Planer kinetics of rigid body

• There are three types of rigid body planar motion: in order of increasing complexity, there are– Translational

• This type of motion occurs if every line segment on the body remains parallel to its original direction during the motion.

• Two type of translation:– Rectilinear translation– curvilinear translation

– Rotation about a fixed axis • when a rigid body rotates about a fixed axis, all the particles of

the body, except those which lie on the axis of rotation, move along circular paths

– General plane motion • when a body is subjected to general plane motion, it undergoes a

combination of translation and rotation, Figure 8.1(d). The translation occurs within a reference plane, and the rotation occurs about an axis perpendicular to the reference plane.

Translation

• When a rigid body undergoes a translation, all the particles of the body have the same acceleration, so that aG= a . Angular acceleration, α=0, gave ΣMG=0.

Rectilinear translation– Particles of body travel along straight line path– Equation of motion

– If point A chosen ( lies at a perpendicular distance d from action of maG. Moment equation applies

0

)(

)(

G

yGy

xGx

M

amF

amF

dmaMMM GAAKA ;

Free body diagram

Kinetic diagram

Curvilinear translation– All particles of the body travel along parallel curved paths.– Equation of motion

– If the moment equation ΣMG=0 is replaced by a moment summation about the arbitrary point B, the required moment become

0

)(

)(

G

tGt

nGn

M

amF

amF

])([][; nGtGBBKB amhameMMM

Procedure for analysisFree body diagram– Establish x, y or n, t inertial coordinate system and draw the free-

body diagram in order to account for all the external forces and couple moments that act on the body.

– The direction and sense of the acceleration of the body’s mass center aG should be established.

– Identify the unknowns in the problem.– If it is decided that the rotational equation of motion ΣMP=Σ(MK)P is to

be used in the solution, then consider drawing the kinetic diagram, since it graphically accounts for the components m(aG)x, m(aG)y , or m(aG)t, m(aG)n and is therefore convenient for “visualizing” the terms needed in the moment sum Σ(MK)P .

Equation of Motion– Apply the three equations of motion in accordance with the

established sign convention.– To simplify the analysis, the moment equation ΣMG=0 can be replaced

by the more general equation ΣMP=Σ(MK)P , where point P is usually located at the intersection of the lines of action of as many unknown forces as possible.

– If the body is in contact with a rough surface and slipping occurs, use the frictional equation F=KN. Remember, F always acts on the body so as to oppose the motion of the body relative to the surface it contacts.

Kinematics– Use kinematics if the velocity and position of the body are to be

determined.– For rectilinear translation with variable acceleration, use

– For rectilinear translation with constant acceleration, use

– For curvilinear translation, use

dtdsv

dvda

dtdva

GG

vGGSGG

GG

/

,

,/

200

02

02

0

2

1)()(

],)([2)(

,)(

tatvss

ssavv

tavv

GGGG

GGGGG

GGG

tGGGGtG

GtG

GnG

advvdsa

dtdva

va

)(,)(

,/)(

,/)( 22

Example 8.1

• The car shown has a mass of 2 Mg and a centre of mass at G. Determine the car’s acceleration if the “driving” wheels in the back are always slipping, whereas the front wheels freely rotate. Neglect the mass of the wheels. The coefficient of kinetic friction between the wheels and the road is = 0.25.

• Solution– Free body diagram

– Method 1 : Equation of motion

therefore

)3(....................0)75.0()3.0(25.0)25.1(;0

)2....(....................0)81.9(2000;)(

)1.(....................)2000(25.0;)(

mNmNmNM

NNNamF

akgNamF

BBAG

BAyGy

GBxGx

kNNkNNsma BAG 7.12;88.6;/59.1 2

• Method 2 : Free body and kinetic diagrams– Moment equation applied at A

)3.0()2000()25.1()81.9(2000)2(; makgmNmNMM GBAKA

Free body diagram

Kinetic diagram

Solve it with equation …(1)

2/59.17.12

245256002

0200025.0

smakNN

aN

aN

GB

GB

GB

=

Example 8.2: curvilinear translation

The 100-kg beam BD is supported by two rods having negligible mass. Determine the force created in each rod if at the instant =30o and (angular velocity)=6rad/s.

free body diagram– Moves with curvilinear translation (B,D and G move along circular

path)– Angular motion of rod AB

• Tangential component of acceleration acts downward to the left due to the clockwise direction of α

• Therefore 222 /18)5.0()/6( smmsradra nG

• Solution– Using Equation of motion

)3..(..........0)4.0)(30cos()4.0)(30cos(;0

)2.(....................)(10030sin981;)(

)1).......(/18(10030cos981;)( 2

mTmTM

akgamF

smkgNTTamF

DBG

tGtGt

DBnGn

2/90.4

32.1

sma

kNTT

tG

DB

Solving three equation

Rotation about a fixed axis

• When rigid body constraint in rotation the body centre mass G move in circular path – The acceleration is represent by its tangential and normal component

– tangential component of acceleration has a magnitude of and must act in a direction which is consistent with the body’s angular acceleration α

– The magnitude of the normal component of acceleration is

GtG ra

GnG ra 2

• The equation can be write as

• Considering the IGα vector acts in the same direction as α and have a magnitude of IG (body moment inertia about axis perpendicular to page and passing through G), therefore moment equation will be

• Will rewrite if the vector passes through O

• By adding parallel theorem axis

GtGt

GnGn

rmamF

rmamF

)(

)( 2

GG IM

2GGo mrIM

2mdII Go

oo IM

Example 8.3

The 30-kg uniform disk shown in Figure is pin-supported at its center. If it starts from rest, determine the number of revolutions it must make to attain an angular velocity of 20 rad/s. Also, what are the reactions at the pin? The disk is acted upon by a constant force F=10N , which is applied to a cord wrapped around its periphery, and a constant couple moment M=5 Nm. Neglect the mass of the cord in the calculation.

• Solution– Free body diagram

• Mass centre is not subjected to acceleration however the disk have a clockwise angular acceleration

• The moment of inertia of the disk about the pin is

• The three unknowns are Ox, Oy and α.– Equation of motion

222 6.0)2.0)(30(2

1

2

1kgmmkgmrI o

2

2

/7.11

)6.0(5)2.0(10;

304

0103.294;)(

0;

srad

kgmNmmNIM

NO

NNOamF

OamF

oo

y

yyGy

xxGx

– Kinematics• Since is constant and is clockwise, the number of radians the disk

must turn to obtain a clockwise angular velocity of 20 rad/s is

• Hence

rad

sradsrad

oco

1.17

0)/7.11(20)/20(

222

22

revrad

revrad 73.2

2

11.17

The 20-kg slender rod shown in Figure 8.9 is rotating in the vertical plane, and at the instant shown it has an angular velocity of = 5 rad/s . Determine the rod’s angular acceleration and the horizontal and vertical components of reaction at the pin at this instant.

• Solution– Free-body and Kinetic Diagrams

• As shown on the kinetic diagram, point G moves in a circular path and so has two components of acceleration.

• It is important that the tangential component act downward since it must be in accordance with the angular acceleration of the rod.

– Method 1: Equation of motion

])3)(20(12

1[60)5.1(;

)5.1)()(20()81.9(20;

)5.1()/5)(20(;

2

22

mkgNmmOIM

mkgNOrmF

msradkgOrmF

tGG

tGt

nGn

– Method : sum moment at point O

• Also, since for a slender rod, we can apply

oko MM

)5.1)](5.1)((20[])3)(20(12

1[)5.1()81.9(2060 2 mmkgmkgmNNm

2/90.5 srad

2

3

1mlI o

;oo IM ])3)(20(

3

1[)5.1()81.9(2060 2mkgmNNm

2/90.5 srad

=

Momentum and impulse

• From 2nd Newton Law,

• Linear momentum for an object – given as the output of multiplication of object mass with the velocity

of the object, i.e. p=mv, .– Momentum is a vector quantity and the SI unit is Ns @ kg m/s.

mumvFtt

uvm

maF

Where :Ft is the impulse

is the momentum change mumv

Linear momentum conservation principle– Moving object may transfer or lose some of its momentum to another

object. – the total of momentum does not change provided that there is no

external force imposed to them.Total of momentum before impact = Total of momentum after impact.

– Types of impacts• Elastic collision• Inelastic collision

• Elastic collision– two object is moving with individual velocity collides, and then separates and

each object moves with different velocity. – Total of kinetic energy and momentum is conserved.

• Inelastic collision– two objects moving with individual velocity collides and attached

together, moving with different initial velocity.– Total of momentum is conserved but the total of kinetic energy for this

system is not conserved.– Equation:

• Impact for elastic body– Collision between two elastic bodies for two spherical that have the

same size and elasticity behaviour but with different materials. From Newton elasticity law

1212211 )( vmmumum

10..................21

12

xeuu

vv

– If e = 0, the material is not elastic– If e=1, the material is fully elastic.– If e = 0, inelastic collision, – If e = 1, elastic collision,

vvv 21 vmmumum )( 212211

2112 uuvv

Example 8.6

A sphere with a mass of 6kg is moving with a velocity of 5 m/s and collides with another spherical with a mass of 4 kg moving with a velocity of 3 m/s. Determine the velocity of spherical after the collision if both special moving in (a) same direction (b) different direction. Given e = 0.5.

Work

• What does WORK mean to you?• Are you doing WORK when…

– Lifting weights?– Walking with a big bag of grocery in your hand?– Completing your homework assignment?– Writing an essay?

Concept of work• WORK is done only when a constant force applied on an object, causes the

object to move in the same direction as the force applied.

• Work or W or U is defined as the output of force, F multiply by the distance taken in the direction (parallel distance) of force, s.

sFW

S2S1 s

F

SI unit is N.m or Joule, J.

• When F is not parallel to s, then we must take the component of F which is parallel to s .

s

Fh= F cos θFv =F sin θWk =Fhs= F (cos θ)s

• The force imposed on a body gives:-– Positive work: If the direction of force parallel to its movement– Negative work: If the direction of force opposite to its movement– Zero work: If the direction of force perpendicular to its movement

s

F1

F4

F3

F2

U3=0

U4=-F4s

U1=F1s

U2=-(F2cos )s

Example 8.7

A force of 4N pulls an object at the slope of in horizontal line. Calculate the work been done if the object was moved in 3m.

Solution:Know ; F = 4 N; s = 3 m; = 60o

Need ; Wuse ; W = F.s parallel = F cos x s

W= 4 x cos 60o x 3 = 6J

4N

Direction of movement

60o

Energy – Quick Re-cap

• Energy is the capacity to do work• SI Unit: Joule (J)• Many forms• Common ones:

– Kinetic– Potential– Electric– Chemical– Solar– Nuclear

Kinetic Energy

• A form of energy that a body in motion possess due to movement.• A body a rest, will it possess any Kinetic Energy?• Examples:

– Bullet shot out from pistol– Helicopter flying at 120km/h

• The amount of Kinetic Energy of a moving body depends on:– Mass of body (kg)– Velocity (ms-1)

• When either mass or velocity of moving body is increased, Kinetic Energy will also increase.

• Formula:

• SI Unit: Joule [ J ] … same as Work Done

Kinetic Energy = x Mass x (Velocity)2

K = x m x v2

Units: [ J ] [kg] [ms-1]2

2

1

2

1

Mass = m kg

Velocity, V

Examples of Kinetic Energy

• Find the KE of an empty van of mass 1000kg moving at 2m/s.

• Find the KE of van when it is loaded with goods to give a total mass of 2000kg, and moving at 2m/s.

• Find KE of unloaded van when it speeds up to 4m/s.

KE of van at 2m/s = ½ x 1000 x (2)2

= 2000 J = 2 kJ

KE of van at 2m/s = ½ x 2000 x (2)2

= 4000 J = 4 kJ

KE of van at 2m/s = ½ x 1000 x (4)2

= 8000 J = 8 kJ

Relationship work and Kinetic Energy

• The work done by the net force acting on a body is equal to the change in the body’s kinetic energy

This relationship is true as long as there is no change in vertical position.

EnergykineticW

energykineticmumvsF

a

uvmasF

smasF

WsF

22

22

2

1

2

1.

)2

(.

..

.

Potential Energy

• Potential energy is the energy possessed by an object as a result of its POSITION or CONDITION.

• Two common kinds:– Gravitational Potential Energy– Elastic Potential Energy

• Gravitational Potential Energy (GPE)– Energy that can be possessed by an object due to its POSITION.– Any object that is at ground level has ZERO GPE.– If object is lifted a certain height above ground, its GPE has increased– Examples:

• When a chair lifted from ground a distance of 1m and You sitting on the 3rd storey of this building

• Elastic Potential Energy– Energy that can be possessed by an object due to its CONDITION. Examples:– Examples:

• Spring … when stretched or compressed• Rubber band … when stretched

Gravitational Potential Energy

• Can be calculated with:GPE = F x distance

=mass gravitational height above acceleration ground level

= m g hUnits: [J] [kg] [m/s2] [m]

SI Units of GPE : Joule [J]

Ground,0 GPE

Distance fromground, h

Object on top ofbuilding, of mass, mg

earth

Elastic Potential Energy

• Using a Hooke’s Law– force which required to stretch or compressed a spring is

proportionate to stretching or compression of a spring if the elastic limit is not exceeded

• If a force, F increased with the fixed rate to stretch a spring to x m, potential energy stored in the spring is

)/(tan............ mNtconsspringkwherekxF

xF

2

2

)(21

21

21@2

1

xk

kx

FsFxU

45

• Energy always remain same or fixed in quantity!• Energy of an object can be thought of as the sands in an hourglass!• But this sand can change position, from the top to bottom and bottom to

top! Likewise energy can change in form– eg. From Kinetic Energy Potential Energy

Conservation of Energy

46

• Conversion of energy is the term used to denote change in energy from one form to another.

• Eg.– Burning candle: Chemical Heat, Light– Fuel: Chemical Heat KE Electricity– Nuclear explosion: Nuclear Heat, light– Spring: Elastic PE KE

• For O-Levels, we are only concerned with:

• KE GPE (E.g: roller coaster, Falling object)

• And such situations are only found when a moving object is at the same time undergoing changes in height

47

• Free falling object– An object in free fall means the object is falling freely, under the

influence of gravity

When the object is at the highest position, the GPE is at maximum and KE is zero.

When the object is falling, the GPE decreases as it loses height, and the KE increases

At the lowest position, the KE is at maximum and GPE is zero.

Example 8.8

Figure shows a pendulum swing where the pendulum mass is 5 kg. The pendulum is pushed from point A which is located 15cm higher than B.a) What is potential energy stored in pendulum at point A.b) Maximum velocity in the pendulumc) Kinetic energy in the pendulum when reached at point B.

40 c

m

15cm

solution

A stone with a mass of 0.1 kg is throwing straight up with a velocity of 12 m/s. Determine the potential energy at maximum peak and kinetic energy when reached the ground with assumption that there is no restriction.

Solution:-

Power

• Power is the rate that we use energy.

• The units for power :– J/s– Kg m2 / s2 /s– N m /s

• Efficiency,

Fvt

sF

t

WP

Time

EnergyorWorkPower

.

powerimput

poweroutput

Remarks

• Energy is the ability to move• Potential is stored energy (Statics)

– Dependant on height• Kinetic is moving energy (Dynamics)

– Dependant on velocity• Springs store energy dependant on distance and constant

Example 8.10

A lorry is moving with a constant velocity of 30 m/s along its way. If the applied friction force is 800N, calculate the engine power.

Solution;

kW

vFpowerEngine

NF

FFmaF g

2430800

800

Example 8.11

A force is applied to compress a spring to the wall. If the initial length of the spring is 50 cm and the last length is 30cm while the last force applied to the spring is 20N,

(a) Calculate the work done,(b) Kinetic elastic energy by the spring

solution

50 cm 30 cm

20 N

Before compression After compression

kJ

xFWork

4)3050(202

1

)(2

1

2

2

a) b) 4kJ the work done by the compression of the spring

Example 8.12

a) A car weigh 1 tan moves on the horizontal road with a constant velocity of 10 m/s and a total of 400N friction force is applied on it. Calculate the pulling power of the engine.

b) If the car riding up the hill with the slope of at horizontal plan and assume that friction force imposed constant at 400N, what is the engine power required to push the car at the velocity of 15 m/s.

kW

vFP

NF

mawhere

maFF g

4

10400

400

0

(a)

NFg 400

mg

F

(b)

5

NFs 400

kWx

vFP

N

F

mgFF

maFF

g

g

83.18151255

1255

5sin)81.9(1000400

05sin

From 2nd Newton Law,

Example 8.13

A mass of 200g been pushed to a spring and compressed to 15 cm from equilibrium. The mass was shot to the right. If the friction is ignored, calculate the speed of the mass that moves to the left when pushed

Solution

K=400 N/m

200 g

From 2nd Newton Law:-

smv

v

mvxk

/71.6

2.0)15.0(400

2

1)(

2

1

22

22