Particle motion
Hiroki Okubo
z
3
P
When a particle P moves along a straight line, the
position of P at any instant of time t can be specified by its distance s measured from some convenient reference point O fixed on the line. At time
t + ∆t the particle has moved to P ′ and its coordinate becomes s + ∆s. The change in the position
coordinate during the interval ∆t is called the displacement ∆s of the particle. The displacement
would be negative if the particle moved in the negative s-direction.
z
R
y
φ
r
θ
x
y
4
x
Introduction
Kinematics is the branch of dynamics which describes the motion of bodies without reference to
the forces. It is often described as the geometry
of motion. A thorough working knowledge of kinematics is a prerequisite to kinetics, which is the
study of the relationship between motion and the
corresponding forces.
2
Velocity and acceleration
The average velocity of the particle during the interval ∆t is the displacement divided by the time
interval or vav = ∆s/∆t. As ∆t becomes smaller
and approaches zero in the limit, the average velocity approaches the instantaneous velocity of the
particle, which is
Figure 1: Coordinates
1
Rectilinear motion
v = lim
∆t→0
∆s
ds
=
= ṡ
∆t
dt
(1)
The velocity is the time rate of change of the position coordinate s. The velocity is positive or negative depending on whether the corresponding displacement is positive or negative.
The average acceleration of the particle during
the interval ∆t is the change in its velocity divided
by the time interval or aav = ∆v/∆t. As ∆t becomes smaller and approaches zero in the limit, the
average acceleration approaches the instantaneous
acceleration of the particle, which is
Particle motion and choice
of coordinates
dv
d2 s
∆v
=
= v̇ = 2 = s̈
∆t→0 ∆t
dt
dt
(2)
The position of particle P at any time t can be
described by specifying its rectangular coordinates The acceleration is positive or negative depending
x, y, z, its cylindrical coordinates r, θ, z, or its on whether the velocity is increasing or decreasing.
spherical coordinates R, θ, φ as shown in Fig. 1.
Note that the acceleration would be positive if the
a = lim
1
2. A particle moves along the x-axis with an initial velocity vx = 50 m/s at the origin when
t = 0. For the first 4 seconds it has no acceleration, and thereafter it is acted on by a retarding force which gives it a constant acceleration
ax = −10 m/s2 . Calculate the velocity and the
x-coordinate of the particle for the conditions
of t = 8 s and t = 12 s and find the maximum
positive x-coordinate reached by the particle.
Solution. The velocity of the particle after
t = 4 s is computed from
Z t
Z vx
dt
dvx = −10
particle had a negative velocity which was becoming less negative. If the particle is slowing down,
the particle is said to be decelerating.
Eliminating the time dt from Eqs. (1) and (2), we
obtain a differential equation relating displacement,
velocity, and acceleration.
vdv = ads
5
or
ṡdṡ = s̈ds
(3)
Examples
1. The position coordinate of a particle which is
confined to move along a straight line is given
by s = 2t3 − 24t + 6, where s is measured
in meters from a convenient origin and t is in
seconds.
vx
(a) Determine the time required for the particle to reach a velocity of 72 m/s from
its initial condition at t = 0.
Solution. The velocity is obtained by
successive differentiation of s with respect
to the time.
v = 6t2 − 24 m/s
= −10(t − 4)
= −10t + 90 m/s
(7)
At the specified times, the velocities are
vx
= −10(8) + 90 = 10 m/s
(8)
vx
= −10(12) + 90 = −30 m/s
(9)
The x-coordinate of the particle at any time
greater than 4 seconds is the distance traveled during the first 4 seconds plus the distance
traveled after the discontinuity in acceleration
occurred. Thus,
Z t
x = 50(4)+ (−10t+90)dt = −5t2 +90t−80 m
(4)
Substituting v = 72 into Eq. (4) gives us
72 = 6t2 − 24, from which t = ±4 s. Because the negative value is of no physical
interest, the desired result is 4 s.
(b) Determine the acceleration of the particle
when v = 30 m/s.
Solution. The acceleration is obtained
by successive differentiation of v with respect to the time.
2
4
50
vx − 50
4
(10)
For the two specified times,
x = −5(8)2 + 90(8) − 80 = 320 m (11)
(5)
x = −5(12)2 + 90(12) − 80 = 280 m(12)
Substituting v = 30 m/s into Eq. (4)
gives 30 = 6t2 − 24, from which the positive root is t = 3 s, and the corresponding
acceleration is a = 12(3) = 36 m/s2 .
(c) Determine the net displacement of the
particle during the interval from t = 1
s to t = 4 s.
Solution. The net displacement during
the specified interval is
When the velocity is zero, Eq. (7) is 0 =
−10t + 90, from which t = 9 s. The maximum
positive x-coordinate is
a = 12t m/s
xmax = −5(9)2 + 90(9) − 80 = 325 m
References
[1] J. L. Meriam and L. G. Kraige, (2001), Engineering Mechanics, Volume 2, Dynamics, 5th
edition, Wiley
∆s = s(4) − s(1)
= [2(4)3 − 24(4) + 6]
−[2(1)3 − 24(1) + 6]
= 54 m
(13)
(6)
2