Relative velocities, Kinematics of Machines (2131906)

1. Kinematics of Machines (2131906)
Active Learning Assignment
Topic Name:-“Relative velocities”
Guided By:- Prof. Samir Raval
Name:- Jani Parth U. (150120119051)
Branch:- Mechnical
Div:- A-3

2.  Contents
1. Introduction
2. Relative Velocity
3. Relative Velocity Equation
4. Case 1 : Rotation Of Rigid Link About A Fixed Centre
5. Case 2 : Relative Velocity Equation Of Two Point On A Rigid Link
6. Example Of Relative Velocity

3.  Introduction
The Study Of Motion Of Various Parts Of A Machine Is Important For
Determining Their Velocities And Accelerations At Different Point And
At Different Moments.
The Acceleration And Velocity In A Mechanism Are Due To Dynamic
Forces On Various Parts.
The Velocities And Accelerations Can Be Determined By Analytical
Methods.
In This Presentation We Are Study About The Relative Velocity Method

4.  Relative Velocity
 Consider Two Point A And B Moving In a Plane Having VA And
VB As Their Absolute Velocities Respectively. As Shown In Fig.

5. If The Both Point Get Same Motion, There Is No Change In Their
Relative Velocity. Let -VA Velocity Is Given To Each Point As Show
In Fig.
Due To -VA Velocity To Point A, The Velocity At Point A Becomes
Zero And At Point B It Becomes VB – VA.
The Vector Expression Of Velocity At Point B Can Be Expressed As
VB And – VA .
It Is Defined As The R.V Of Point B With Respect To A And Written
As VBA

6. • Thus It Can Be Written In Vector Form As VB = VA + VBA

7. Relative Velocity Equation
 The rate of change of displacement with respect to time is
called velocity.
 In mechanism we consider a velocity of any selected pint, not
for a line, volume, coordinate system of collection of points.
 The velocity of each point of mechanism is different.
 The principle of relative velocity method may be understood
by deriving the general equation for relative velocity.

9.  Case 1 : Rotation Of Rigid Link About A Fixed Centre
Consider A Rigid Link D Rotating About Fixed Centre O .Point A
On Link B Is Distance R From O. Line OA Makes Angle With The X
Axis Let And Be The Coordinates Of Point A In X And Y Direction
Respectively. Thus,
𝑥1= 𝑅𝑐𝑜𝑠𝜃
𝑦1= 𝑅𝑠𝑖𝑛𝜃
Differentiation of x1 and y1 with respect to time gives the velocity
𝑑𝑥1
𝑑𝑡
= 𝑅 −𝑠𝑖𝑛𝜃
𝑑𝜃
𝑑𝑡
= −𝑅𝜔𝑠𝑖𝑛𝜃
and
𝑑𝑦1
𝑑𝑡
= 𝑅 𝑐𝑜𝑠𝜃
𝑑𝜃
𝑑𝑡
= 𝑅𝜔𝑐𝑜𝑠𝜃

10. 𝑑𝑥1
𝑑𝑡
= 𝑉𝐴
𝑋
= velocity of point a in x-direction
𝑑𝑦1
𝑑𝑡
= 𝑉𝐴
𝑦
= velocity of point a in y-direction
𝜔 =
𝑑𝜃
𝑑𝑡
= Angular Velocity Of OA
Equation Can Be Written As
𝑉𝐴
𝑋
= −𝑅𝜔𝑠𝑖𝑛𝜃
𝑉𝐴
𝑦
= 𝑅𝜔𝑐𝑜𝑠𝜃
Total Velocity Of Point A Is Given By
VA = (−𝑅𝜔𝑠𝑖𝑛𝜃) 2 + 𝑅𝜔𝑐𝑜𝑠𝜃 2 = R 𝜔

11. • Case 2 : Relative Velocity Equation Of Two Point On A Rigid
Link
Suppose There Are Two Points A And B On A Rigid Link C.Let Ab=r And It
Males An Angle With The X-axis, () And () Be The Coordinates Of Point A
And B Respectively.
From Fig, The Coordinates Of Point B In X And Y Direction Can Be Written
As
XB=Xa+ Rcosθ
yB=ya+Rsinθ
Differentiating equation with respect to time
𝑑𝑋 𝐵
𝑑𝑡
= 𝑉𝐵
𝑋
=
𝑑
𝑑𝑡
𝑋𝐴 − 𝑅𝜔𝑠𝑖𝑛𝜃
𝑑𝑦 𝐵
𝑑𝑡
= 𝑉𝐵
𝑦
=
𝑑
𝑑𝑡
𝑦 𝐴 + 𝑅𝜔cosθ

12. 𝑉𝐵
𝑋
=𝑉𝐴
𝑋
-𝑅𝜔𝑠𝑖𝑛𝜃
𝑉𝐵
𝑦
=𝑉𝐴
𝑌
+ 𝑅𝜔cosθ
WHERE 𝑉𝐵
𝑋
=
dxB
dt
=THE VELOCITY OF POINT B IN THE X-DIRECTION
𝑉𝐵
𝑌
=
d𝑌B
dt
=THE VELOCITY OF POINT B IN THE Y-DIRECTION
𝑉𝐴
𝑋
=
dxA
dt
=THE VELOCITY OF POINT A IN THE X-DIRECTION
𝑉𝐴
𝑌
=
d𝑌A
dt
=THE VELOCITY OF POINT B IN THE Y-DIRECTION
VECTORIAL ADDITION OF TWO COMPONENTS OF EQUATION GIVES THE TOTAL
VELOCITY OF POINT B

13. VB = VB
X
→ VB
Y
= VA
X
− VA
Y
→ (- Rωsinθ →Rωcosθ)
HERE (VA
X
→ VA
Y
)=VA =TOTAL VELOCITY OF POINT A
AND (- Rωsinθ →Rωcosθ) = Rω
THUS EQUATION CAN BE WRITTEN AS
VB = VA →R𝜔 = VA VBA
Where VBA Is The Velocity Of Point B With Respect To A This Equation Is
The Relative Velocity Equation For Two Point Lying On A Rigid Link

14.  In Slider-crank Mechanism, The Stroke Of The Slider Is 200 Mm And The Obliquity Ratio
Is 4.5. The Crank Rotates Uniformly At 400 Rmp Clockwise, While The Crank Is
Approaching The Inner Dead Centre And The Connecting Rod Is Normal To The Crank,
Find The Velocity Of Piston And The Angular Velocity Of The Connecting Rod.
Stroke= 200 mm
N=obliquity ratio=4.5
N= crank speed=400rpm, uniform, clockwise
Crank and connecting rod are perpendicular to each other, when
the crank is approaching i.d.c
To find: Vp and ωcp
Now, r= radius of crank=
200
2
=100
l= length of connecting rod= n.r= 450 mm
ωoc = angular velocity of crank =41.89rad/sec,
clockwise

15. Velocity Magnitude (mm/s) Direction Vector Scaled length (mm)
VC 4189 ⊥ to DC oc 41.89
VP/C unknown ⊥ to PC cp Unknown
VP unknown Along path of P op unknown

16. Since ocp is a right angled triangle, the crank angle 𝜃 is given by
𝑡𝑎𝑛𝜃=
𝑙
𝑟
𝜃=77.47velocity analysis
ωoc = angular velocity of crank =41.89rad/sec
Vc =(oc) ωoc = =4189 mm/sec
From velocity polygon ocp. We get
Op=43 mm, pc= 9mm
Linear velocity of the piston is given by
Vp =4300 mm/sec
Vp/c =900 mm/sec
and angular velocity of connecting rod is given by
ωcp =2 rad/sec