 6°incline and coasts through a distance of 150 m to point B. The brakes are then applied, causing the automobile to come to a stop at point C, 20 m from B. Knowing that slipping is impending during the braking period and neglecting air resistance and rolling resistance, determine (a) the speed of the automobile at point B, (b) the coefficient of static friction between the tires and the road

6°

w

N

F

VA = 0; Vc = 0; w = 19620 N

U (A-B) = TB – TA

19620 (150) Sin 6 = ½ (2000) (VB2)

VB =17.5 m/s

U (A-C) = TC – TA

19620 (170) Sin 6 – F (20) = 0

19620 (170) Sin 6 – 19620 (µ) cos 6 (20) = 0

µ = 0.89


Car B is towing car A with 4.6 m cable at a constant speed of 9 m/s on an uphill grade when the brakes of car B are fully applied causing it to skid to a stop. Car A, whose driver had not observed that car B was slowing down, then strikes the rear of car B. Neglecting air resistance and rolling resistance and assuming a coefficient of kinetic friction of 0.9, determine the speed of car A just before the collision.

mg

F = 0.9 N

N = mg cos 5

0

F = 0.9 mg cos 5

=> d = 4.2 m

N

mg

Car B

Car AFor Car A: initial to contact

(-mg sin 5) (4.6+4.2) = ½ m VA2 – ½ (m) 92 => VA = 8.22 m/s


Conservative forces

Potential energy

PE due to displacement; U (1-2) = Wy1 – Wy2 = V1-V2

PE due to spring deformation; U (1-2) = ½ k(x1)2 – ½ k (x2)2

Forces whose work done is independent of path followed; depends only on initial and final positions => conservative forces

• Concept of potential energy can be applied if the work of the force is independent of the path followed by its point of application.

( ) ( )22211121 ,,,, zyxVzyxVU −=→

For any conservative force applied on closed path

∫ F. dr = 0


• Elementary work corresponding to displacement between two neighboring points,

( ) ( )( )zyxdV

dzzdyydxxVzyxVdU

,,

,,,,

−=

+++−=

du = -dv (x, y, z)The elementary work of a conservative force is exact differential

Vz

V

y

V

x

VF

dzz

Vdy

y

Vdx

x

VdzFdyFdxF zyx

grad−=

∂∂

+∂∂

+∂∂

−=

∂∂

+∂∂

+∂∂

−=++

r

Differential of a function of several variables

Fx = -∂v/∂x; Fy = -∂v/∂y; Fz = -∂v/∂zDepends only on position of point of application

F = -grad V For conservative force


Conservation of energy

U (1-2) = V1 – V2; V – POTENTIAL ENERGY

U (1-2) = T2 – T1; T – KINETIC ENERGY

V1 – V2 = T2 – T1

T1+V1 = T2+V2

When a particle moves under the action of conservative forces, the sum of kinetic energy and potential energy of particle remains constant

T + V = CONSTANT = MECHANICAL ENERGY, E


At A1; T1 = 0; V1 = Wl; T1 + V1 = Wl

At A2; T2 = ½ (W/g) v22 = ½ (W/g) (2gl) = Wl; V2 = 0

T2 + V2 = Wl

E = T + V = CONSTANT

Energy: A1 = only potential; A2 = Only kinetic

Only PE depends on elevation and not KE. So, speed is same in A, A’


Particle will have same speed at A, B, C as long as weight of particle and normal reaction of path are the two forces acting; without friction

A B C

Weight of particle, forces exerted by spring – conservative forces

Friction force => Non-conservative force; work done by friction force can not be expressed as change in potential energy as it depends on path followed by point of application

A mechanical system involving friction will have decrease in total mechanical energy; dissipated as heat etc.


A 20 N collar slides without friction along a vertical rod as shown. The spring attached to the collar has an undeflected length of 4 cm and a constant of 3 N/cm. If the collar is released from rest at position 1, determine its velocity after it has moved 6 cm. to position 2.

Position 1: ( )( )

0

0cmN24

cmN24cm 4cm 8cm/N3

1

1

2212

121

=

+⋅=+=

⋅=−==

T

VVV

kxV

ge

e

PE:

KE:

Position 2: ( )( )( )( )

22

222

12

2

2212

221

1020

21

cmN 6612054

cmN 120cm 6N 20

cmN54cm 4cm 01cm/N3

vmvT

VVV

WyV

kxV

ge

g

e

==

⋅−=−=+=

⋅−=−==

⋅=−==

PE:

KE:


Conservation of Energy:

cmN 66cmN 240 22

2211

⋅−=⋅+

+=+

v

VTVT

↓= sm5.92v


A force P is slowly applied to a plate that is attached to two springs and causes a deflection x0. In each of the two cases shown, derive anexpression for the constant , ‘ke’ in terms of k1 and k2 , of the single spring equivalent to the given system, that is, of the single spring whichwill undergo the same deflection x0 when subjected to the same force P.

13.55

Series connection: Force in both springs is the same = P

X0 = X1 + X2

P/Ke = P/K1 + P/K2

=> Ke = (K1.K2)/(K1+K2)

Parallel connection: Deflection in both springs is the same = X0

P = P1 + P2 = K1 X0 + K2 X0

P = Ke X0 => Ke = K1 + K2


A 750-g collar can slide along the horizontal rod shown. It is attached to an elastic cord with an undeformed length of 300 mm and a spring constant of 150 N/m. Knowing that the collar is released from rest at A and neglecting friction, determine the speed of the collar (a) at B, (b) at E.

13.59


2.7 kg

K = 2627 N/m

A 2.7 kg collar can slide without friction on a vertical rod and is held so it just touches an undeformed spring. Determine the maximum deflection of the spring (a) if the collar is slowly released until it reaches an equilibrium position, (b) if the collar is suddenly released.

a) Collar in equilibrium

ΣF = 0 => -26.5+2627X = 0

Xmax = 0.01 m

b) Collar in suddenly released

T1+V1 = T2+V2

0+0 = 0 + (-Wh+1/2x2627xh2)

h = 2 (26.5)/2627 = 0.02 m


A spring is used to stop a 90.7 kg package which is moving down a 20° incline. The spring has a constant k = 22 kN/mand is held by cables so that it is initially compressed 15 cm. Knowing that the velocity of the package is 2.4 m/s when it is 7.6 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringingthe package to rest.

Position 1 is at the top of the incline; position 2 is when the spring has maximum deformation

T1 + V1 = T2 + V2

½ (90.7) (2.4) + [1/2 (22000) (0.15)2 + 890 (7.6+X) sin 20] =

0 + [0 + ½ (22000) (X+0.15)2]

Where x = Deformation of the spring

Solving this, X = 0.367 m


Method of impulse and momentum

• Method of impulse and momentum: directly relates force, mass, velocity, and time.

F = d (mv) / dt

Fdt = d (mv)

∫ F dt = mv2 - mv1t1

t2

mv1 + ∫F dt = mv2t1

t2

Integral => Impulse of force, F (Imp1-2)

Vector quantity; Unit : N.s

Imp1-2 = i ∫FX dt + j ∫Fy dt + k ∫Fz dt, from t1 to t2

FX

t

t1 t2


mv1 + ∫F dt = mv2t1

t2

mv1 + Imp1-2 = mv2

The final momentum of the particle can be obtained by adding vectorially its initial momentum and the impulse of the force during the time interval.

Note: KE, work => scalar quantities; Momentum, impulse => vector quantities


Several forces acting on one particle mv1 + Σ Imp1-2 = mv2

Two or more particles Σ mv1 + Σ Imp1-2 = Σ mv2

• Impulse of action and reaction forces exerted by particles cancel out; Only the impulses of external forces need be considered

• If sum of external forces is zero, then Σ mv1 = Σ mv2. This says that total momentum of the particles is conserved. (discussed later)


Impulsive motion

Force acting on a particle during a very short time interval that is large enough to cause a significant change in momentum is called an impulsive force and resulting motion is called impulsive motion

When impulsive forces act on a particle,

21 vmtFvm =∆+ ∑

When a baseball is struck by a bat, contact occurs over a short time interval but force is large enough to change sense of ball motion

Non-impulsive forces are forces for which (F ∆t) is small and therefore, may be neglected; Eg., weight of body, spring force etc.


Impulse motion of several particles

v∆ 21 mtFvm =+ ∑∑ ∑

Second term involves only impulse and external forces

If all the external force are non-impulsive, then Σ m v1 = Σ m v2

This indicates that total momentum of particle is conserved, andnot energy; Eg., Two particles moving freely collide each other


A 0.5 kg baseball is pitched with a velocity of 80 m/s. After the ball is hit by the bat, it has a velocity of 120 m/s in the direction shown. If the bat and ball are in contact for 0.15 s, determine the average impulsive force exerted on the ball during the impact.

x

y

x component equation:

y component equation:

( ) ( ) ( )

N58.42

40cos1209.81

5.015.080

9.815.0

40cos21

=

°=+−

°=∆+−

x

x

x

F

F

mvtFmv

( ) ( )

N21.26

40sin1209.81

5.015.0

40sin0 2

=

°=

°=∆+

y

y

y

F

F

mvtF

F = 64.03 N; Θ = 24.16°


The initial velocity of the block in position A is 9 m/s. Knowing that the coefficient of kinetic friction between the block and the plane is 0.30, determine the time it takes for the block to reach B with zerovelocity, if (a) θ = 0, (b) θ =20°.

A) θ = 0

t = 3.05 s


b) θ = 20

t = 0.96 s


18Mg 13Mg

72 km/hrA light train made of two cars travels at 72 km/h. The mass of car A is 18 Mg and the mass of car B is 13 Mg. When the brakes are suddenly applied, a constant braking force of 19 kN is applied to each car. Determine (a) the time required for the train to stop after the brakes are applied, (b) the force in the coupling between the cars while the train is slowing down.

(a) Entire train: 72 km/h = 20 m/s


b) Car A: mA = 18 Mg = 18x103 kg; t1-2 = 16.32 s


Direct Central Impact

Oblique Central Impact

• Impact: Collision between two bodies which occurs during a small time interval and during which the bodies exert large forces on each other.

• Line of Impact: Common normal to the surfaces in contact during impact.

• Central Impact: Impact for which the mass centers of the two bodies lie on the line of impact; otherwise, it is an eccentric impact..

• Direct Impact: Impact for which the velocities of the two bodies are directed along the line of impact.

• Oblique Impact: Impact for which one or both of the bodies move along a line other than the line of impact.

Impact


Before impact

At maximum deformation

After impact

VA’ VB’

• Bodies moving in the same straight line, vA > vB .

• Upon impact the bodies undergo a period of deformation, at the end of which, they are in contact and moving at a common velocity.

• A period of restitution follows during which the bodies either regain their original shape or remain permanently deformed.

• Wish to determine the final velocities of the two bodies. The total momentum of the two body system is preserved,

• A second relation between the final velocities is required.

mAvA + mBvB = mAvA’ + mBvB’

(In scalar, same direction)

Direct central impact


• Period of deformation: umPdtvm AAA =− ∫

• Period of restitution: AAA vmRdtum ′=− ∫

During period of deformation, P is the impulse force exerted on A by B

During period of restitution, R is the impulse force exerted on A by B

In general, R and P are different; Impulse of R < Impulse of P

10 ≤≤

−

′−==

=

∫∫

e

uv

vu

Pdt

Rdt

nrestitutio of tcoefficien e

A

A

Particle A


Particle B

• A similar analysis of particle B yieldsB

B

vu

uv e−−′

=

• Combining the relations leads to the desired second relation between the final velocities. ( )BAAB vvevv −=′−′

The relative velocity of two particles after impact can be obtained by multiplying relative velocity of two particles before impact with ‘e’.

=> This property is used to find ‘e’ of particles experimentally


( )BAAB vvevv −=′−′ + ve sign = right motion

- ve sign = left motion

FINAL:


mAvA + mBvB = mAvA’ + mBvB’ ( )BAAB vvevv −=′−′

1. Perfectly plastic impact, e =0 => vB’ = vA’

=> No period of restitution; Particles stay together after impact

Let vB’ = vA’ = v’ => mAvA + mBvB = (mA+ mB) v’ ;

=> v’ can be solved

2. Perfectly elastic impact, e =1 => vB’ - vA’ = vA - vB

=> Relative velocities before and after impact are equal; Impulses received by each particle during period of deformation and restitution are same; Particles move with same velocity but opposite in nature

=> Total energy and total momentum conserved (not in general case)

=> vA’ and vB’ can be solved


VB’

Oblique central impact

VA’Velocities of particles are not directed along the line of impact => Oblique impact

Final velocities VA’ and VB’ and directions

are un-known

n axis => along line of impact; t axis => common tangent

Frictionless, smooth surface; Impulses due to internal forces directed along the line of impact – n axis

-F ∆t

F ∆t


1) Tangential component of momentum for each particle considered separately is conserved.

( ) ( ) ( ) ( )tBtBtAtA vvvv ′=′=

2) Normal component of total momentum of the two particles is conserved.

( ) ( ) ( ) ( )nBBnAAnBBnAA vmvmvmvm ′+′=+

3) Normal components of relative velocities before and after impact are related by the coefficient of restitution.

( ) ( ) ( ) ( )[ ]nBnAnAnB vvevv −=′−′

We have 4 independent equations to solve for components of velocities of A and B after impact

Particles can move in space; no constraints


One or two particles have constraints

• Block constrained to move along horizontal surface.

• Three unknowns: i) Final velocity of ball in direction and magnitude and ii) unknown final block velocity magnitude. Three equations required.

• Impulses from internal forcesalong the n axis and from external forceexerted by horizontal surface and directed along the vertical to the surface.

FF −andextF


1) Tangential momentum of ball is conserved.

( ) ( )tBtB vv ′=

2) Total horizontal momentum of block and ball is conserved.

( ) ( ) ( ) ( )xBBAAxBBAA vmvmvmvm ′+′=+

3) Normal component of relative velocities of block and ball are related by coefficient of restitution.

( ) ( ) ( ) ( )[ ]nBnAnAnB vvevv −=′−′

Not extension of central impact case

mava - ∫ Pdt cos θ = mau; mAu - ∫ Rdt cos θ = mAvA’

Period of deformation Period of restitution


10 ≤≤

−

′−==

=

∫∫

e

uv

vu

Pdt

Rdt

nrestitutio of tcoefficien e

A

A

Multiply all velocities by cos θ to obtain their projections on line of impact

e = un – (vA’)n / (va)n - un

Similar to the central impact case, ‘e’ can be derived between initial and final velocities


A 20 Mg railroad car moving at a speed of 0.5 m/s to right collides with a 35 Mg car which is at rest. If after collision 35 Mg car is observed to move to the right at a speed of 0.3 m/s. find the coefft. of restitution between the two cars


(20 x 0.5) + 0 = 20 x vA’ + (35 x 0.3)

vA’ = -0.025 m/s

( )BAAB vvevv −=′−′ e = 0.3 – (-0.025)/0.5-0 = 0.65


The magnitude and direction of the velocities of two identical frictionless balls before they strike each other are as shown. Assuming e = 0.9, determine the magnitude and direction of the velocity of each ball after the impact.

( ) sm0.2630cos =°= AnA vv ( ) sm0.1530sin =°= AtA vv

( ) sm0.2060cos −=°−= BnB vv ( ) sm6.3460sin =°= BtB vv

( ) ( ) ( ) ( )tBtBtAtA vvvv ′=′=We know that, Find vA’ t, vB’ t

( ) ( ) ( ) ( )nBBnAAnBBnAA vmvmvmvm ′+′=+

( ) ( ) ( ) ( )[ ]nBnAnAnB vvevv −=′−′ Find vA’n & vB’n


Two steel blocks slide without friction on a horizontal surface;immediately before impact their velocities are as shown. Knowingthat e = 0.75, determine (a) their velocities after impact, (b) the energy loss during impact.

a) Put given input values and solve these two equations; find final velocities

b) ∆E = T1 – T2


• Three methods for the analysis of kinetics problems:

- Direct application of Newton’s second law

- Method of work and energy

- Method of impulse and momentum

Application of three methods

• Select the method best suited for the problem or part of a problem under consideration.


(b) Pendulum A from A1 to A2: Apply conservation of energy principle, find vA2 at A2

(c) A hits B: Total momentum of two pendulums is conserved; use relation between relative velocities; find vA3, vB3 after impact

(d) B from B3 to B4: Apply conservation of energy principle to pendulum B

(b) (c) (d)


A 7.9 kg sphere A of radius 11.4 cm moving with a velocity v0 of magnitude 1.8 m/s strikes a 0.73 kg sphere B of radius 5 cm which was at rest. Both spheres are hanging from identical light flexible cords. Knowing that the coefficient of restitution is 0.8, determine the velocity of each sphere immediately after impact.

Ball AF∆t

mAvA

+θ

=mAvA’

mAvA - F∆t cos θ = mAvA’

Ball B

F∆t θ =

θ mBvB’

F∆t = mBvB’

vA’ cosθ – vB’ = e (-v0 cos θ)Solve three equns,

vA’ = +1.56 m/s; vB’ = 2.76 m/s

0.164

0.064θ

Θ =22.9°


System of particles

Motion of large number of particles considered together

Application of Newton’s second law for system of particles

• Newton’s second law for each particle Pi in a system of n particles,

( )

forceeffective

forcesinternalforceexternal

1

1

=

==

×=×+×

=+

∑

∑

=

=

ii

iji

iii

n

jijiii

ii

n

jiji

am

fF

amrfrFr

amfF

Repeat this for ‘n’ number of particles & will get ‘n’ equations. The vectors miai

are called effective forces of the particles; The external forces Fi and internal forces fij acting on the various particles form a system equivalent to thesystem of the effective forces miai.


The external forces Fi and internal forces fij acting on the various particles form a system equivalent to the system of the effective forces miai.

• Summing over all the elements,

( ) ( ) ( )∑∑ ∑∑

∑∑ ∑∑

== ==

== ==

×=×+×

=+

n

iiii

n

i

n

jiji

n

iii

n

iii

n

i

n

jij

n

ii

amrfrFr

amfF

11 11

11 11


• Since the internal forces occur in equal and opposite collinear pairs, the resultant force and couple due to the internal forces are zero,

( ) ( )∑∑∑∑

×=×

=

iiiii

iii

amrFr

amF

The system of the external forces acting on the particles and the system of the effective forces of the particles are equipollent

NOTE: Equipollent system of vectors: ΣF = ΣF’; ΣMo = ΣMo’


Linear & angular momentum

Linear momentum of the system of particles,

∑∑

∑

==

=

==

=

n

iii

n

iii

n

iii

amvmL

vmL

11

1

Angular momentum about fixed point O of system of particles,

ΣF = L; ΣMo = Ho

Resultant of the external forces is equal to rate of change of linear momentum of the system of particles; Moment resultant about fixed point O of the external forces is equal to the rate of change of angular momentum of the system of particles

n( )

( ) ( )

( )∑

∑∑

∑

=

==

=

×=

×+×=

×=

n

iiii

n

iiii

n

iiiiO

iiiiO

amr

vmrvmrH

vmrH

1

11

1 collinear


Motion of the mass center of a system of particles

Mass center G of system of particles is defined by position vector which satisfies

Gr

n

∑=

=i

iiG rmrm1

Differentiating, ∑=

=i

iiG rmrm1

n

∑ ===

Lvmvmn

iiiG

1

Differentiating, ∑== FLam G

This defines motion of mass center G of the system of particles; The mass center of a system of particles moves as if the entire mass of the system and all the external forces were concentrated at that point; MOTION OF EXPLODING SHELL

Similarly angular momentum of a system of particles about mass center can be obtained

Where ‘m’ is the total mass Σmi of the

particles


Conservation of momentum for a system of particles

• If no external forces act on the particles of a system, then the linear momentum and angular momentum about the fixed point O are conserved.

constantconstant

00

==

==== ∑∑

O

OO

HL

MHFL

• Concept of conservation of momentum also applies to the analysis of the mass center motion

constant==G

Constant=GvvmL

Similarly, HG = Constant


Kinetic energy of system of particles

KE of system of particles is defined as sum of KEs of various particles of the system

∑=

=n

iiivmT

1

221

It is convenient to consider separately the motion of mass center G of the system and the motion of the system relative to a moving frame attached to G

∑=

′+=n

iiiG vmvm

1

2212

21T

Kinetic energy is equal to kinetic energy of mass center ‘G’plus kinetic energy relative to the centroidal frame Gx’y’z’.


Work-energy principle for system of particles

Principle of work and energy can be applied to each particle Pi ,

2211 TUT =+ →

where represents the work done by the internal forces, fij and the resultant external force Fi acting on Pi .

21→U

Principle of work and energy can be applied to the entire system by adding the kinetic energies of all particles and considering thework done by all external and internal forces

Although fij and fji are equal and opposite, the work of these forces will not, in general, cancel out.

If the forces acting on the particles are conservative, the work is equal to the change in potential energy and

2211 VTVT +=+

which expresses the principle of conservation of energy for the system of particles.


Principle of impulse & momentum for system of particles

ΣF = L; ΣMo = Ho

21

12

2

1

2

1

LdtFL

LLdtF

t

t

t

t

=+

−=

∑ ∫

∑ ∫

21

12

2

1

2

1

HdtMH

HHdtM

t

tO

t

tO

=+

−=

∑ ∫

∑ ∫


21

12

2

1

2

1

LdtFL

LLdtF

t

t

t

t

=+

−=

∑ ∫

∑ ∫

21

12

2

1

2

1

HdtMH

HHdtM

t

tO

t

tO

=+

−=

∑ ∫

∑ ∫

• The momenta of the particles at time t1 and the impulse of the forces from t1 to t2 form a system of vectors equipollent to the system of momenta of the particles at time t2 .

With no external forces acting, L1 = L2; H1 = H2

The linear momentum & angular momentum of system of particles are conserved


• Kinetics principles established so far were derived for constant systems of particles, i.e., systems which neither gain nor lose particles.

• A large number of engineering applications require the consideration of variable systems of particles, e.g., hydraulic turbine, rocket engine, etc.

• For analyses, consider auxiliary systems which consist of the particles instantaneously within the system plus the particles that enter or leave the system during a short time interval. The auxiliary systems, thus defined, are constant systems of particles.

Variable systems of particles


• System consists of a steady stream of particles against a vane or through a duct.

• The auxiliary system is a constant system of particles over ∆t.

• Define auxiliary system which includes particles which flow in and out over ∆t.

( )[ ] ( )[ ]BiiAii

t

t

vmvmtFvmvm

LdtFL

∆+=∆+∆+

=+

∑∑∑

∑ ∫ 21

2

1

dm ( )AB vv

dtF −=∑

Steady stream of particles

Air flow through duct or blower


• Fluid Flowing Through a Pipe

• Jet Engine

• Fan

• Fluid Stream Diverted by Vane or Duct

• Helicopter

Applications


Systems gaining or losing mass

• Define auxiliary system to include particles of mass m within system at time t plus the particles of mass ∆m which enter the system over time interval ∆t.

• The auxiliary system is a constant system of particles.

21

2

1

LdtFL

t

t

=+ ∑ ∫

( ) ( ) vmvvmvmtF a ∆∆+−∆+∆=∆∑

( )[ ] ( )( )vvmmtFvmvma

∆+∆+=∆+∆+ ∑

negligibleu = va - v


m)vmtF (∆-∆=∆∑ u

ΣF = m (dv/dt)-u (dm/dt)

ΣF + u (dm/dt) = ma

In the case of losing mass case, rate of change of mass is negative


A 20 N projectile is moving with a velocity of 100 m/s when it explodes into 5 and 15 N fragments. Immediately after the explosion, the fragments travel in the directions θA = 45o and θB = 30o. Determine the velocity of each fragment.

x

y


• Write separate component equations for the conservation of linear momentum.

x components:

( )1002030cos1545cos5 =°+° BA vv

y components:

030sin1545sin5 =°−° BA vv

• Solve the equations simultaneously for the fragment velocities.

sm6.97sm207 == BA vv

( ) ( ) ( ) 0

0

20155 vgvgvg

vmvmvm

BA

BBAA

=+

=+


Grain falls onto a chute at the rate of 240 N/s. It hits the chute with a velocity of 20 m/s and leaves with a velocity of 15 m/s. The combined weight of the chute and the grain it carries is 600 N with the center of gravity at G. Determine the reactions at C

and B.

The sum Σmivi of the momenta of the particles supported by the chute is same at ‘t’ and ‘t+∆t’

kg/s24 /sm10

N/s2402==

∆∆

t

m


( ) ( ) ( ) °∆−=∆+−+∆− 10sinByA vmtBWCvm

( ) °∆=∆ 10cosBx vmtC

=+ ∑ ∫ 21 LdtFL

Apply impulse-momentum principle,

In x-axis:

In y-axis:

( ) ( )( ) ( ) °∆−°∆=

∆+−+∆−

10sin1210cos6

1273

BB

A

vmvm

tBWvm

=+∑∫ 2,1, CCC HdtMH

Put input values in three equations and solve for B, Cx, Cy

( )584 .75 N 354 .5 432 .75 NB C i j= = +r r r


A bullet is fired with a horizontal velocity of 500 m/sthrough a 3-kg block A and becomes embedded in a 2.5-kg block B. Knowing that blocks A and B start moving with velocities of 3 m/s and 5 m/s, respectively, determine (a) the mass of the bullet, (b) its velocity as it travels from block A to block B

(a) the mass of the bullet

mv0+mA(0)+mB(0) = mvB+mAvA+mBvB

m = mAvA+mBvB / (v0-vb) = (3) (3) + (2.5) (5) / (500-5) = 43.434 x10-3 kg

(b) velocity as it travels from block A to block B

mv0+mA(0) = mv1+mAvA => v1 = mv0 - mAvA / m =

= 43.434 x10-3 (500) – (3) (3) / 43.434 x10-3 = 292.8 m/s


A system consists of three particles A, B, and C. We know that WA = WB = 17.79 N and Wc = 124.55 N and that the velocities of the particles, expressed in m/s are, respectively, vA = 42i + 63j, vB = −42i + 63j, and vC = −9j − 6k. Determine the angular momentum HO of the system about O.

H0 = ZERO


In a game of pool, ball A is moving with a velocity v0 of magnitude v0 = 4.57 m/s when it strikes balls B and C, which are at rest and aligned as shown. Knowing that after the collision the three balls move in the directions indicated and assuming frictionless surfaces and perfectlyelastic impact (that is, conservation of energy), determine the magnitudes of the velocities vA, vB and vC. v0

vA

vB

vC

j

i

X-axis:

y-axis:

Find VB and Vc in terms of V0 and VA

Put VB and Vc


The nozzle shown discharges a stream of water at a flow rate Q= 1.8 m3/min with a velocity v of magnitude 18.29 m/s. The stream is split into two streams with equal flow rates by a wedge which is kept in a fixed position. Determine the components of the force exerted by the stream on the wedge.

Q = 1.8 m3/min = 0.03 m3/s

Impulse-momentum principle∆m (v) + F ∆t = ∆m/2 (v1) + ∆m/2 (v2)

F = ∆m / ∆t (1/2 v1+1/2 v2 – v)

j

i

v1

v2

Velocity vectors:


F

F = - 117.17 i - 56.8 j N


Sand falls from three hoppers onto a conveyor belt at a rate of 40 kg/s for each hopper. The sand hits the belt with a vertical velocity v1 = 3 m/sand is discharged at A with a horizontal velocity v2 = 4 m/s. Knowing that the combined mass of the beam, belt system, and the sand it supports is 600 kg with a mass center at G, determine the reaction at E.

FE

a a a a

42

h

W = mg = 600 x 9.81 = 5886 N

R = 4040 N (+)

kinetics of particles study of the relations existing between the

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