Động lực học công trình Tài liệu tiếng Anh 1
Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (6.61 MB, 18 trang )
Chapter 1
Particle Kinematics
Part I. Rectilinear Motion
A.
B.
C.
D.
Rectilinear Motion of Particles
Uniformly Accelerated Rectilinear Motion
Relative Motion
Constrained Motion
A. Rectilinear Motion of Particles
Position, x
The vector connecting the origin and the particle.
For the rectilinear motion, the direction may be
replaced by a plus/minus sign.
Displacement, x
The vector connecting the position of a particle
at time t to that at time t + t . Note that if
x0 = 0 then x = x .
Velocity, v
x dx
=
t0 t
dt
v = lim
Chapter 1, Page 1
Rectilinear Motion of Particles
Accelration
Acceleration, a
v dv d 2 x
a = lim
=
=
t0 t
dt dt 2
also
dv dx
dv
a=
=v
dx dt
dx
Chapter 1, Page 2
Problem 1.1
(Sample Problem 11.1, Page 609)
The position of a particle is defined by
x = t 3 6t 2 15t + 40 (m)
Determine
(a) the time at which the velocity will be zero,
(b) the position and distance traveled by the particle at that time,
(c) the acceleration of the particle at that time, and
(d) the distance traveled by the particle from t = 4 s to t = 6 s.
Velocity:
dx
v=
= 3t 2 12t 15 (m/s)
dt
Acceleration:
a=
dv
= 6t 12 (m/s 2 )
dt
(a) t = 5 s
(b) x5 = 60 m
d05 = 100 m
(c) a5 = 18 m s 2
(d) d46 = 18 m
Chapter 1, Page 3
Determination of Motions
Key Equations
v=
a=
1. Knowing a = f (t)
dv
= f (t)
dt
dv = f (t)dt
dx
dt
dv
dt
dv
a=v
dx
v
v0
dv =
v v0 =
v=
t
0
t
t
0
0
f (t) dt
f (t) dt
dx
= v(t)
dt
dx = vdt
x
x0
dx =
x x0 =
x=
t
v(t) dt
0
t
v(t) dt
0
t
v(t) dt + x
0
0
f (t) dt + v0
Chapter 1, Page 4
Problem 1.2
(Sample Problem 11.2, Page 610)
dv
= 9.81
dt
dv = 9.81dt
Ball tossed with 10 m/s vertical
velocity from window 20 m
above ground. Determine
(a) velocity and elevation above
ground at time t,
(b) highest elevation reached by
ball and corresponding time,
and
(c) time when ball will hit the
ground and corresponding
velocity.
(a) v = 10 9.81t (m/s) , y = 20 + 10t 4.905t 2 (m)
(b) y = 25.1 m when t = 1.019 s
(c) v = 22.2 m s () when t = 3.28 s
v
10
dv =
t
9.81dt
0
v
t
v = 9.81t
10
0
v 10 = 9.81t 0
v = 10 9.81t (m/s)
dy
= 10 9.81t
dt
y
20
dy =
t
(10 9.81t) dt
0
y = 20 + 10t 4.905t 2 (m)
Chapter 1, Page 5
Determination of Motions
Key Equations
v=
2. Knowing a = f (x)
dv
= f (x)
dx
vdv = f (x)dx
v
dx
dt
dv
a=
dt
a=v
dv
dx
dx
= v(x)
dt
v
v0
v dv =
2
v 2 v0
=
2
2
x
x0
x
x0
x
f (x) dx
f (x) dx
v 2 = v02 + 2 f (x) dx
x0
dx
v
dt =
t
0
dt =
t=
x
x0
x
x0
dx
v
dx
v
Solve for x,
x = x(t)
Chapter 1, Page 6
Problem 1.3
Simple Harmonic Motion
A block of mass m is attached to
a spring of constant k. The
spring is stretched to a length of
R and then released. Express
the velocity and position of the
block with time t.
a=
k
x , x0 = R , v0 = 0
m
v
v
0
dv
k
= x
dx
m
v dv =
x
R
(
k
v = (R 2 x 2 )
m
Let
R
dx
= R sin
dt
d(Rcos )
= dt
R sin
d
= dt
k
x) dx
m
2
x = Rcos ,
SIMULATION
k
= 2
m
then
v = R sin
t
0
dt =
1
0
d
= t
x = Rcos t
v = R sin t
t=
x
Chapter 1, Page 7
Determination of Motions
Key Equations
3. Knowing a = f (v)
Method 1:
dv
dt
dv
dx
dv
f (v)
dt =
t
0
a=v
v
dv
= f (v)
dt
dx
v=
dt
a=
Method 2:
dt =
t=
v
dv
f (v)
v0
dv
f (v)
v
v0
dv
= f (v)
dx
dx =
v
x = x0 +
v
x
vdv
f (v)
x0
dx =
v0
v0
vdv
f (v)
vdv
f (v)
Chapter 1, Page 8
Problem 1.4
(Sample Problem 11.3, Page 611)
Brake mechanism used to reduce
gun recoil consists of piston
attached to barrel moving in fixed
cylinder filled with oil. As barrel
recoils with initial velocity v0,
piston moves and oil is forced
through orifices in piston, causing
piston and cylinder to decelerate at
rate proportional to their velocity.
Determine v(t), x(t), and v(x).
Knowing
a = kv , v0 , x0 = 0
dv
= kv
dt
dv
= kdt
v
v
v0
t
dv
= k dt
0
v
v
ln v = kt
v0
ln
dx
= v0 e kt
dt
dx = v0 e kt dt
x
0
dx =
t
ve
0
0
kt
dt
t
v
x = 0 e kt
k
0
v
x = 0 (1 e kt )
k
v
= kt
v0
v = v0 e kt
Chapter 1, Page 9
Problem 1.4
By eliminating t, we obtain
a velocity in terms of x.
v = v0 e kt
e
kt
x=
Alternative Method
Knowing a = kv
dv
= kv
dx
dv = kdx
v
=
v0
v
we have
v0
(Continued)
kt
(1 e )
k
v
v
= 0 (1 )
v0
k
v
v0
x
dv = k dx
0
v = v0 kx
v = v0 kx
Chapter 1, Page 10
B. Uniformly Accelerated Rectilinear Motion
Key Equations
v=
dx
dt
a = constant
dv
=a
dt
v
v0
a=
dv
dt
dv
a=v
dx
dv
=a
dx
vdv = adx
v
t
dv = a dt
0
v = v0 + at
dx
= v0 + at
dt
1
x = x0 + v0t + at 2
2
v
v0
x
v dv = a dx
x0
1 2
(v v02 ) = a(x x0 )
2
v 2 = v02 + 2a(x x0 )
Velocity of freefall
v = 2gh
Chapter 1, Page 11
Problem 1.5
(Problem 11.36, Page 624)
(a) y = y1 + v1t +
1 2
at
2
1
0 = 89.6 + v1 (16) + (9.81)(16)2
2
v1 = 72.9 m/s
(b) v 2 = v12 + 2a( y y1 )
0 = (72.9)2 + 2(9.81)( ymax 89.6)
ymax = 360 m
Chapter 1, Page 12
C. Relative Motion
Retilinear Motion
x B A = x B x A or x B = x A + x B
A
v B A = v B v A or v B = v A + v B
A
aB A = aB a A or aB = a A + aB
A
Key Concept
Physical meanings of x B A , v B A ,
and aB A : motion of B observed
from A.
Chapter 1, Page 13
Problem 1.6
(Sample Problem 11.4, Page 620)
Ball thrown vertically upward from 12 m
level in elevator shaft with initial
velocity of 18 m/s. At same instant,
open-platform elevator passes 5 m level
moving upward at 2 m/s.
Determine (a) when and where ball hits
elevator and (b) relative velocity of ball
and elevator at contact.
Ball
v B = 18 9.81t
y B = 12 + 18t 4.905t 2
Elevator
vE = 2
y E = 5 + 2t
(a) y B = y E , t = 3.65 s
(b) v B E = v B v E = 16 9.81t
vB E
t =3.65 = 19.81 m/s
Chapter 1, Page 14
D. Constrained Motion
Key Concept
Degrees of freedom =
number of moving parts number of constraints
x A + 2x B = constant
2x A + 2x B + xC = constant
x A1 + 2x B1 = x A2 + 2x B2
2x A1 + 2x B1 + xC1 = 2x A2 + 2x B2 + xC 2
x A + 2x B = 0
2x A + 2x B + xC = 0
v A + 2v B = 0
2v A + 2v B + vC = 0
a A + 2aB = 0
2a A + 2aB + aC = 0
Chapter 1, Page 15
Problem 1.7
(Sample Problem 11.5, Page 621)
Pulley D is pulled down with a constant velocity of 75
mm/s. At t = 0, collar A starts moving down from K
with a constant acceleration and no initial velocity.
Knowing that velocity of collar A is 300 mm/s as it
passes L, determine the displacement, velocity, and
acceleration of block B when block A is at L.
SIMULATION
Collar A:
300 = a At
1 2
a t
200
=
2 A
t=4 3 s
a A = 225 mm/s 2 ( )
3 moving parts - 1 cable
= 2 DOFs
x, v, a
Kinematic constraints:
x A + 2x D + x B = const
x A + 2x D + x B = 0
v A + 2v D + v B = 0
a A + 2aD + aB = 0
At time t = 4 3 s
a A = 225 ( ), aD = 0
aB = 225 mm/s 2 ()
v A = 300 ( ), v D = 75 ( )
v B = 450 mm/s ( )
x A = 200 ( ), v D = 75 ×
4
= 100 ( )
3
x B = 400 mm ( )
Chapter 1, Page 16
Problem 1.8
(Problem 11.49, Page 628)
3 moving points - 2 cables
= 1 DOFs
Kinematic constraints:
xC + 2x E = 0
x E + xW = 0
vC + 2v E = 0
v E + vW = 0
(a, b) v E = 4.5 m/s ( )
vC = 9.0 m/s ( )
vW = 4.5 m/s ( )
(c) vC
(d) vW
E
E
= vC v E = 13.5 m/s ( )
= vW v E = 9.0 m/s ( )
Chapter 1, Page 17
