lecture 3 : rigid body kinematicsdynlab.mpe.nus.edu.sg/mpelsb/me3112/l3n.pdf · lecture 3 : rigid...

Lecture 3 : Rigid Body Kinematics

or describing the motion of finite sized objects

Gerard Leng, ME Dept, NUS

3. Rigid Body Kinematics

3.1. Preliminaries

3.1.1. What is a rigid body ?

A rigid body is an object that doesn’t change it’s shape..

In other words ...

1) The distance between any two points on a rigid body remains

the same.

2) You need to check its orientation as it moves.


3.1.2. How do you describe the motion of a rigid body ?

In general, a rigid body can

a) translate in three dimensions

and

b) change its orientation (i.e. rotate)

about three axes.

Animate


A rigid body has a maximum of 6 degrees of freedom.

Why ?

3.1.3. How many degrees of freedom are there ?

How many degrees of freedom does a particle have at most ?


3.1.4. What happens if the rigid body motion is restricted to a

plane ?

For example if the rigid body lies

on the XY plane, it can also rotate

about the Z axis.

X

Y

Hence there are only at most 3 degrees of freedom

Good news : In this course we will restrict the study of rigid

body motion to this planar case.


3.2. Rigid body kinematics

3.2.1. How to relate the motion of two points on a rigid body

1. The motion of any two points A and B on a rigid body cannot

be independent.

How to find the relation ?

2. The trick is to embed a reference frame on the rigid body

and let the reference frame rotate with the rigid body.


A

Br

w

VB = VA + VB/A

= VA + d rB/A /dt

= VA + ( rB/A ’ + w x rB/A )

= VA + w x rB/A

embedded reference

frame

i

j


VB = VA + w x rB/A

Formula : Hence we can relate the velocity of any two points

on a rigid body as follows :


3.2.2. What about the accelerations of any two points on a

rigid body?

Similarly, accelerations of any two points are not independent.

Using the same trick ...

aB = aA + aB/A

= aA + d2 rB/A /dt2

= aA + ( rB/A ’’ + w x (w x rB/A )

+ w ' x rB/A + 2 w x rB/A ’ )

= aA + w x (w x rB/A ) + w ' x rB/A


Formula : Accelerations of any two points on a rigid body are

related as follows :

aB = aA + w x ( w x rB/A ) + w ' x rB/A


3.2.3. Simplification for planar motion

If the rigid body is moving in a plane

then the angular velocity vector w is

perpendicular to rB/A .

Saves you the trouble of taking two cross products !

w x ( w x rB/A ) = - w2 rB/A rB/A

w

Holds only for planar motion !


Example : A sliding rod

1. Rod AB, length L m

Step 1 : What is the first step for any dynamics problem ?

X

Y

Answer : Define a suitable reference frame

2. Point A sliding down at VA m/s

3. Find VB for given


Y

X

Step 2 : What is the principle involved ?

Answer : Relate velocities of two points

on a rigid body

VB = VA + w x rB/A

Step 3 : Write the vectors out

rB/A =

VA =

VB =

w =


Putting it together ....

VB = VA + w x rB/A

Step 4 : Are the no. of equations equal to the no. of unknowns ?

Solve the 2 equations for 2 unknowns VB and w

i component :

j component :


Learning how to teach yourself ....

What about accelerations ?

Can we find aB given aA ?

Do we need any additional information ?


Example : Crank-slider - The systematic vector approach

Given :

1) AB = R m BP = L m

2) ’ = rad/s (constant)

Find :

1) angular acceleration of BP

2) acceleration of point P

for a given value of (ie at a given moment in time)


R L

NB : If R, L and are given then is a known quantity. Why ?

Answer :


I

J

2. Consider rigid body AB (crank)

VB = VA + w x rB / A

VA = w =

rB / A =

=

1. Define a reference frameR


aB = aA - 2 rB/A + w ' x rB/A

aA = w =

rB / A =

=

Similarly for acceleration of point B

In other words ....

We can find the velocity and acceleration of point B


3. Consider rigid body BP (link). What’s special about VP ?

Answer :

rP / B = w =

Using VP = VB + w x rP / B

we have two equations for the two unknowns VP , w

L

I

J


Similarly using

aP = aB - w2 rP /B + w’ x rP /B

We have two equations for the two unknowns aP and w’.

w’ =aP =

I

J


3.3 Instantaneous centre of rotation

If the rigid body moves on a plane then at each instance in time

there is a point about which the rigid body seems to rotate.

VB

VAA

B

w

C

Given velocity of point A we can

find a point C such that

VA = w x rA/C 1)

For any other point of the rigid body

VB = VA + w x rB/A

2)


Substitute 1) in 2)

VB = w x rA/C + w x rB/A

= w x ( rB/A + rA/C )

= w x rB/C

Hence every other point of the rigid body rotates about point C too.

Note that

i) The point C is the instantaneous centre of rotation for plane motion.

ii) Its location changes with time.

iii) It need not lie on the rigid body.


Example : A sliding rod – using instantaneous centre

1. Rod AB, length L m

2. Point A sliding down at VA m/s

3. Find VB and w for given

Solution

1. The instantaneous centre C

lies on a line perpendicular

to VA passing through A.

2. Point C must also lies on a

line perpendicular to VB

passing through B.

VA

VB

C


VA

VB

C From the geometry

VA = w L sin

VB = w L cos

These are the same equations

obtained by the previous method.

Hence

w = VA / ( L sin)

VB = VA / tan

Warning : Do not apply this method for computing accelerations !


Example : Crank slider – using instantaneous centre

R L

I

J

O

A

B

Given VA and find

i) VB and

ii) w the angular speed of AB


R L

I

J

O

A

B

Question : Where is the instantaneous centre ?

Answer :.

VA

VB

From the geometry

1. Find OB

2. BC =

3. OC =

4. AC =


Using the concept of instantaneous centre

VA =

VB =

2 equations for the 2 unknowns w and VB or solve graphically.

lecture 3 : rigid body kinematicsdynlab.mpe.nus.edu.sg/mpelsb/me3112/l3n.pdf · lecture 3 : rigid...

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