ME2115 Cheatsheet PDF

Preview

Summary

1 KinematicsFor constant velocity motionx=x 0 +vtFor constant acceleration motionv=v 0 +atx=x 0 +v 0 t+ 1 2at 2v 2 =v 02 + 2 a(x−x 0 )1 Circular Motion Particle acceleration is given by⃗a(t)=rα(t)⃗eθ(t)−rω 2 (t)⃗er(t)⃗v=⃗ω×r⃗⃗a=⃗α×⃗r+⃗ω×(⃗ω×r⃗)1 Non-Circular Motion ⃗a=( ́r−rθ ́ 2 )⃗er+(rθ ́+ 2 ́rθ ́...

Description

1

Kinematics

P

For a fixed point

on a rigid body and a point

Q

 α M G= I G 

Moment

moving on a rigid

body For constant velocity motion

A

For constant acceleration motion

A

1 x = x 0 + v 0 t+ at 2 2

is given by

B

Q

in frame

A

is given by

Principle of Work and Energy U 1 →2=T 2−T 1

v A = ω × r A /C a A = α × r A / C

2 1 1 T = m|v G | + I G ω2 2 2

Circular Motion

F ≤ μs N

a = α × r + ω ×(  ω ×r)

Rolling with sliding

If the body slips at

a =( ´r −r θ´ 2 )  er + ( r θ´ +2 r´ ´θ )  eθ

,

This is also the case if

Kinematics of Rigid Bodies P

and

Q

is found, velocity at any other

v B= ω ×r B / C

General Motion

For two fixed points

C

point is

3

α 

Mass Properties

3.1

r Q =r P + r Q / P

( I O )AA' = ( I G )BB' + M d 2

v Q =v P +v Q/ P =v P+ ω × r Q/ P

d (T +V )=0 dt For a rolling body without slipping, the frictional forces does not do any work because there is no displacement associated with it. In this case, conservation of energy is still applicable.

3.2

Simple Harmonic Motion

The equation of motion for both spring-mass and simple pendulum can be written in a general form as

´u +ω2nu=0

Radius of gyration

I G =m k

aQ =a P +a Q/ P =a P + α × r Q / P + ω × ( ω × r Q/ P)

Free Vibration Without Damping

6.1

Parallel Axis Theorem

happens to be the instantaneous centre of rotation

V 1+T 1=V 2+T 2

6

, on a rigid body with angular

A

If only conservative forces exist in the system,

Instantaneous Centre of Rotation

Once the instantaneous centre of rotation

2

Where

u

is the displacement of the system, and

angular frequency.

2.2

Rotating Frame

For a vector A

q

that is varying in the rotating frame

B  dq d q = +ω  ×q dt dt

4

Plane Motion Linear momentum Force Angular Momentum

, its kinetic energy is given by

of the rigid body.

F=μk N 2.4

A

1 2 T= IAω 2

2

aθ =rθ´ +2 ´r θ´

acceleration

v s

v A =v s + ω × r A /C

Non-Circular Motion

ar =r´ −r θ´

For a rigid body rotating about a fixed point

Point of contact C only has centripetal acceleration C is the instantaneous centre of rotation

v = ω × r

2.1

is given by

For a rigid body in 2D planar motion,

Rolling Motion

a ( t )=rα ( t )  e θ ( t ) −r ω 2( t )  er ( t )

2

5

B B aQ = Aa P +  ω × v Q + aQ α × r Q / P + ω × ( ω × r Q /P) +2 

Particle acceleration is given by

1.2

A

 M G=rA /G ×  FA

Rolling without sliding

2

Moment about centre of mass due to a force acting at

v Q = v P + ω ×r Q / P+ v Q

2.3

v =v 0+2 a ( x − x 0 ) 1.1

A

in frame

A

The acceleration of

v =v 0 +at

2

Q

The velocity of

x = x 0 +vt

 L=M v G  F =M aG  H G=I G  ω

The general solution is given by

u= A sin ( ωn t+ϕ)

ωn

is the natural

2

A= u 0+

√

v0 ωn

2

ϕ=tan

( )

(

v =A ωn sin ( ωn t+ϕ) + Period of Oscillation

Natural Frequency

[ Hz ]

u 0 ωn v0

−1

2 n

(( ) )

π a=A ω sin ωn t+ϕ+π 2 2π τn= ωn 1 ωn f n= = τn 2 π

6.1.1 Spring-Mass System

m ´y + ky=0 k =2 π f n ω n= m

Equation of Motion

√

Natural Angular Frequency

[ rad / s]

ω n= Natural Angular Frequency

[ rad / s]

Where

cos θ ≈ 1 2 θ cos θ ≈ 1− 2

We can identify a special point

√

6.4

Springs in Parallel Springs in Series

Equation of Motion

−αt

xm

can be expressed in terms of maximum velocity

´θm =θm ωn

or

C1 =

θm

7.1

Under-Damped Vibration

( )

c 2 k −...