Dynamics
LINEAR MOTION
(1)
s
(2)
V
(3) a
(4) v,
= X ·X
1 0
-x1. Xo
-t
V1 • V0
-
-t
-Vo +
-
!J.x
-
-
6t
!J.v
-
-
!J,t
at
(5) s -V
0 t + .l..
a
t2
-
2
(6)
v2=
1 V 2 +
0 2as
(7) s -1 ( V
0 + V
1) t
--
2
Variable Symbol Units
Position X m
Displacement s m
Velocity (initial) Vo m/s
Velocity (final) v, m/s
Acceleration a m/s2
Time t s
F=ma
F=ks
Force (Total) F N
Mass M kg
Spring stiffness k N/mm
a
C = c.;2 r
F
C = mCJ2 r
Centrifugal Accel. a, m!s'
Centrifugal Force Fe N
W=FS
PE=mgh
KE=0.5 m v2
SE= 0.5 k x2
P=W/t
P=Fv
Work w J
Potential Energy PE J
Kinetic Energy KE J
Spring Energy SE J
Power p 'N
Conservation of Energy
(11)
(12)
(13)
(14)
(15)
(16)
ROTARY MOTION
0 = 0 -0
1 0
(l) = 01 -00 60
-
-
t 6t
a.=
(01 • (00 6ro
-
-
t 6t
(l) =
1 (l) +
0 a.t
0 -coo t + .l..
a. t2
-
2
(l) 2
=
1 (l) 2+2a.0
0
(17)
0 -1 ( ro0 +
ro1) t
--
2
Rot Variable Symbol Units
Angle 0 rad
Angular Displacement 0 rad
Angular Velocity (Initial) COo rad/s
Angular Velocity (Final) (01 rad/s
Angular Acceleration a. rad/s2
Time t s
T=la,
Torque T Nm
Mass Moment of Inertia I kgm2
Torsional Spring stiffness /Not used/ (Not used/
a
C = v2 / r
F
C = mv2 / r
Centrifugal Acceleration a, m!s'
Centrifugal Force Fe N
W=T8
-
KE= 0.5 I CJ2
-
P=W/t
P=TCJ
Work w J
(No circular gravity) . .
Kinetic Energy (rotary) KE J
(Nor used) . .
Power p vv
PEI+ KEl + SEl + ,vin -,vout
= PE2 + KE2 + SE2
Mo1llentn1ll = Ill v
l1llpnlse =Ft= Change of Illomentn1ll = Ill(v1-v0)
Coefficient of Restitution c = v1/v0
1 ft= 0.3048 m Prefix Sy1nbol
1 in = 25 .4 Illlll ' G g1ga
1 Rev = 21T rad Illega M
1 rad= 180/1T = 57.3 ° kilo k
1 1ll/s = 3.6 k1n/h 1llilli Ill
1 RPM= 21T/60 rad/s 1ll1cro µ
Value
109
106
103
10-
3
1
o-
6
To solve a linear or rotary problem:
(1) Pick 3 known + 1 unknown variable
(2) Choose formula with this set of 4 variables
(3) Re-arrange to solve for unknown variable
Trigonometric Rules
a2
= b2 + c2 -2bc cos A
a
sin A
= b -
-C
sin C sin B
Newton's Laws of Motion
(1) If no net force then constant velocity
(2) F = ma
(3) Every action has equal & opposite reaction
Friction on Inclined Plane
How to do it (typical)
(1) Resolve into normal and parallel components
(2) If in equil, then I normal = 0 and I parallel = 0
(3) Apply Fr= µF
nonce you have 2 unknowns
Friction Force: Fr= µF0
Weight Force: F w = m g
Normal Force: F = F cos 8
n w
Parallel Force: F = F sin 8
p w
µ= Coefficient of friction
4>= Angle of friction 8= Angle of inclination
MOMENTS OF INERTIA I
thin hoop or ring of thick ring of inner radius solid cylinder or disc
radius R & mass M: R1 , outer radius R2, and of radius R and
mass M: mass M:
M*R•2 M*(R1•2 • R2•2)/2 (M*R•2)/2
solid sphere of radius thin-walled hollow sphere slender rod of length L
R and mass M: of radius R & mass M: and mass M, spinning
around center:
(2/5)*M*R•2 (2/3)*M*R•2 (M*L • 2)/12
N = RPM
Shaft Power
P=Tm= 21rNT
60
T = Torque (Nm)
V Belts Pitch
diameter
�
= e(µetsin /J)
Velocity Ratio
VR =SE/ SL
Sheave
outside
diameter
SE = Distance moved by effort
SL = Distance moved by load
Mechanical Advantage
MA= FL / FE
FE = Force of effort
FL = Force of load
Wedge
angle Belt
ride-out
Groove
depth
C
flat plate with sides of
length A and B and
mass M:
GVG/Pl)/l,1
slender rod of length L
and mass M, spinning
around end:
(M*L • 2) /3
Ft
Ff
Ltt.r--t.11r25N
VR of block and tackle fc =100 Ni
Efficiency Number of ropes
7/ =MA/ VR attached to the load h=10 cm
