Dynamics AVISO
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Transcript of Dynamics AVISO
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Dynamicsis that branch of mechanics which deals
with the state of motion of particles and bodiesunder the action of forces.
A particleis a physical body or portion of a physical
body which has mass, the dimensions of which arenegligible in terms of its surroundings and its motion.
A rigid body is any quantity of matter, the particles of
which do not move relative to each other.
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Motion of a Particle
A) Rectilinear translation- a particle constrained
moving along a straight-line path.
B) Curvilinear translationa particle
constrained moving along a curved path.
P
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General Rectilinear Motion
The linear displacement of a particle at any
time is its change in position with reference to
same fixed point.
Q S
R
Linear displacement from Q
is represented by the vector
QS.
-S +S
0 P
Sp
a b
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Linear Velocity
Defined to be the first derivative of displacement
with respect to time, which can be thought of as the
time rate of change of displacement or the time rate
of travel.
-S +S
0 P
Sp
a bP
S
Sp
Velocity is therefore a vector quantity, while speed is a scalar quantity.
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Case I. Given Displacement as a Function of Time, s=f(t)
The velocity and acceleration functions can be
obtained by successive differentiation of s with
respect to time. Therefore,
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Case II. Given Velocity as a Function of Time, v = f(t)
Substitution of the function into v = ds/dt or f(t) = ds/dt,
which can be solved by separation of variables. Therefore,
or
The acceleration can be obtained be differentiating the given
function with respect to time. Therefore,
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Case III. Given Acceleration as a Function of Time, a = f(t)
Substitution of the function into a= dv/dt or f(t) = dv/dt,
which can be solved by separation of variables. Therefore,
or
The displacement function can be obtained by integrating v=
ds/dt which would be
or
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Case IV. Given Velocity as a Function of Displacement, v = f(s)
Substituting the function into v= ds/dt gives f(s) =
dv/dt, which can be solved by separation of variables.
Therefore,
This would yield t as a function of s; it would therefore
be necessary to solve for s as a function of t to obtains = g(t). Since s is now a function of t, the velocity and
acceleration functions can be obtained as in Case I
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Case V. Given Acceleration as a Function of Velocity, a = f(v)
Substituting the function into a= ds/dt or f(v) = dv/dt,which can be solved by separation of variables.Therefore,
This would give t as a function of v and it would benecessary to solve for v as a function of t to obtain v
=g(t). Since v is now a function of t, the displacementand acceleration functions can be obtained as in CaseII.
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Example
A particle moves along a straight line so that after t seconds itsdisplacement s in meters from a fixed reference point O on the
line is given by The particle is 4m to
the left of the origin at t =2s. Determine :
a)The acceleration of the particle when t = 3s
b)The displacement during the interval from t =2s to t =5s,
c)The total distance travelled during the interval from t = 2s to
t =5s and ;
a)The average velocity during the interval from t = 2s to t = 5s.
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Applying case I
12 m4
t=2s
+S
When t = 2s,
Therefore, the displacement is positive to the left of the
origin, and all measurements will be positive when directed
to the left.
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(a) The acceleration when t = 3s can be obtained
from the acceleration expression. Therefore,
(b) When t=2s;
When t=5s;
The displacement during this interval is :
mS 412)2(48)2(30)2(4 232
mS 2212)5(48)5(30)5(4 23
5
leftthetomsss 1842225
rightthetodirectedsm
sm
a
2
2
/12
/12
60)3(24
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(c) For rectilinear translation, the total distance is
equal to the magnitude of the displacement
unless the particle reverses direction of motionduring the time interval. A particle reverses
motion during an interval if velocity becomes
zero during the time interval. In this example,the time when velocity is zero can be obtained
from the velocity expression :
sorst
tt
tt
tt
4,1
0)4)(1(
045
04860122
2
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(c) cont. During the interval from t=2s to t=4s,
the magnitude of the velocity is positive; that is,
the particle is moving to the left, and from t=4sto t=5s, the velocity is negative; that is the
particle is moving to the right. When t =4s, the
position of the particle is :mS 4412)4(48)4(30)4(4 234
,62
2240
4422444
4524
SSSSd
The total distance travelled is the sum of the
distance travelled in the two directions; that is,
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(d) The average velocity during t =2s to t=5s can be
obtained by using :
25
25;tt
ssv
t
Sv aveave
leftthetos
m625
422
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Example
The horizontal motion of the shaft and plunger is arrested by the
resistance of the attached disk. The disk moves through the oilbath. The velocity of the plunger is vo in the position P where s
=0 and t =0, and the velocity varies accordingly to the relations
v=vo-ks where k is a constant. Derive the expressions for the the
positions coordinates s, velocity v, and acceleration a in terms ofthe time t.
dt
dsksv
dt
dsv
o
;
disk
shaft
plunger
S
v
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tvk
ksvk
tksvk
tcuk
dtuk
duk
duds
kdsdu
ksvuLet
cudu
u
RECALL
dtksv
ds
o
ts
oo
s
o
t
o
s
o
t
oo
)ln1
()ln(1
ln1
ln1
)(
ln1
:
0
0
0
kt
o
kt
o
kto
o
kt
o
o
kt
o
kt
oo
kt
o
o
u
u
o
o
ekva
evv
ek
v
vevk
s
kvevks
evksv
ev
ksv
ue
ueRECALL
ktv
ksv
1
1
1
)ln(
;:
)ln(
ln
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Example
The resistance to motion of a particle in air is
approximately proportional to the square of its velocityv for speeds not exceeding 150 m/s. Thus, the
deceleration is given by the expression a = -kv2, where k
is taken to be a constant whose numerical value
depends on the prevailing air conditions and the shape,
roughness, and mass. If a particle which moves in a
horizontal straight line is fired with an initial velocity of
50m/s for a condition where k =1/1000m-1, in whatdistance and elapsed time after firing will the velocity
be reduced to 8 m/s?
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dt
dvv
dt
dva 2
100
1;
50
15050100
1001
5050
100
1
50
11
5011
1001
1
100
1
100
1
500
50 20
vtv
tv
v
tv
vt
vt
v
dvdt
vt
vt
tv
tv
vtv
vtv
vtv
vtv
2
100
)2(100
2100
2100
)50(2
50
15050100
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Displacement therefore ;
mt
s
ts
ts
dtt
ds
dt
ds
t
dt
dsv
ts
ts
,2
2ln100
2ln100)2ln(100
)2ln(100
2
100
2
100
00
00
When speed has been reduced to 8 m/s
s5.1012
1008
Displacement in this time is : ms 26.1832
5.102ln100
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Rectilinear Motion with Uniform Acceleration
(Equation 1.6)
(Equation 1.7)
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Three Fundamental Equations in Rectilinear
Motion with Uniform Acceleration
For a Free falling body:
Neglect the frictional resistance effects of air on the particles
Acceleration due to gravity varies for different locations It can be observed that the value of g decreases as the
elevation above sea level increases
Assume that g has a constant value of 9.8m/s^2 directed
downward
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Case 1 Case 2
a =-g
So =0
Vo
+S
So =0
Vo
+S
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Case 3
a =-g
So =0
Vo
+S
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Example
A point moving with constant acceleration travels 33m
in the half second which elapses after the second of itsmotion and 198m in the eleventh second of its motion.
Find its initial velocity and acceleration.
Solution:
t=0s t=2s t=2.5s t11=11s
S11
S10 198 m
S2 33 m
S2.5
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Example
A stone is dropped from a balloon which rises vertically
at a constant rate of 4 seconds from the ground. Thestone reaches the ground in 10 seconds. Find the
velocity and the height of the balloon when the stone is
dropped
Solution:
a =0
So =0
Vo
+S h
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a =g
So =0
Vo
+S
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Flight of a Projectile