Chapter5

Chapter 5

Planar Rigid Body Kinetics

Kinetics: Four-Step Problem Solving Method

The suggested plan of action for solving kinetics problems:

1. Free body diagram(s). Draw appropriate free body diagrams (FBDs) for the problem. Yourchoice of FBDs is problem dependent. For some problems, you will draw an FBD for eachbody; for others, you will draw an FBD for the entire system. An integral part of your FBDsis your choice of coordinate systems. For each FBD, draw the unit vectors corresponding toyour coordinate choice.

2. Kinetics equations. At this point, you will need to choose what solution method(s) thatyou will need to use for the particular problem at hand. In this section of the course wewill study four basic methods: Newton/Euler, work/energy, linear impulse/momentum andangular impulse/momentum. Based on your choice of method(s), write down the appropriateequations from your FBD(s) from Step 1.

3. Kinematics. Perform the needed kinematic analysis. A study of the equations in Step 2above will guide you in deciding what kinematics are needed to find a solution to the problem.

4. Solve . Count the number of unknowns and the number of equations from above. If you donot have enough equations to solve for your unknowns, then you either: (i) need to drawmore FBDs, OR (ii) need to do more kinematic analysis. When you have su�cient equationsfor the number of unknowns, solve for the desired unknowns from the above equations.

Chapter5

A. Rigid Body Kinetics: The Newton-Euler Equations

Background

In our earlier studies of the kinetics of particles, we have used the following set of equations for asingle particle i (Newton’s Second Law and the angular momentum equation):

~Fi = mi~ai

~MOi =d

dt

⇥~ri/O ⇥ (mi~vi)

⇤; O is a FIXED point

We also saw that for a SYSTEM of particles, the above equations become:

⇣X~F⌘

ext

= m~aG

⇣X~MA

⌘

ext

=d

dt

X

i

⇥~ri/A ⇥

�mi~vi/A

�⇤+ ~rG/A ⇥ (m~vA)

where G is the center of mass for the system, A is an arbitrary point in the system, m =P

imi

is the total mass of the system, and⇣P ~F

⌘

ext

and⇣P ~MA

⌘

ext

are the total external forces and

moments (about point A), respectively, acting on the system.

A continuous body can be thought of as a collection of an infinite set of particles with each particlehaving an infinitesimal mass. For a continuous body, we can replace the summations over themasses by an integral over the mass of the body:

X

i

( • ) mi !Z

vol

( • ) dm

Objectives

In these lectures our goal is develop and use the set of Newton-Euler equations in solving kineticsproblems dealing with planar motion of rigid bodies.

9¥O

Chapter4

A

!ri/ A

i

G

system of N particles

We will use this relationship in the next chapter when we develop the dynamics equations for rigidbodies.

Note on internal forces : Iii ' F'ji , along a straight line I '%.iii.Fili

ri,*

→

Talia

EMA,int- tis

.

ith's, i-- Iris. - fix. ) xiii= Fig x Fi; -- o - because Ty; H Fi ;

Lecture Material

From the equations in the Background section above, we have the following dynamical equationsfor a system of particles:

⇣X~F⌘

ext

= m~aG

⇣X~MA

⌘

ext

=d

dt

X

i

⇥~ri/A ⇥

�mi~vi/A

�⇤+ ~rG/A ⇥ (m~vA)

A

!ri/ A

i

G

system of N particles

A

!ri/ A

i

G

system of N particles

!vi/ A

To produce an equivalent set of equations for a rigid body, we need to:

• enforce a rigid connection between all points in the system. For this we will use the rigidbody velocity equation between the velocity of A and particle i:

~vi/A = ~vi � ~vA = ~! ⇥ ~ri/A

• envision a rigid body as an infinite set of particles of infinitesimal size for which:

X

i

( • ) mi !Z

vol

( • ) dm

To this end, we substitute the rigid body velocity kinematics equation into the first term on theright-hand side of the angular momentum equation producing:

~ri/A ⇥�mi~vi/A

�= mi~ri/A ⇥

�~! ⇥ ~ri/A

�

From the figures on the following page, we see that the right hand side of the above equationbecomes:

~ri/A ⇥�mi~vi/A

�= mi

��~ri/A��2 ~!

I'A3

d×NM" = dm = pd * dy

D dy

¥

Rigid body : Ti -- VI. thx Fila

-

= f l . , edu

-

Chapter5

Rigid body kinetics: Newton-Euler equations V-7 ME274

Aside:

• Since !ri /A lies in the xy-plane and

!! is in the z-direction, the vector

!! " !ri /A also

lies in the xy-plane perpendicular to !ri /A and has a magnitude of !

!ri /A .

• Therefore, !ri /A !

!" ! !ri /A( ) is in the z-direction and has a magnitude of !

!ri /A2 .

• From this we can write: !ri /A !

!" ! !ri /A( ) =" !ri /A

2 k̂ = !" !ri /A2

!! " !ri /A

!ri /A

!!

!! " !ri /A

!ri /A

!ri /A !

!" ! !ri /A( )

Therefore, the angular impulse-momentum equation becomes:⇣X

~MA

⌘

ext

=d

dt

h⇣Xmi

��~ri/A��2⌘~!i+ ~rG/A ⇥ (m~aA)

=d

dt[IA~!] + ~rG/A ⇥ (m~aA)

= IA~↵+ ~rG/A ⇥ (m~aA)

where

IA =X

mi

��~ri/A��2= mass moment of inertia about point A of a SYSTEM OF PARTICLES

If we now represent the rigid body as a system of particles [Pi

( • ) mi !R

vol

( • ) dm], the

expression for IA becomes:

IA =X

mi

��~ri/A��2 !

Zr2dm = mass moment of inertia about point A of a RIGID BODY

=-

-

Recall in ME 270 : Syeda - polar area moment

A

!ri/ A

i

G

system of rigidly ! connected particles

A

!r

dm

G

continuous rigid body

Some Special Forms of the Euler Equation:

The moment equation derived above for a rigid body:

⇣X~MA

⌘

ext

= IA~↵+ ~rG/A ⇥ (m~aA)

is known as “Euler’s Equation”. Consider the following observations related to the choice of pointA for Euler’s Equation:

• If you choose A to be the center of mass, G, then ~rG/A = ~0. For this case, the above equationreduces to:

X~MG = IG~↵

• If you choose A to be a fixed point on the body, O, then ~aO = ~0. For this case, the aboveequation reduces to:

X~MO = IO~↵

• If you choose a point A which has an acceleration vector that is parallel to ~rG/A (and therefore,

~rG/A ⇥ ~aA = ~0), then the above equation reduces to:

X~MA = IA~↵

For all other points, we need to use the full form of the Euler Equation:

X~MA = IA~↵+ ~rG/A ⇥ (m~aA)

-full form of Euler Equation

• If a body is in translation →I'⇒

Chapter5

CHALLENGE QUESTION: The preceding discussion points out the importance of wiselychoosing point A to enable the use of the short version of Euler’s Equation:

X~MA = IA~↵

Consider a wheel rolling without slipping on a stationary surface, as shown below. Can you use ano-slip contact point C for this moment equation?

B

! A

O

b

C

O

!aC

!vO , !aO

no slip

ANSWER: Recall that the acceleration of the no-slip point C has an acceleration that pointsdirectly toward the center of the wheel. If the wheel is homogeneous (O is the center of mass G),then ~rG/C is parallel to ~aC , and as a result, we can use

P ~MC = IC~↵ for our Euler equation. Onthe other hand, if the wheel is inhomogeneous with O not being the center of mass, then point Ccannot be used. Also, if the wheel slips at C, then the acceleration of C no longer points towardO. For this case, point C cannot be used in the short version of Euler’s Equation.

Example 5.A.1

Given: A crate of mass m slides to the right on a rough surface (with a kinetic coe�cient offriction of µk).

Find: Find the reactions at contact points A and B on the crate.

Rigid body kinetics: Newton-Euler equations V-13 ME274

Example V.1.1 A crate of mass m slides to the right on a rough surface (kinetic coefficient of friction isµk ). Determine the reactions at contact points A and B on the crate.

G

B A

sliding motion

rough surface

b b

d

G

B A

Slides with no external tones .-

Need to find Jia.

I,

Translation ⇒o

IF,= MA Gx

⇒ - fa - fi -- Maa× mg

-Uk Ha -MkNb = man, I-2kg -- M Aa , =o #Tn

, £7,4,

⇒ Nat Nb - MS = o

FA -

- Nk NA , fB = Mk Nbchoose G :

IMG = Ig 2 = o

⇒ µ, b - fad - N,b - fb d '- o

⇒Hab -Mk Nad - Nbb -Mic Nbd -0

⇒ NA . Ms if , fb

Chapter5

Example 5.A.4

Given: The figure below.

Find: For what acceleration of the frame does the uniform slender rod maintain the orientationshown in the figure? Consider all surfaces to be smooth.

Rigid body kinetics: Newton-Euler equations V-16 ME274

Example V.1.4 For what acceleration of the frame will the uniform slender rod maintain the orientation shown in the figure? Consider all surfaces to be smooth.

a

!

A

B

L

KmgNA G

-

K-TNB

Translation ⇒ ji -- o

Efx -- ma ⇒ Naima

Ify -

- o ⇒ NB- mg=o

choose G : -2mg = o ⇒ Nb Ease -Watz Sino -- o

⇒ a = ?

Chapter5

Example 5.A.6

Given: The collar O has a constant acceleration of a to the left. A thin, homogeneous bar oflength L and mass m is pinned to the collar.

Find: Find a value of a such that the bar is held at a constant angle of ✓ = 25�.

θ

m, L

a

O

Oy

0×9 fry*ToG

D= Const ⇒ £-0 I AGx= a, Agy -- o

Efx -

- Maan, -- ma ⇒ Ox = MA

Eff Maaya ⇒ Oy - mg 0

Choose Gi

-2mg - Ig 2=0 ⇒ -Q, Iago - Oy { since -_ o

⇒ - mafioso - mg 's Sino -o

⇒ a -Stano

- Independent of M, <