slides about mechanics

1. PRESANTATION
MECHANICS
 SUBMITTED TO:
ENGR. SHAHZAD
 SUBMITTED BY:
2019-IM-122 (HUSSAIN AFTAB RANJHA)
2019-IM-125 (M. AFAQ SAHI)
2019-IM-128 (MEHRAN HASSAN WAHLA)
2019-IM-131 (TANVEER FAISAL CHEEMA)

2. DYNAMICS
 KINETICS OF SYSTEM OF PARTICLES.
 RIGID BODY.
 CONSERVATION OF ENERGY AND MOMENTUM.
 PLANE KINETICS OF RIGID BODIES.
 ROTATION.

3. • Dynamic consist of :
i. Kinematics
 analysis of the geometry of motion without concern for the forces causing the
motion; which involving the quantity such as displacement, velocity,
acceleration, and time.
ii. Kinetics
 the study of motion and the forces associated with motion; it involves the
determination of the motion resulting from given forces

4. Displacement and Distance:
• Distance is a scalar quantity and displacement is a vector
quantity.
• To travel from point A to point B, say, there are 3 paths with
different distance, but there have the same displacement. Path
3 has longer distance than Path 1 or 2 but the entire three path
have the same displacement.
• Displacement is the basic parameter values to determine
velocity and acceleration
3
2
1
A B

5. Speed and velocity
• Speed is a scalar quantity and velocity is a vector quantity.
• Speed – change of distance per unit of time, SI unit is meters per
second (m/s).
• Velocity – rate of change of displacement with respect to time and
has the same unit as speed.
• Velocity Resultant – combination of two or more velocity and can be
determined using triangle law or parallel rectangular law to
determine resultant vector.
t
s
ervalTime
ervalceDis
vSpeed



int
inttan
,

6. Acceleration
• Acceleration
• change of velocity increment and decrement of velocity is known as de-
acceleration. It is expressed in terms of displacement/time/time. SI units are .
or
• Ration velocity
• difference between two velocities and can be determined by using
triangle and parallel rectangle method.
t
v
ervalTime
ervalVelocity
aonAccelerati



int
int
, v
dt
ds
where
ds
ds
dt
dv
a 

7. Graph Displacement-time (s-t):
M
s
t
D
B
A
C
L
t (s)
s (m)
OL is the graph s-t for a body moves with
constant velocity.
Therefore OA = constant acceleration
Gradient for OL,
Gradient for OL,
The gradient is the velocity of the body.

8. Graph velocity-time (v-t):
• AB = A body moves with constant velocity
• PQ=constant acceleration
• Distant = Area below the graph
• Area (1) = Area of ABCO
• Area (2) = Area of OPQS
• Gradient, m = acceleration for the body
• Graph 1,
• Graph 2,

9. Type of motion
• Type of motion
• Rectilinear motion
• the particle or body moves in a straight lines and does not rotate about its center of
mass
• Circular motion
• a particle follows the path of a perfect circle.
• General plane motion
• simultaneouslya particle may follow a path that is nether straight nor circular. This also
applies to a body have both rotating and rectilinear motion

10. Rectilinear motion
• The three equation that relate the variables of time, displacement,
velocity, and acceleration to each other are
where
atuvorasvv
atuvoratvv
attvs



222
0
2
0
2
0
2
2
1
stimet
smonacceleratitconsa
smvelocityfinalv
smvelocityinitialuv
mntdisplacemes
,
/,tan
/,
/,)(
,
2
0






11. To use these equations, the following points should
bear in mind;
• Acceleration, although it may be in any direction, must be constant.
Constant velocity is a special case in which acceleration is constant at
zero.
• A free-failing body has an acceleration of
• Designate the direction that is to be positive. The direction of the
initial velocity or displacement is often used as the positive direction.
• An object that decelerates or slows down in the positive direction is
treated as having a negative acceleration (de-acceleration).
2
/81.9 smga 

12. Vertical motion
(motion under gravitational effects)
• When an object falls to the earth, it will move very fast i.e., increased
speed.
• The increased motion which is free-fallings (air resistance is ignored) close to
the earth surface is known as gravitational acceleration.
• Gravitational acceleration is same to all bodies, regardless of size or
chemical composition at any location.
• Where motion upward is negative and motion downwards is positive.
s = maximum height

13. Straight motion on slope plane
• Involved the effect of gravitational acceleration like vertical motion.
• However, the calculated value of gravitational acceleration is the component of
gravitational acceleration which is parallel to the direction of motion.
• Acceleration component which is perpendicular to the motion is ignored because
it does not gave any effect to the particle motion.
• Particles rolling from B to A, gravitational acceleration can be divided into
component of rectangular, i.e.,
• At the direction of particle motion
• At the direction parallel to the plane
asvv
atvv
attvs
2
2
1
2
0
2
0
2
0



singa 

14. Newton Motion Law
• First Newton Law
• states that a body in resting state will be in resting sate and a body moving in
constant velocity will move in straight line if no external force acting on it.
• Second Newton Law
• states that momentum change rate will be proportionate to its resultant
force.
2
2
/
)/()()(
1
tan
smkgN
smaxkgmNF
kwheremaF
tconskwherekmaF
maF





The equation shows that:-
Force, F should be acting continuously and
constant acceleration produced constant
acceleration, a.
Force, F acting on the body is resultant
force.
Direction for the force, F and acceleration,
a is the same.

15. Third Newton Law
states that every single acting force will have another force which
have the same reaction but with different direction
T
N
W W
a b.
A body imposed an acting force,
i.e., the weight of the object, W
on the table surface and the
table imposed a reaction force
i.e., normal reaction, N upon the
body
A body imposed a reaction force
i.e., the weight of the object, W
upon the tie rod and the tie rod
imposed a reaction force i.e.,
tension upon the body.

16. Application ( Example using Newton Law)
• Two object connected by light tie
• Constant Tension (T) along tie rod
• Similar acceleration but tie rod in elastic
• Use Second Newton Law
• Resultant acceleration with the statement that the body imposed the same or
opposite direction
)()2..(....................
)()2.....(..........
)()()1....(..........
1
11
22
iiforamT
iforamgmT
iiandiforamTgm
maF





17. Newton ‘s Law Gravitational attraction
• Newton postulated the law governing the gravitational
attraction between any two particles. Stated mathematically,
Where
F = force of gravitation between the two particles
G = universal constant of gravitation; according to
experimental evidence, G = 66.73 x 10-12 m3/kgs2.
m1,m2=mass of each of the two particle
r = distance between the two particles
2
21
r
mm
GF 

18. Friction force
• Friction force
• the force that reacting when 2 surfaces rubbing each other and
the force reacting in different direction.
• Advantage of friction force enables a body to initiate motion
and stop. N
F
W
Rough floor
CASE 1;
F < friction force
(The object will be in the resting
condition and static).
Friction force = static force
mgWNwhereNFs  
CASE 2;
F > friction force
(the object is moving in the
direction of F and kinetic force exist)
NFk 

19. • From the second Newton Law
F= ma
F-Q = ma
Analysis involving kinetic friction
• Direction of friction force is opposite to direction of object’s
movement
• The kinetic friction force value is proportionate to normal reaction
force upon rubbing surface.
• The kinetic friction force is not depending upon the object’s velocity
or net body velocity (if both rubbing objects are moving with
different velocity
a
Q F
A body on smooth floor

20. • F and friction force from the contact surface i.e.,
• Since is proportionate to the normal reaction, N hence we can write;
μk is kinetic friction constant. The value is in the interval of 0  μk 1.0
• From 3rd Newton Law, N=mg , therefore we can substitute the value of
N in (1),
• Therefore the equation in (a ) become
a
F
N
W
Object on rough floor
forcefrictiontheisFwithaFFF gg ).......(..........
)1....(..........NFNF gg 
mgF kg 
mgFF k

21. Particle vs. rigid body mechanics
• What is the difference between particle and rigid body
mechanics?
• Rigid body – can be of any shape
• Block
• Disc/wheel
• Bar/member
• Etc.
• Still planar
• All particles of the rigid body
move along paths equidistant
from a fixed plane
• Can determine motion of
any single particle (pt)
in the body
particle
Rigid-body (continuum of
particles)

22. Types of rigid body motion
• Kinematically
speaking…
• Translation
• Orientation of AB
constant
• Rotation
• All particles rotate
about fixed axis
• General Plane Motion
(both)
• Combination of both
types of motion
B
A
B
A
B
A
B
A

23. Kinematics of translation
• Kinematics
• Position
• Velocity
• Acceleration
• True for all points in R.B.
(follows particle kinematics)
B
AABAB rrr /


AB vv


AB aa


x
y
rB
rA
fixed in the bodySimplified case of our relative motion of particles
discussion – this situation same as cars driving
side-by-side at same speed example

24. Rotation about a fixed axis – Angular Motion
• In this slide we discuss the motion of a line or
body  since these have dimension, only they
and not points can undergo angular motion
• Angular motion
• Angular position, θ
• Angular displacement, dθ
• Angular velocity
ω=dθ/dt
• Angular Acceleration
• α=dω/dt
Counterclockwise is positive!
r

25. Angular velocity
Magnitude of ω vector = angular speed
Direction of ω vector  1) axis of rotation
2) clockwise or counterclockwise rotation
How can we relate ω & α to motion of a point on the body?
angular velocity vector always
perpendicular to plane of rotation!

26. Relating angular and linear velocity
• v = ω x r, which is the cross product
• However, we don’t really need it because θ = 90° between our ω and r vectors we
determine direction intuitively
• So, just use v = (ω)(r)  multiply magnitudes

27. Important quantities
i
ii Rm
2
i
iivm


i
i
i
ii
m
Rm
Moment of inertia, I. Used in t=I. So, I divided by
time-squared has units of force times distance
(torque).
Total linear momentum is the sum of
the individual momenta.
Center-of-mass is a distance. Has to
have units of meters.

28. Rotational Relationships
2
2
1

t






IK
I
IL
t
t
















Position
Velocity
Acceleration
Momentum
Force/Torque
Kinetic Energy
t
tt




0
2
0
2
1
Master equations:
One-to-one
correspondence of
rotational equations to
linear equations.

29. Torque
 
 
 



t
I
mr
rrm
rma
Fr





2

30. Force and Torque Combined: What is the
acceleration of a pulley with a non-zero moment of
inertia?
R
a
IITR
R
a
TR
I






t
t
Torque relation for pulley:
maTmg 
Force Relation for Mass
Put it together:









2
2
1
mR
I
mamg
R
a
IT
maTmg














2
1
1
mR
I
ga
NOTE: Positive down!

31. How high does it go?
Use energy conservation.
Initial Kinetic Energy:
Initial Potential Energy:
Final Kinetic Energy:
Final Potential Energy:
Zero
Zero
22
2
1
2
1
ImvKI 
mgHUF 
Use v=R, and then set Initial Energy = Final Energy.








2
2
2
2
2
1
2
2
1
2
1
mR
I
g
v
H
mgH
R
v
Imv

32. Angular Momentum
Angular momentum is “conserved” (unchanged), in the
absence of an applied torque.
t
L





t


IL 

33. Comparison of Linear and Angular Momentum
m
P
K
t
P
F
VmP
2
2








I
L
K
t
L
IL
2
2








t

Linear momentum is conserved in absence of an applied force.
Angular momentum is conserved in absence of an applied torque.
(translational invariance of physical laws)
(rotational invariance of physical laws)
It’s time for some demos…..

34. L is conserved!
if  
but
ffii IIL  
The final moment of
inertia is less, so
How about the kinetic energy?
f
f
i
i
I
L
K
I
L
K
2
2
2
2


Since the inertia decreases and L stays the
same, the kinetic energy increases!
Q: Where does the force come from to do the
work necessary to increase the kinetic energy?
A: The work is done by the
person holding the
weights!

35. Gyroscopes show change in L from applied
torque
Try the bicycle wheel demo!

36. Static Equilibrium
Static equilibrium is achieved when both the NET FORCE and NET
TORQUE on a system of objects is ZERO.
Q: What relationship must hold between M1 and M2
for static equilibrium?
A: M1 g X1 = M2 g X2

37. Static balance.
What are the forces F1 and F2?

38. Walking the plank.
How far can the cat walk safely?

39. Which mass is heavier?
1. The hammer portion.
2. The handle portion.
3. They have the same mass.
Balance Point
Cut at balance point

40. Atwood Machine with Massive Pulley
Pulley with moment of inertia I,
radius R. Given M1, M2, and H,
what is the speed of M1 just
before it hits the ground?
Strategy: Use conservation of mechanical energy.
Initial kinetic energy is 0. Initial potential energy is M1gH.
Final kinetic energy is translational energy of both blocks plus
rotational energy of pulley.
Final potential energy is M2gH.
Set final energy equal to initial energy.
Three steps: 1. Write down initial kinetic and potential energy.
2. Write down final kinetic and potential energy.
3. Set them equal (no friction).
HINT:

41. Static balance and a strange yo-yo.
The mass M of the yo-yo is known.
The ratio of r and R is known.
What is the tension T1 and T2, and
mass m?
Strategy:
1. Write down torque equation for yo-yo.
2. Write down force equation for yo-yo
and mass m.
3. Eliminate unknowns.

42. The case of the strange yo-yo.
0
0
0
21
2
21



MgTT
mgT
RTrT
Equations to solve:
Torque
Force on mass m.
Force on yo-yo.
mgT
r
R
TT


2
21
Mg
M is known, R/r is known.
Step 1: Rearrange. Step 2: Substitute.
Mg
r
R
T
MgT
r
R
T








1
0
2
22
1
1
1
1
2






a
M
m
a
a
MgT
a
Mg
T
Step 3: Solve (a=R/r).