unit 5 rigid body dynamics

ENGINEERING MECHANICSUnit – V

Rigid Body Dynamics

S.Thanga Kasi RajanAssistant ProfessorDepartment of Mechanical EngineeringKamaraj College of Engineering & Technology,Virudhunagar – 626001.Tamil Nadu, India

Email : stkrajan@gmail.com

S. ThangaKasiRajan, stkrajan@gmail.com 2

Kinematics of Rigid BodiesA rigid body has size that is not negligible and does not deform (distance between two points on body is constant). (Idealisation)

Rigid body motion involves translation and/or rotationTypes of Rigid Body Plane Motion

Translation: - No rotation of any line in the body- All points in body have same velocity and acceleration- No relative motion between any two particles

Rectilinear translation

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Translation

Every line segment on the body remains parallel to its original direction during the motion

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Fixed-axis rotation:- All points move in circular paths about axis of rotation

Curvilinear translation

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Rotation about fixed axis

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

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General plane motion

Combination of translation and rotation

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General plane motion- Both translation and rotation occur- Distances between particles are fixed

Note: We will consider plane motion only.

- Relative motion of one particle to another will always be a circular motion

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General Plane Motion is the summation of a Translation and a Rotation

Consider the motion of the rigid bar AB:

General Motion

Rotation about AA2

We could break this motion down another way:

General Motion

Translation with B

Rotation about BA’1

B1 B’1

Translation with A

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Rigid Bodies:

Why are Rigid Bodies so different from Particles?

- Size negligible compared to motion

Particles:

- All forces act through center of gravity

- Neglect rotation about center of gravity

- Points of application, and lines of action of forces are important

- Rotation and Moments about center of gravity are important

Rigid Bodies Vs Particles

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Types of rigid body planar motion

Translation – only linear direction

Rotational about fixed axis – rotational motion

General plane motion – consists of both linear and rotational motion

Rigid-Body Motion

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Example

Rectilinear translation Rotation about a fixed axis

Curvilinear translationGeneral plane motion

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Translation

ABAB /rrr Position

constdtd

Velocity

Acceleration

All points move with same velocity and acceleration14/12/2014

Summary• Time dependent acceleration

dvvdsa

• Constant acceleration

tavv c 0

1 tatvss c

)(2 020

2 ssavv c

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Rotation About a Fixed axis

Angular Position ( q )

Defined by the angle q measured between a fixed reference line and rMeasured in rad

Angular DisplacementMeasured as dqVector quantityMeasured in radians or revolutions1 rev = 2 p rad

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qq dtd

Angular velocity ( )“the time rate of change in the angular position”

Angular acceleration“the time rate of change of the angular velocity”

= f(q)

q dd 14/12/2014

Constant Angular Acceleration

)(2 02

02 qq c

1 tt cqq

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Comparison

dtdsv )(ts

dvvdsa

tavv c 0

1 tatvss c

)(2 020

2 ssavv c

dtdq 2

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Motion of Point P

Position :

qrs The arc-length is

Is defined by the position vector r

tvdtds

( )qrdtd

rVelocity

“tangent to the path”

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Acceleration

( )rdtd

rr 2)(

Direction of an is always toward O

“rate of change in the velocity’s magnitude”

“rate of change in the velocity’s direction”

a 22rt aa ( ) ( )222 rr 42 r

Motion of Point P

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2211 rrS qq

2211 rr

2211 rra

s , v, a

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Restat = 4t m/s2

ra tP )(

)2.0()4(

tdtd 20

210 tdtd

Problem 1

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Problem 2

Cable C has a constant acceleration of 225 mm/s2 and an initial velocity of 300 mm/s, both directed to the right.

Determine (a) the number of revolutions of the pulley in 2 s, (b) the velocity and change in position of the load B after 2 s, and (c) the acceleration of the point D on the rim of the inner pulley at t = 0.

SOLUTION:• Due to the action of the cable, the

tangential velocity and acceleration of D are equal to the velocity and acceleration of C. Calculate the initial angular velocity and acceleration.

• Apply the relations for uniformly accelerated rotation to determine the velocity and angular position of the pulley after 2 s.

• Evaluate the initial tangential and normal acceleration components of D.

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Problem 2SOLUTION:• The tangential velocity and acceleration of D are equal to the

velocity and acceleration of C. ( ) ( )( )

( )srad4

smm300

( )( )

( ) 2srad33

rasmmaa

• Apply the relations for uniformly accelerated rotation to determine velocity and angular position of pulley after 2 s.

( )( ) srad10s 2srad3srad4 20 t

( )( ) ( )( )rad 14

s 2srad3s 2srad4 22212

( ) revs ofnumber rad 2

rev 1rad 14

N rev23.2N

( )( )( )( )rad 14mm 125

srad10mm 125

m 75.1sm25.1

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Problem 2• Evaluate the initial tangential and normal acceleration

components of D.

( ) 2smm225CtD aa

( ) ( )( ) 2220 smm1200srad4mm 57 DnD ra

( ) ( ) 22 smm1200smm225 nDtD aa

Magnitude and direction of the total acceleration,

( ) ( )22

1200225

nDtDD aaa2smm1220Da

( )( )

2251200

4.7914/12/2014

Problem 3

The double gear rolls on the stationary lower rack: the velocity of its center is 1.2 m/s.

Determine (a) the angular velocity of the gear, and (b) the velocities of the upper rack R and point D of the gear.

SOLUTION:• The displacement of the gear center in

one revolution is equal to the outer circumference. Relate the translational and angular displacements. Differentiate to relate the translational and angular velocities.

• The velocity for any point P on the gear may be written as

Evaluate the velocities of points B and D.

APAAPAP rkvvvv

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Problem 3SOLUTION:• The displacement of the gear center in one revolution is

equal to the outer circumference.

For xA > 0 (moves to right), < 0 (rotates clockwise).

p 122rx

Differentiate to relate the translational and angular velocities.

m0.150sm2.1

( )kk srad8

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Problem 3• For any point P on the gear, APAAPAP rkvvvv

Velocity of the upper rack is equal to velocity of point B:

( ) ( ) ( )( ) ( )ii

rkvvv ABABR

sm8.0sm2.1m 10.0srad8sm2.1

( )ivR sm2

Velocity of the point D:

( ) ( ) ( )iki

rkvv ADAD

m 150.0srad8sm2.1

( ) ( )sm697.1

sm2.1sm2.1

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Slider Crank Mechanism

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Slider Crank Mechanism consists of 1. Crank shaft – Pure Rotation2. Connecting rod – Both Translation and

Rotation3. Piston – Pure Rotation

The motion of Connecting rod depends on motion of crank shaftSimilarly the motion of piston depends on motion of connecting rod.

Slider Crank MechanismSlider Crank Mechanism

S. ThangaKasiRajan, stkrajan@gmail.com 2914/12/2014

Slider Crank MechanismMotion of Crank AB

VB = VA + VB/A here VA = 0 because A is fixed

therefore

VB = VB/A = rAB . ωAB

Slider Crank MechanismMotion of Connecting Rod:

When crank rotates in clockwise direction, connecting rod rotates in anticlockwise direction.Also VC/B is perpendicular to the axis of the connecting rod

Apply sine and cosine rule to find the magnitude and direction the velocity of each component

Problem 4

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In the reciprocating engine shown in the figure, the crank AB has a constant angular velocity of 2000 rpm. For the crank position indicated determinei). Angular velocity of Crank ABii). Angular Velocity of the Connecting Rod BCiii). Velocity of Piston

Problem 4

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Problem 4

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S.ThangaKasiRajan, stkrajan@gmail.com 35

References1. Ferdinand P Beer & E.Russell Johnston “VECTOR MECHANICS FOR ENGINEERS STATICS & Dynamics”, (Ninth Edition) Tata McGraw Hill Education Private Limited, New Delhi.2. Engineering Mechanics – Statics & Dynamics by S.Nagan, M.S.Palanichamy, Tata McGraw-Hill (2001).

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S.ThangaKasiRajan, stkrajan@gmail.com 36

Thank you

Any Queries contact

S.Thanga Kasi RajanAssistant ProfessorDepartment of Mechanical EngineeringKamaraj College of Engineering & Technology,Virudhunagar – 626001.Tamil Nadu, India

Email : stkrajan@gmail.com

02/01/2017

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