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Trang 1Chapter 1
Particle Kinematics
Part I Rectilinear Motion
A Rectilinear Motion of Particles
B Uniformly Accelerated Rectilinear Motion
C Relative Motion
D Constrained Motion
Trang 2A 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
v = lim
t0
x
t =
dx dt
Trang 3Rectilinear Motion of Particles
Acceleration, a
a = lim
t0
v
t =
dv
dt = d2x
dt2
also
a = dv
dx
dx
dt = v dv
dx
Accelration
Trang 4Problem 1.1 (Sample Problem 11.1, Page 609)
The position of a particle is defined by
x = t3 6t2 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:
v = dx
dt = 3t2 12t 15 (m/s)
Acceleration:
a = dv
dt = 6t 12 (m/s2)
(a) t = 5 s (b) x5 = 60 m
d05 = 100 m
(c) a5 = 18 m s2
(d) d46 = 18 m
Trang 5Determination of Motions
1 Knowing a = f (t)
dv
dt = f (t)
dv = f (t)dt
v0 dv
v
= f (t) dt
0
t
v v0 = f (t) dt
0
t
v = f (t) dt
0
t
+ v0
Key Equations
v = dx
dt
a = dv
dt
a = v dv
dx
dx
dt = v(t)
dx = vdt
x0 dx
x
0
t
x x0 = v(t)dt
0
t
x = v(t)dt
0
t
+ x0
Trang 6Problem 1.2 (Sample Problem 11.2, Page 610)
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
dv
dt = 9.81
dv = 9.81dt
10dv
v
0
t
v 10
v
= 9.81t 0
t
v 10 = 9.81t 0
v = 10 9.81t (m/s)
dy
dt = 10 9.81t
20dy
y
= (10 9.81t)dt
0
t
y = 20 + 10t 4.905t2 (m)
(a) v = 10 9.81t (m/s), y = 20 +10t 4.905t2 (m)
(b) y = 25.1 m when t = 1.019 s
(c) v = 22.2 m s () when t = 3.28 s
Trang 7Determination of Motions
2 Knowing a = f (x)
v
dv
dx = f (x) vdv = f (x)dx
v0 v dv
v
x0
x
v2
2 v0
2
2 = f (x) dx
x0
x
v
2 = v02 + 2 f (x) dx
x0
x
dx
dt = v(x)
dt = dx
v
0dt
t
= dx
v
x0
x
t = dx
v
x0
x
Solve for x,
x = x(t)
Key Equations
v = dx
dt
a = dv
dt
a = v dv
dx
Trang 8Problem 1.3
a = k
m x , x0 = R , v0 = 0
v dv
dx = k
m x
0 v dv
v
= ( k
m x) dx
R
x
v
2 = k
m (R
2 x2) Let
x = Rcos ,
k
m =2
then
v = R sin
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.
dx
d(Rcos)
d
0dt
t
= 1
t =
= t
x = Rcost
v = R sint R
SIMULATION
Simple Harmonic Motion
Trang 9Determination of Motions
3 Knowing a = f (v) Method 1:
dv
dt = f (v)
dt = dv
f (v)
dt
0
t
f (v)
v0
v
f (v)
v0
v
Method 2:
v
dv
dx = f (v)
dx = vdv
f (v)
dx
x0
x
f (v)
v0
v
x = x0 + vdv
f (v)
v0
v
Key Equations
v = dx
dt
a = dv
dt
a = v dv
dx
Trang 10Problem 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 v 0,
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
dt = kv dv
v = kdt
dv v
v0
v
0
t
ln v v0
v
= kt
ln v
v0 = kt
v = v0e kt
dx
dt = v0e kt
dx = v0e kt dt
0 dx
x
= v0e kt dt
0
t
x = v0
k e
kt
0
t
x = v0
k (1 e kt)
Trang 11Problem 1.4 (Continued)
By eliminating t, we obtain
a velocity in terms of x.
v = v0e kt
e kt = v
v0
we have
x = v0
k (1 e kt)
= v0
k (1 v
v0)
v = v0 kx
Alternative Method
Knowing a = kv
v dv
dx = kv
dv = kdx
v0 dv
v
0
x
v = v0 kx
Trang 12B Uniformly Accelerated Rectilinear Motion
a = constant dv
dt = a
v0dv
v
0
t
v = v0 + at
dx
dt = v0 + at
x = x0 + v0t + 1
2at
2
v dv
dx = a vdv = adx
v0v dv
v
x0
x
1
2(v
2 v02) = a(x x0)
v2 = v02 + 2a(x x0)
Key Equations
v = dx
dt
a = dv
dt
a = v dv
dx
Velocity of freefall
v = 2gh
Trang 13Problem 1.5 (Problem 11.36, Page 624)
(a)
y = y1 + v1t + 1
2 at
2
0 = 89.6 + v1(16)+ 1
2(9.81)(16)2
v1 = 72.9 m/s
(b) v2 = v12 + 2a( y y1)
0 = (72.9)2 + 2(9.81)( y 89.6)
Trang 14C 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
a B A = a B a A or a B = a A + a B A
Key Concept
Physical meanings of x B A, v B A, and a B A : motion of B observed from A.
Trang 15Problem 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 4.905t2
Elevator
v E = 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 v B E t=3.65 = 19.81 m/s
Trang 16D Constrained Motion
x A + 2x B = constant
x A1 + 2x B1 = x A2 + 2x B2
x A + 2x B = 0
v + 2v = 0
2x A + 2x B + x C = constant
2x A1 + 2x B1 + x C1 = 2x A2 + 2x B2 + x C 2
2x A + 2x B + x C = 0
2v A + 2v B + v C = 0
Key Concept
Degrees of freedom = number of moving parts - number of constraints
Trang 17Problem 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.
3 moving parts - 1 cable
= 2 DOFs Kinematic constraints:
x A + 2x D + x B = const
x A + 2x D + x B = 0
+ 2v + v = 0
Collar A:
300= a A t
200= 1
2a A t2
t = 4 3 s
a A = 225 mm/s2 ( )
At time t = 4 3 s
a A = 225 ( ), a D = 0
a B = 225 mm/s2 ()
v A = 300 ( ), v D = 75 ( )
v B = 450 mm/s ( )
4
x, v, a
SIMULATION
Trang 18Problem 1.8 (Problem 11.49, Page 628)
3 moving points - 2 cables
= 1 DOFs Kinematic constraints:
x C + 2x E = 0
x E + x W = 0
v C + 2v E = 0
v E + v W = 0
(a, b) v E = 4.5 m/s ( )
v C = 9.0 m/s ( )
v W = 4.5 m/s ( )
(c)
v C E = v C v E = 13.5 m/s ( )
(d)
v W E = v W v E = 9.0 m/s ( )