© Sharif University of Technology - CEDRA By: …meghdari.sharif.edu/files/Session3A.pdfBy:...

By: Professor Ali Meghdari© Sharif University of Technology - CEDRA

By: Professor Ali Meghdari

Rfx

):( slippagenowhenR

f x

):( beginsslippagewhenR

fx

Example: THE FALLING STICK PROBLEM

Y

P

θm

g

G

R

f

x

X

(x,y

)

2ℓ

μ = Coefficient of

Friction

© Sharif University of Technology - CEDRA


amF

From Kinematics:

sincoscossin 2 xxx (1)

From Kinetics:

)cossin(sincos 2 yyy (2)

)sincos( 2 mxmfx

In x-direction: (3)

)]cossin([

)cossin(

2

2

gmR

mymmgR

In y-direction:

(4)



Y

P

θ

mg

G

R

fx X

(x,y)

2ℓ

μ = Coefficient of Friction



PP

PP IIdt

dHM )(

:R

fx

Let us discuss about the ratio

,First, we will try to eliminate the derivatives of θ

(i.e., ) in Equations (3) and (4):

Applying the Moment of Momentum Equation about the point

“P”, or the mass center “G”, we have:

sinsinsin2

2

P

PPk

gmgmkImg

(5)

)( GyrationofRadiusm

Ik P

Pwhere:



)( GyrationofRadiusm

Ik P

Pwhere:

kP is a measure of length, which defines the mass distribution.

Radius of a thin ring having the same mass as the body and with

the same moment of inertia about its axis of symmetry as the body

has about axis P (It is the effective radius of a body as far as

rotation is concerned, as if all the mass is concentrated at kp).

Equations (3), (4), and (5) form 3-independent dynamic equations.

Applying Energy Equation (Total Energy is Constant), we have:

T + V = constant



Gθ

P

Datu

m

ℓ

ℓ

h=(1-cos

θ))cos1( mgmgh

When θ = 0 (at the top):

V = P.E. = 0

T = K.E. = 0

Then :

V =

22

22 GG

Ivm T =

22

2

1

2

1

PPII

Or, when P is fixed (before slippage;

no dissipation of energy), then:

T =



Hence, Energy Equation can be written as (fx ; is non-working):

)(0.)cos1(2

2

sconventionbyconstmgI

P

T + V =

{Integrating Equation (5) with respect to time, would also

result in Equation (6)}

Substituting Eqs. (5) and (6) into Eqs. (3) and (4) results in:

)cos1(2

2

2 P

k

g (6)



)2cos3(sin2

2

P

xk

mgf

(7)

)]1)(cos1cos3(1[2

2

P

kmgR

(8)

P

k

R

fxThis is an expression for ,as a function of

“θ” and ( ), which is also independent of massand g.

)]3

1)(cos1(cos3)[(

)3

2(cossin3

2

P

x

kR

f(9)



22

3

4)2(

3

1 mmI

P

Now, let us analyze the problem:

Consider a homogeneous stick with evenly distributed mass, where;

154.13

22 PPP

kmkIsince:

154.1P

k

What is the physical meaning of this?

It means that if we take all the mass of the body and put it all at:

the rod dynamically would behaves the same as the distributed mass (equivalent

to the original form). But if we put all the mass at the mass center, the rod

dynamically would not behave the same as the distributed mass.



G

k0

Also:

{if all mass is at the mass center G} {if all mass is stretched to both ends of the stick}

212

P

P

kk

Therefore, from equation (10), we have:

21 P

k)(

R

fx

Now, if we fix the value of ( ) and then plot

vs.

θ , we have:

22

2

G

GP

Pk

m

mI

m

Ik

Note that:

(G = mass center) (10)



(Plot of vs. θ): )(R

fx



)]3

1)(cos1(cos3)[(

)3

2(cossin3

2

P

x

kR

f(9)

Note that: From equation (9), when:

03.483

2cos

R

fx

Meaning that slip starts for sure when: θ ≤ 48.3



1

P

P

kk

Special Case: consider a point mass with all the mass at the

mass center, therefore;

tan

)3

2(coscos3

)3

2(cossin3

R

fx

, and:

Similar to the case of a particle on a slope

Where slippage begins when: μ=tanθ

θ

μ


Purpose:

Define and compute kinematical quantities in various coordinate

systems.

Topics:

Path Variables Description (Intrinsic Coordinates)

Cartesian (Rectangular) Coordinates

Orthogonal Curvilinear Coordinates (i.e. Cylindrical, Spherical,

Elliptical, etc.)

Coordinate Transformations


Expectations:

It is expected that all students to be proficient in:

A successful analyst must be able to select and/or to

transform to the most appropriate coordinate system

that conforms best to the motion.


Path Variables Description (Intrinsic Coordinates):

Motion of a particle (or a point) is described in terms of

the properties of its path (i.e. speedometer & odometer in

cars, and road map mileages).

Intrinsic-Coordinate; since any change in the basic

parameters is associated with the properties of the path

(i.e. path dependent).


Consider a particle traveling through the path “ ”:

s : arc length

rP/O = rP/O(s) = Position Vector at time “t” , where: s = s(t)



dt

edvev

dt

evd

dt

vda t

t

t )(

Accélération: since et = et(s) , and s = s(t) , we have :

)( RuleChaindt

ds

ds

ed

dt

ed tt But,

Velocity:

)(;: /

//

PathtoTangentds

rdewhere

evesdt

ds

ds

rd

dt

rdv

OP

t

tt

OPOP

P

(3.1)

hence ;ds

edveva t

t

2



n

t

nntt

t

t

t

eds

ede

dseeded

ede

eddsds

1

dθρ

dθ

(s+ds)ne

(s+ds)te

(s)ne

(s)te

Center of Curvature = C

(s+ds)te

(s)te

det

S ds


)(1

figurefromeds

edn

t

But,

where en ┴ et , and normal to the path.

ta = tangential acc., and

na = normal/centripetal acc.

nnttnt eaeaev

eva

2

(planar motion/path)

(3.2)


For a particle traveling on a path or a curve in 3-dimension (x,y,z)

coordinate so that its path is described by the position vector “r”

as a function of the parameter “t within a possible range”, we have:

Then, the radius of curvature “ρ” is computed from the following relation:

321 )()()( etzetyetxr (3.3)

2/122

3])())([(

)(

11rrsrr

s

(3.4)

2/12222/1 ])()()[()( zyxrrs

where: a dot denotes differentiation with respect to t, and;

(3.5)

)(])()()[(0

2/1222 LengthArcdtzyxst

t

and,

(3.6)


)()(

)]()([)(

3

2

4

vectorunitbinormalrrs

eee

rrrsrsds

dt

dt

ed

ds

ede

s

r

ds

dt

dt

rd

ds

rde

ntb

ttn

t

However, for a planar curve or path like “y=f(x), and z=0, so that t=x,

equation (3.4) reduces to:

2/1222/1222/1 ])()[(])()[()( dydxdsyxrrs and (3.8)

2/322/32

2

2

])(1[])(1[

1

y

y

dx

dy

dx

yd

(3.7)

dxdx

dys

x

x 0

2/12 ])(1[ (3.9)


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