EGR232 Constrained Motion Dynamics

8/12/2019 EGR232 Constrained Motion Dynamics

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EGR 232 Engineering Dynamics Fall 2012

Lecture 32: Constrained Plane Motion

Today:

ome!or" #uestions

Read C$a% 1& 'ection (ome!or": C$a%) 1&* Pro+lems ,(* (-* -,* 10-

Lecture .utcomes:

Follo!ing today/s lecture you s$ould +e a+le to

a%%ly e!ton/s 2ndLa! to rigid +odies

+e a+le to set u% %ro+lems !$ere t$e rotation is a+out a noncentroidal %oint

+e a+le to set u% %ro+lems or rolling motion)

Constrained Plane Motion:

Constrained plane motion refers to the fact that many times, locations on a rigid body will beconstrained to move in a prescribed fashion. Where a pin moves along a slot or a body rotates

about a pin at one end, or wheel must roll along the curve of the road, certain points on the bodywill have locations required of them.

The dynamics of these constrained bodies may be found by applying Newton's 2ndaws as inprevious problems, but these constraints may have some effect on how the accelerations !both

linear and angular" are related to each other.

Common types of Constrained #lane $otion%Noncentroidal &otation% &olling $otion of Wheel%

! centroid at center of wheel"

ctive Constraint equations% ctive Constraint (quations%

Gt G / O

a r= G G / C a r=

Gn G / Oa r=

2

aGt

F2

F1

G

aGn

..

aG

C

F

aCn

C

G


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)eneral procedural steps for solving these types of problems%

*" +ind the inematic relationships%

a" ocate the relative positions of contact points and centroids. b" -dentify effect of any constraints !such as pulley problems"

c" etermine the angular velocities of each rigid body.

d" etermine the relationships between the angular acceleration and thecentroidal acceleration of the rigid bodies.

2" pply Newton's 2ndaw to each rigid body% a" raw the +/ and cceleration iagram of each rigid body

b" Write out the equations to match the diagrams

xx G

F m a = yy G

F m a =

P G GM I r ma = + r r

0" pply friction equations if appropriate

a" for impending motion%

f sF N= b" for friction below the point of impending motion%

f sF N


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(ample 3%

n 45 g dis is released at rest from the position shown.

etermine the hori6ontal and vertical support reactions at .

)iven%

m 7 45 g &adius% r 7 *.8 m

$ass moment of inertia of a dis about its centroid% I mR= 21

2

9elocity nalysis% &eleased from rest

A Gv v= = 0 and = 0

cceleration analysis%

Aa = 0 G A G / A G / Aa a r r = + 2

G

a k ri i r j = + = +

r0 0 0

+orce:$oment analysis% +/

x GxF m a = Ax GxF ma=

AxF = 0

y GyF m a =

Ay GyF W ma =

Ay GyF W ma mg mr m( g r ) = + = + = +

GM I =

AyF r I = AyI mr mr

Fr r

= = = 1

2

2

2

therefore%

mrmg mr + =

2

g

r=

2

3

then the reaction force may be found as

Aymr mr g

F mgr

= = =

2 1

2 2 3 3

so%

AxF = 0

and AyF mg ( kg )( . m / s ) N= = =21 1

80 9 81 2623 3

G

aGy

aG4

5r 6 1)7 m

F5y

8

F54

G

5r 6 1)7 mm

G


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(ample 4%

long slender bar of mass, m, and length, , is inclined at

an angle of ; with respect to the flat hori6ontal plane. Thebar is not to slide as it is released from rest, determine

a" the support reactions N and +f.

b" the required coefficient of friction.c" the acceleration of the centroid ).

)iven% no slipping ength,

$ass, m

ngle, ;

$ass moment of inertial for long slender rod% I mL= 21

12

)eometry G / AL L r !s i sin j = +

r

2 29elocity analysis% /ody released from rest% v7 v)75

cceleration analysis%

Aa = 0 G A G / A G / Aa a r r = +

2

GL L a k !s i sin j

= + + +

r0 0

2 2

L L !s j sin i = 2 2

+orce:$oment nalysis%

x GxF m a =

fL

F m sin = 2

y GyF m a =

LN W m !s

= 2 N mg mL !s = +

1

2

GM I =

fL L

N !s F sin I + = 2 2

9

L

m

G

5

9

m

G

aGy

aG4

5

8

F

9

L

G


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therefore%

L L

mg mL !s !s mL sin sin I + + =

1 1

2 2 2 2


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E4am%le -:

force # of *4 N is applied to the cord which wraps around the central hub of the wheel shown.

>ub radius is 45 mm and wheel radius is *?5 mm. The mass of the wheel and hub is 8 g and ithad a radius of gyration of *25 mm.


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Force 5nalysis:

x GxF m a = f GP F ma =

y GyF m a = N W = 0 N W mg= =

GM I = #&' f "#$$% Pr F r I =

'ol


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Case 2: 'li% at 5:

5t %oint 5: At Point G:

A Ax n a a i a j= + G G a a i j= + 0

5cceleration analysis:

G A G / A G / Aa a r r = + 2

( )G Ax n "#$$% "#$$% a i j a i a j k r j r j + = + + 2

0

G Ax n "#$$% "#$$% a i j a i a j r i r j + = + 20

i: G Ax "#$$% a a r= =: n "#$$% a r= 2

Force 5nalysis:

x GxF m a = y GyF m a = f GP F ma = N W = 0

N W mg= =

GM I = Friti!n +,

#&' f "#$$% Pr F r I = f kF N=

Com+ine to ind a54* * * and F

N W mg . . N= = = =5 9 81 491f kF N . . . N= = =0 15 49 1 7 365

#&' f "#$$% Pr F r I = #&' f "#$$% Pr F r

I

=

( . ( N mm kg m / s mm

mm kg N m

=

2

2

18 60 7 365 120 1 1000

72000 1 1

. ra* / s= 22725

f Ax "#$$%P F ma mr = fAx "#$$%P F

a r

m

= +

( ). N

. m( . ra* / s . m / skg

= + =2 2

18 7 3650120 2 725 2 454

5

t$ereore t$e dis" does sli% and t$e acceleration o %oint G is gi


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EGR 232 Eng) Dynamics 8 'et 32 Fall 2012

Pro+lem 1&:,(

uniform slender rod of length 7 55 mm and mass m 7 1 g is suspended from ahinge at C. hori6ontal force # of magnitude 38 N is applied at end /.

Dnowing that r 7 228 mm, determine a" the angular acceleration of the rod,

b" the components of the reaction at C.


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Pro+lem 1&: (-

Two uniform rods, /C of mass 0 g and C( of mass 1 g, are connected by a pinat C and by two cords / and /(. The T shaped assembly rotates in a vertical plane

under the combined effect of gravity and a of a couple $ which is applied to rod /C.

Dnowing that at the instant shown the tension is 4 N in cord /, determinea" the angular acceleration of the assembly, b" the couple $.


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Pro+lem 1&:-,

homogeneous sphere


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Pro+lem 1&:10-

Two uniform diss and / each of weight 1 lb, are connected by a 0 lb rod as shown. counterclocwise couple $ of moment *.8 lbAft is applied to dis . Dnowing that the diss roll

without sliding, determine a" the acceleration of the center of each dis b" the hori6ontal

component of the force eerted on dis / by pin .

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EGR232 Constrained Motion Dynamics