McGill MECH 314 Final Exam Crib Sheet

Directions
Rolling vs . Slipping Acceleration Analysis Vector

-
z X TAB/VAlB =
VAB/AAB
VELOCITY Ups =T pz + Vos12 general RIGID BODY

-
Morion
Ap
=
AQ + Apa + Ana CW >
-
↓
slipping 312 Vie # 0 : = 0 - -
Cow ↑
Apa X Rpa ( + Real
>
-

= <
rolling : pse =
0
Cri ↑
w"Rpa
-
A pa = -
(1/ Rea opposite
ACCELERATION CC w
direction ( - ↓
slipping :
Eps Apz +A =
siz +12 + 212 NONRIGID MOTION

Aps1z 252 54312 Pa
,
>
-
= x
Ap =
Aq + A +
Apa Useful crib Identities

Ap31z V3312 pedist between
(l-sin20)"
2wXVp , a ( + Vpa)
Cost
-
=
1
Apa
- =

, circle centers =

p
cos(x
Asiz sinxsin
F
y)
Apa coSXCoSY
[
rolling Epy
=


of P if Q is fixed
: +
=

Apa -
trajectory
sin(X[y) =
SinXcosy 1 cosXsiny
-

2 Pr
, Ar 212 Apa = Ap +
Apa los2X = los2X-sin'X
and Alsiz
·-
I
& Sin 2x = 2 Sinx cosX


opposite direction
↳
Ana Var Vanishes if Vera
Ap
= -




>12 r is straight
O
r
---




Oy = 00
Analytical +
Computational Algebraic Velocity Analysis 0- - 02

T
Y Oz(t) , Oy(t) , r. (t) :

RBA +
RcB =
RcA
1 .

loop closure
Write equation :


N Vi
↑++5 F 0
B
- =


reCoSE2
----
*: + ry20S8z = V,

2
2 . X and y components :
·




y
:
raSinOc +
FySinGy = O
↑ COSO ,
+ r2C0S82 + r> COSO1 -


ruCOSOy =
O
3

& 02
C
+ (x1) for 82 83 i ,
↑ SinO , + r2SinOz + ResinOs-rySinOz = O


&
,
8
> X
↑ differentiation Bo- 1__
& dr(x2) for
E2 Es
-



11II 3 time
TA r
_
.




,
I , ,




Force Analysis · (racosol =
-Rosidon
E


* X :
VBECOS O-$)
F =
0 : [FY =
0 [F" = 0 Emo = 0
, ,




Crcos) =
BeSin(o-P)
:
Y
Rising
- -

cost
-F Fi
F2
-




~ [
T




T
> E,
F Es . solve
4
systems of equations
F ,
E IC of velocity
Inertial Force +
Torque :
Imaginary Reaction of coincident diff
↳ location a
pair of points on 2 rigid bodie



Fo , hig ! 2923
↳] As Replace / Fi
=
o ,
1
,
!1
,
uP


Fi = -

. si
M
Mi = hFi -h
=2 eX :


· Pis Possible P's

/
12 13
:




15/6
Need P:

Pis 132 134
Gix(i)
6O
+




Ex
,




↑
=
#
= 2242526

Fin
-




3433 36 Need P, s :



(2D)
My 20ix
B




:
1
= -
; O
·
4546
153 156
P35 P> Pis +




/
,
C y
5/
0
- >
-
h
Fin + [F , =
Fi T
+ Piz
Wheel 2 :
purely sliding
i
i
2
Wheel 3 : purely rolling
·

=>



5 P
>

My 0
3 j

+ =
Joi
Pis

Friction Principle of Superposition: Accel . 4kP Example
1) Perform kinematic analysis


"I!? P Ey 2) Make complete static force analysis of the
mechanism (using known values of applied F and M) D3
A
Eps Ad As +
03p 3) Calculate F in and M in by taking inertial forces as
↑ Apy + +
e = =
M

M(given) newly applied forces & torques but ignoring other X Vps12
F
I
F loads already used in static analysis > 12 = 2 W2



jam
B

(m)
" ,4 +


f = tan · a M 4) Add results obtained in individual steps, gives slip btwn
V312
A =
-
+
A
2 2 and 3



F_F
↳ always F En cost
= P

Upsiz RpB
-
W2 CCW

F Fysiny siz
Fin
= +
from Fn
14
*
always goes against 3


+A+ A
·
Vesiz :
direction bi

Fin IMF,
=




staric friction : given velocity ·
An =
Ac
I 3
A212 Il Di
friction :
Fin MFi
B


sliding
+

= rolling btwn

2A 2 ands


c