Dynamics of particles , Enginnering mechanics , murugananthan

Dynamics of Particles
K.Murugananthan,
Assistant professor,
Kamaraj college of engineering & Technology.
Virudhunagar.

Displacement
 Difference between an object’s final position and its starting position.
Does depend on direction.
 Displacement = final position – initial position
 x = xfinal – xinitial

Velocity
 Rate at which an object is moving.
 speed = distance / time

Acceleration
Acceleration is defined as the rate of change
of velocity.

Relative Motion
 The motion of any object depends on the
frame of reference – or point of view – of
the observer.
 Pedestrian sees car moving along street.
Driver sees car at rest.
 Both observers are correct about the motion
of the car.

6
Newton’s Second Law of Motion
• If the resultant force acting on a particle is not
zero, the particle will have an acceleration
proportional to the magnitude of resultant and
in the direction of the resultant.
• We must use a Newtonian frame of reference, i.e., one that is
not accelerating or rotating.
• If no force acts on particle, particle will not accelerate, i.e., it will
remain stationary or continue on a straight line at constant
velocity.
F ma
r r
• If particle is subjected to several forces:
F ma
r r

7
Linear Momentum of a Particle
   
dv
F ma m
dt
d d
mv L
dt dt
 
 

rr r
rr
L mv
r r
Linear momentum
F L
r r&Sum of forces = rate of change of linear momentum
0F 
r
If linear momentum is constant
Principle of conservation of linear momentum

8
Sample Problem 1
An 80-kg block rests on a horizontal plane. Find
the magnitude of the force P required to give the
block an acceleration of 2.5 m/s2 to the right. The
coefficient of kinetic friction between the block and
plane is mk = 0.25.
SOLUTION:
• Draw a free body diagram
• Apply Newton’s law. Resolve
into rectangular components

9
Sample Problem 1
80 9.81 785
0.25k
W mg N
F N Nm
   
 
:maFx 
  cos30 0.25 80 2.5
200
P N  

:0 yF
sin30 785 0N P  
 
sin30 785
cos30 0.25 sin30 785 200
N P
P P
  
    
534.7P N
Pcos30
Psin30
Solve for P and N
1052.4N N

10
Sample Problem 2
The two blocks shown start from rest.
The horizontal plane and the pulley are
frictionless, and the pulley is assumed to
be of negligible mass. Determine the
acceleration of each block and the
tension in the cord.

11
x A AF m a  1 100 AT a
y B BF m a
 
 
2
2
2
300 9.81 300
2940- 300
B B B
B
B
m g T m a
T a
T a
 
  

y C CF m a 02 12  TT
x
y
O
• Kinematic relationship: If A moves xA
to the right, B moves down 0.5 xA
1 1
2 2B A B Ax x a a 
Draw free body diagrams & apply Newton’s law:
  12940- 300 2 0Ba T   2940- 300 200 0B Aa a 
 2940- 300 2 200 0B Ba a  
2
4.2 /Ba m s
2
8.4 /Aa m s 1 840T N 2 1680T N

12
Sample Problem 3
The 12-lb block B starts from rest and slides
on the 30-lb wedge A, which is supported by
a horizontal surface.
Neglecting friction, determine (a) the
acceleration of the wedge, and (b) the
acceleration of the block relative to the
wedge.
Block
Wedge

13
N1
aBn
aBt
WBcosq
WBsinq
N1
WB
N1cosq
N1sinq
sinB B BtW m aq 
212
12 0.5 16.1 /
32.2
Bt Bta a ft s   
aA
1 sin A AN m aq  1
30
0.5
32.2
AN a
1 2cos AN W Nq  1 cosB B BnN W m aq 
But sinBn Aa a q  Same normal acceleration (to maintain contact)
1 cos sinB B AN W m aq q   1
12 0.5
10.39
32.2
AN a

  
2
5.08 /Aa ft s 2
2.54 /Bna ft s 
Draw free body diagrams for block & wedge

14
N1
aBn
aBt
WBcosq
WBsinq
N1
WB
N1cosq
N1sinq
2
cos sin 12.67 /Bx Bt Bna a a ft sq q    
aA
2
sin cos 10.25 /By Bt Bna a a ft sq q     /B A B Aa a a 
r r r
   / 12.67 10.25 5.08
17.75 10.25
B Aa i j i
i j
   
  
r r rr
r r
20.5
30°

15
Sample Problem 4
The bob of a 2-m pendulum describes an arc of a circle in a
vertical plane. If the tension in the cord is 2.5 times the
weight of the bob for the position shown, find the velocity
and acceleration of the bob in that position.

16
Sample Problem 4
Resolve into tangential and normal components:
:tt maF 


30sin
30sin
ga
mamg
t
t
2
sm9.4ta
:nn maF 
 

30cos5.2
30cos5.2
ga
mamgmg
n
n
2
sm03.16na
• Solve for velocity in terms of normal acceleration.
  2
2
sm03.16m2 nn av
v
a 

sm66.5v
mgsin30
mgcos30

17
Sample Problem 5
Determine the rated speed of a
highway curve of radius  = 400 ft
banked through an angle q = 18o. The
rated speed of a banked highway curve
is the speed at which a car should
travel if no lateral friction force is to
be exerted at its wheels.
SOLUTION:
• The car travels in a horizontal circular
path with a normal component of
acceleration directed toward the center
of the path.The forces acting on the car
are its weight and a normal reaction
from the road surface.
• Resolve the equation of motion for
the car into vertical and normal
components.
• Solve for the vehicle speed.

18
Sample Problem 5
SOLUTION:
• The car travels in a horizontal circular
path with a normal component of
acceleration directed toward the center
of the path.The forces acting on the
car are its weight and a normal
reaction from the road surface.
• Resolve the equation of motion for
the car into vertical and normal
components.
:0 yF
q
q
cos
0cos
W
R
WR


:nn maF 

q
q
q
2
sin
cos
sin
v
g
WW
a
g
W
R n


• Solve for the vehicle speed.
   

18tanft400sft2.32
tan
2
2
qgv
hmi1.44sft7.64 v

19
Angular Momentum
OH r mv 
r r r
Derivative of angular momentum with respect to time:
O
O
H r mv r mv v mv r ma
r F
M
       
 



r r r r r r r r r& & &
r
r
Sum of moments about O = rate of change of angular momentum
andr mv
r r
L mv
r r
From before, linear momentum:
Now angular momentum is defined as the moment of momentum
OH
r
is a vector perpendicular to the plane
containing
Resolving into radial & transverse components:
2
OH mv r mrq q  &
Moment of about OF
r

20
Equations of Motion in Radial & Transverse Components
 
 qq
q
qq


rrmmaF
rrmmaF rr
2
2





21
Central Force
When force acting on particle is directed
toward or away from a fixed point O, the
particle is said to be moving under a
central force.
Since line of action of the central force passes through O:
0O OM H 
r r&
constantOr mv H  
rr r
O = center of force

22
Sample Problem 6
A block B of mass m can slide freely on
a frictionless arm OA which rotates in a
horizontal plane at a constant rate .0q
a) the component vr of the velocity of B
along OA, and
b) the magnitude of the horizontal force
exerted on B by the arm OA.
Knowing that B is released at a distance
r0 from O, express as a function of r
SOLUTION:
• Write the radial and transverse
equations of motion for the block.
• Integrate the radial equation to find an
expression for the radial velocity.
• Substitute known information into the
transverse equation to find an
expression for the force on the block.

23
Write radial and transverse
equations of motion:
r rF m a
F m aq q




dr
dv
v
dt
dr
dr
dv
dt
dv
vr r
r
rr
r  
 2 2 2 2
0 0rv r rq &
 
1 22 2 2
0 02F m r rq &
 
 
2
0
2
m r r
F m r r
q
q q
 
 
&&&
&& &&
2
r rq &&&
But rv r &
2 r
r
dv
r v
dr
q & 2
r rr dr v dvq &
2
0
r
o
v r
r r o
r
v dv r drq  &
 
1 22 2
0 0rv r rq &

Dynamics of particles , Enginnering mechanics , murugananthan

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