Constant Vx=xo+vtConstant Av=v0+at h g t

x=x0+v0t+12a t2 v g t

v2=v02+2a (x−x0 ) v g h

CIRCULAR MOTION

Tips to solveMake two equations by viewing from different areas for one same valueE.g. one from rotating frame, another from rigid body

Two points on a rigid body

α⃗ is the angular accelerationof the rigid bodyProvided both points are on the same rigid bodyMake P a static point

ROTATING FRAMEUse these only when particle object is moving along rotating rigid body

ω⃗ – angular velocity of frame B (rotating) in frame A (fixed)

Relate rate of change in both framesUse to switch frames

Conversion equation for accelerationAnalyse this by parts1. acceleration term of Q if it is not moving on the rigid body2. acceleration of Q with respect to the (rotating) frame B attached to rigid body3. Coriolis acceleration.

Superscripts – FramesSubscripts – Points

Cosine rule – r2=R2+ I 2−2 RI cosθ r °

MASS PROPERTIES

Moment of inertia

Conversion equation of any free vectors

Conversion equation for velocity

PLANE MOTION – NEWTON’s 2nd LAWLinear momentum

v⃗Q– absolute velocity measured w.r.t. Newtonian frame Angular momentum

IG – need reference point, normally centre of massForce

a⃗G–absolute acceleration measured w.r.t Newtonian frame

Moment

Resultant moment ≠ Rate of change of angular momentumOnly when P is a fixed pointd H⃗ p

dt=M⃗ p

Angular momentum in 3D

Steps to solve1. Kinematics analysis2. Free body diagram3. Sum up forces4. Angular momentum5. Differentiate it

Insert into momentum equation

ENERGY METHOD

If there are only conservative forces

Kinetic energy of a rigid body

SPRING-MASS SYSTEM

md2 xd t2

+kx=0

d2 xd t 2

+ωn2 x=0

x=xm sin (ωn+ϕ )

ωn=√ km=2 π f n

T=2πωn

or T n2=(2π )2(mk )

COMPLEX SPRING MASSIn Series1ke

= 1k1

+ 1k2

k e=k1 k2k 1+k2

In Parallel (both connected directly to mass)k e=k1+k2

FREE VIBRATION OF RIGID BODIESd2θd t2

+ωn2 sinθ=0 – Simple harmonic motion

Maximum angular velocitydθmdt

=ωnθm

Parallel axis theoremI o=IG+md

2 – initial point must be Centre of Gravity

FORCED VIBRATION OF A SPRING MASS SYSTEM

DAMPED VIBRATION

Critical value, cc=2m√ kmccc

=ζ>1(heavy damping)

ζ=1(critical damping)

ζ<1 ( light damping )

DAMPED FORCED VIBRATION

K=1 /k

ζ= λ

2√km= dampingcoefficientcriticaldamping coefficient

BASE EXCITATION

K=1 /k

ANALYSIS OF MECHANISMSLower pairs – 1 degree of freedom

Revoute pairs – change angles Prismatic pairs – linear displacemtns

Higher pairs – >1 degree of freedom

Degrees of freedom – number of independent relative motions

Gruebler’s equationF=3(n-1) - 2L - h

WhereF = total degrees of freedom in the mechanismn = number of links (including the frame)L = number of lower pairs (one degree of freedom)h = number of higher pairs (two degrees of freedom)

rotation + sliding = 2 DOFs

Any structure only 1 frame → treat AFE as one frameTherefore, F only has 2 links attached to it

A, J and G are fixed. AJ is the ground link The ground link is also attached to G. Therefore, the count of contribution of L at G is 2.

FOUR BAR LINKAGES

Grashof crank rockero Shortest link is the input linko Input link has full motiono Output link limited range of motion

Grashof crank cranko Shortest link is the ground linko Input and output have full motion

Grashof double rockero Shortest link is the floating linko Input and output have limited motion

Grashof rocker cranko Shortest link is the output linko Output link has full motiono Input link has limited range of motion

SLIDER CRANK

Moments of inertia

Description Figure Moment(s) of inertia Comment

Point mass m at a

distance r from the axis of

rotation.

A point mass does not have a moment

of inertia around its own axis, but by

using the parallel axis theorem a

moment of inertia around a distant axis

of rotation is achieved.

Two point

masses, M and m,

with reduced mass   and

separated by a distance, x.

—

Rod of length L and

mass m

(Axis of rotation at the end

of the rod)

  [1]

This expression assumes that the rod is

an infinitely thin (but rigid) wire. This is

also a special case of the thin

rectangular plate with axis of rotation at

the end of the plate,

with h = L and w = 0.

Rod of length L and

mass m   [1]

This expression assumes that the rod is

an infinitely thin (but rigid) wire. This is a

special case of the thin rectangular plate

with axis of rotation at the center of the

plate, with w = L and h = 0.

Thin circular hoop of

radius r and mass m

This is a special case of a torus for b =

0. (See below.), as well as of a thick-

walled cylindrical tube with open ends,

with r1 = r2 and h = 0.

Thin, solid disk of

radius r and mass m

This is a special case of the solid

cylinder, with h = 0.

That   is a

consequence of the Perpendicular axis

theorem.

Thin cylindrical shell with

open ends, of radius r and

mass m

  [1]

This expression assumes the shell

thickness is negligible. It is a special

case of the thick-walled cylindrical tube

for r1 = r2.

Also, a point mass (m) at the end of a

rod of length r has this same moment of

inertia and the value r is called

the radius of gyration.

Solid cylinder of radius r,

height h and mass m

  [1]

This is a special case of the thick-walled

cylindrical tube, with r1 = 0. (Note: X-Y

axis should be swapped for a standard

right handed frame)

Thick-walled cylindrical

tube with open ends, of

inner radius r1, outer

radius r2, length h and

mass m

  [1][2]

or when defining the normalized

thickness tn = t/r and letting r = r2,

then 

With a density of ρ and the same

geometry 

 

Tetrahedron of side s and

mass m

 f=ma

—

Octahedron (hollow) of

side s and mass m—

Octahedron (solid) of

side s and mass m—

Sphere (hollow) of

radius r and mass m   [1]

A hollow sphere can be taken to be

made up of two stacks of infinitesimally

thin, circular hoops, where the radius

differs from 0 tor (or a single stack, ,

where the radius differs from -r to r).

Ball (solid) of radius r and

mass m   [1]

A sphere can be taken to be made up of

two stacks of infinitesimally thin, solid

discs, where the radius differs from 0

to r (or a single stack, where the radius

differs from -r to r).

Also, it can be taken to be made up of

infinitesimally thin, hollow spheres,

where the radius differs from 0 to r.

Sphere (shell) of radius r2,

with centered spherical

cavity of radius r1 and

mass m  [1]

When the cavity radius r1 = 0, the object

is a solid ball (above).

When r1 = r2, 

 , and

the object is a hollow sphere.

Right circular cone with

radius r, height hand

mass m

  [3]

  [3

]

—

Torus of tube radius a,

cross-sectional

radius b and mass m.

About a

diameter: 

About the vertical

axis: 

—

Ellipsoid (solid) of

semiaxes a, b, and cwith

axis of rotation a and

mass m

—

Thin rectangular plate of

height h and of

width w and mass m

(Axis of rotation at the end

of the plate)

—

Thin rectangular plate of

height h and of

width w and mass m   [1]

—

Solid cuboid of height h,

width w, and depthd, and

mass m

For a similarly oriented cube with sides

of length  ,  .

Solid cuboid of height D,

width W, and length L, and

mass m with the longest

diagonal as the axis.

For a cube with sides  , 

.

Plane polygon with

vertices  ,  , 

, ...,   and

mass   uniformly

distributed on its interior,

rotating about an axis

perpendicular to the plane

and passing through the

origin.

This expression assumes that the

polygon is star-shaped. The vectors 

,  ,  , ...,   are position

vectors of the vertices.

Infinite disk with

mass normally

distributedon two axes

around the axis of rotation

(i.e. 

Where :   is the

mass-density as a function

of x and y).

—