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Review cheat sheet uoft

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Dominik Danda

Academic year: 2023

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Particle

kinematics?

Kinetics

Carviline

NormTan

Impulse

-Momentum Miscellaneous

T=i=i

-=

mv At =0.3048m

Core Formulas

G

>

a =

2

v

= Rectangular 92

=

9,

+

Fdt

qus =32.2 ft/s2

Sp

=Sot Vot+ Iat I=xi +

yy ner Haute ener

H-

=IqCO

Impulse

wired

·N - Sec

Relative Motion *

I=x4 +y)A a=iei +vien I =Ex

mi

558

- - - x

Ang Velocity

#A

=

FB

+

FAI

-

a=x4 +

jj

torm

sin

in i danus in the

SVdt I

where

I

=Fourora

Circular

MotionsO

actex=

-1+x2

Non-Uniform

Acceleration

o

an=r0=r0==

j Sa

=

(iron) et en = Exma

=E

Mo

smallOe

(since in

S

=Sot rot

+/ Y"alt" dt'dt'

at=

I

=r0 0=

dine

1

Fl

=rmv= mr2w

d

VF=Vot

15 act')

dt' *

er

=-en arc=Or

a

a=acs) -> rdw- ads

Ai= I=

v8e+ ie (10)

=

(H),

+

uodT

=

I e

ris)-

a(s) ds Friction work's

Energy Tr= loco

Remindus

a =

a(r) IFfs) IMsNI

work! dU=

E.

di

Vg

=

mg

Read

Carefully

=Fdr as O Visualize

I: "dt=

t

[VIt3 Iffx)

=

(MN) T

=

2m2 Vs

=

=

bx2 ListEqus

UAI=

1, Efu

ds Havefun

((t')

dt=1s =

s(t)

N=

my

nos O

Coefficient

of Restitution e =

-Cut the

Rigid Body

Kinematics Instantaneous Center

Rigid Body

Kinetics

Rigid

-Body

Diagram

General

Formulas

General

Plane Motion AA Y

=

wxr

as

--

werent

BIAA

aro

IF=maa [Ma=Iq -

=woren Farnet -

Io=Ig+md2 B

can see it

=I

T =

ImY,+ Iqw2

=

wXY

+

x

T =

E

IIcW2

dw=x

dO

wheels

G

=

mr, F,

=

IqWA

E

or=

= WX it. For indo Damped

Oscillation Base Excitation

i

=MSO I=

Emin

Absolute

Motion

Analysis

Relative Motion

Analysis

x, =

wn(-3

+

VEE1)

Xi

sin an amplitude

① Locate Pointon

body ①

Kinematic

Diagram

12

=

Wn(-5-1) Undamped

BE

②Define s=f(0)

(geometry) ②

Cross

product, equate;

Y +2 [Wnx +Wi=0 A i +wix =

Ei(t

Apply

derivatives

Solve. X =

A,e

x,t+AzeAzt Xp =Xsin(rt)

Amplitude

Vibrations? Time Response Frequency:

WavelengthA,

to

reponses-

F

Undamped

Oscillation ONE

Id=

Des +wnx

=0,

an e

fr=

E,

=in

A=

ie Xp

=

Fes

sin(art)

-

-H

Wa

overdamped

Static Deflection

X,

=Acos

(Wut)

+Bsin

(Wnt)

331,

X,,X2Distinct

So

T =

E,

= -

X2=Csin(Wnt +

4)

d SsT

A:

Csin4

B:Cros4 C:

VEB2 Critically Damped

Magnification Amplitude

3=1 x,=Yz M

=

E

=

FTE)

Yo

=Brn=Ccos4Wn Vibration of

Particles =1:

M- 0

(Resonance

4=arctan

(E)

x +2wnx +w =

wt)

Reaches

Equilibrium

fastest

Underdemped

DNF-

C-wnVI-5:

Rotation of

Rigid

Bodies Fo'sDriving I<1, WdI

wn=

Frequency X=csin

(Wat +4) e'eWnt

0 +

wO

=0

o

(2)

Mass Moments of Inertia

Cylinder

Thin Plate z

Ig

=

m

- > radius of gyration

Ix

=

Emr2 Iy,

z==

m(2

+

3r2) Pr

X Ix=72m(b+hz) a Go 3

Iy=iz me hW 7 X

Solid

Sphere

Z Slender Rod C z Thin Disk Iz

=Emb2 -b

I ab z

Ix,y,

z =

Emr2 a. Iy,z

=

Fm2 -"X Ix

==mr2

·

Brick er

Ix=12m y

In,

z =

7mr2 Is it e *a

X

& Determine Values of a forwhich

a) 3=0.5

m=500Ky b) I=1.5

" "atoos. e

at

bartenen

at line in

·

the C

~

reai-1200

kb,where

an atte

Sin X=

I

?

(um] RYA

I

Assumption of X=bsin(t) bedisplacent ic

ecim-s--

amplitude cic cos(X-90)

SFx =mi=-ux-2cx wewant:0+

2zWnO+

wiO=0

- b - e

[MB=IBx =(Iq+md")x

i(mxtw*x

(Fqt

md")a= ky1COSO(a) +cyccos0(2a) ·9

-

20 =

2 ZV

were main re intric nee

I=2asino wa s aO Iradians) ebi I

then since

Wee

ma dYA=asing anapproximationforsince

E

W ~VI "I>X

E =3mV- Velocity,so use the

jc =2a0

I

fact that Wn=w

Ax-were at (

d for critically damped so, " +(CSO)0 +(cosol0=0 A

RI

on and most

e A

*

Ay mv,():IAW2

I

S I mr.(B)=(I,+mra)We

-never

W=mg 93 x (l)e =a Vv

*

*(w=

W=- -50ton+15mc+MvV - >ebfV(z)

are

by

0.94 ->0.9*s

the

y

cute

d

1 e rar, e

IF=ma=T-mg +2=x2+n)

DTc=em,wh

1V=0

=strit--ex

S

·is

LinearUtm: G=mYa

it list in. We

12+('L=y2 Va =VB +WxVa/B

i=a=(X2-i2)I SV=-Wush I VA =VB +W XVA/B

DTc+Vc+Tr+1V=0 11) -ur

It's I 14

=

0 =MIV - Mrs, e

Ho=vao at an

, =

m2=1m homentumconserved( Ho = 2Mo

a t0=0.

X=1m

(73(0.35)kN(3310.25kg

13(0.4)kN -2"-

x=(180-120)

EBDEE

-

200mm - 36 = 83.5-

- e m

=690my M

I

&A

I FoI

**, ie

(0)

putye" to

a=

bot.

E >3060* 1.625m

↓i

c)- - - -A

Iiz"

>40 LX NA W=mg

↓man

ABIA3B

On

Mus

Et

=

E

AngularUtm-Imp:HAs*

EMAdt"

=HAL

We fine a

AB=GA+ AB/A

EFx =maax=

Fats

mur=e-solve

I100 WImg ag=dA+A9/A (0)-(1)

⑳t

M2

N

IMA=

En egrestu - mgcs6o(E)

1.5mg=FdtFA+FB

myg hot the e

⑫A=Ig+ muna

[Fy=0 =NA+NB-134-mg

notrotating!

[Ma=8=13k(0.35)(1.625)-NA(1.625) n=

1, I

[Fx=Ma,=F,ws60-Tws 30 +Ni(1.385)-73k(0.4)(1.355)

[Fy=0 =F,sin60-Tain30-mg

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