Download Lecture L4 - Curvilinear Motion. Cartesian Coordinates and more Study notes Dynamics in PDF only on Docsity!
S. Widnall, J. Peraire 16.07 Dynamics Fall 2009 Version 2.
Lecture L4 - Curvilinear Motion. Cartesian Coordinates
We will start by studying the motion of a particle. We think of a particle as a body which has mass,
but has negligible dimensions. Treating bodies as particles is, of course, an idealization which involves an
approximation. This approximation may be perfectly acceptable in some situations and not adequate in
some other cases. For instance, if we want to study the motion of planets, it is common to consider each
planet as a particle. This simplification is not adequate if we wish to study the precession of a gyroscope or
a spinning top.
Kinematics of curvilinear motion
In dynamics we study the motion and the forces that cause, or are generated as a result of, the motion.
Before we can explore these connections we will look first at the description of motion irrespective of the
forces that produce them. This is the domain of kinematics. On the other hand, the connection between
forces and motions is the domain of kinetics and will be the subject of the next lecture.
Position vector and Path
We consider the general situation of a particle moving in a three dimensional space. To locate the position of
a particle in space we need to set up an origin point, O, whose location is known. The position of a particle
A, at time t, can then be described in terms of the position vector, r, joining points O and A. In general,
this particle will not be still, but its position will change in time. Thus, the position vector will be a function
of time, i.e. r(t). The curve in space described by the particle is called the path, or trajectory.
We introduce the path or arc length coordinate, s, which measures the distance traveled by the particle along
the curved path. Note that for the particular case of rectilinear motion (considered in the review notes) the
arc length coordinate and the coordinate, s, are the same.
Using the path coordinate we can obtain an alternative representation of the motion of the particle. Consider
that we know r as a function of s, i.e. r(s), and that, in addition we know the value of the path coordinate
as a function of time t, i.e. s(t). We can then calculate the speed at which the particle moves on the path
simply as v = s˙ ≡ ds/dt. We also compute the rate of change of speed as at = ¨s = d 2 s/dt 2 .
We consider below some motion examples in which the position vector is referred to a fixed cartesian
coordinate system.
Example Motion along a straight line in 2D
Consider for illustration purposes two particles that move along a line defined by a point P and a unit vector
m. We further assume that at t = 0, both particles are at point P. The position vector of the first particle is
given by r 1 (t) = rP + mt = (rP x + mxt)i + (rP y + my t)j, whereas the position vector of the second particle
is given by r 2 (t) = rP + mt 2 = (rP x + mxt 2 )i + (rP y + my t 2 )j.
Clearly the path for these two particles is the same, but the speed at which each particle moves along the
path is different. This is seen clearly if we parameterize the path with the path coordinate, s. That is,
we write r(s) = rP + ms = (rP x + mxs)i + (rP y + my s)j. It is straightforward to verify that s is indeed
the path coordinate i.e. the distance between two points r(s) and r(s + Δs) is equal to Δs. The two
motions introduced earlier simply correspond to two particles moving according to s 1 (t) = t and s 2 (t) = t^2 ,
respectively. Thus, r 1 (t) = r(s 1 (t)) and r 2 (t) = r(s 2 (t)).
It turns out that, in many situations, we will not have an expression for the path as a function of s. It is
in fact possible to obtain the speed directly from r(t) without the need for an arc length parametrization of
the trajectory.
Velocity Vector
We consider the positions of the particle at two different times t and t + Δt, where Δt is a small increment
of time. Let Δr = r(r + Δt) − r(t), be the displacement vector as shown in the diagram.
translate the velocity vectors, at different times, such that they all have a common origin, say, O �
. Then,
the heads of the velocity vector will change in time and describe a curve in space called the hodograph. We
then see that the acceleration vector is, in fact, tangent to the hodograph at every point.
Expressions (1) and (2) introduce the concept of derivative of a vector. Because a vector has both magnitude
and direction, the derivative will be non-zero when either of them changes (see the review notes on
vectors). In general, the derivative of a vector will have a component which is parallel to the vector itself,
and is due to the magnitude change; and a component which is orthogonal to it, and is due to the direction
change.
Note Unit tangent and arc-length parametrization
The unit tangent vector to the curve can be simply calculated as
et = v/v.
It is clear that the tangent vector depends solely on the geometry of the trajectory and not on the speed
at which the particle moves along the trajectory. That is, the geometry of the trajectory determines the
tangent vector, and hence the direction of the velocity vector. How fast the particle moves along the
trajectory determines the magnitude of the velocity vector. This is clearly seen if we consider the arc-length
parametrization of the trajectory r(s). Then, applying the chain rule for differentiation, we have that,
dr dr ds v = = = etv , dt ds dt
where, s˙ = v, and we observe that dr/ds = et. The fact that the modulus of dr/ds is always unity indicates
that the distance traveled, along the path, by r(s), (recall that this distance is measured by the coordinate
s), per unit of s is, in fact, unity!. This is not surprising since by definition the distance between two
neighboring points is ds, i.e. |dr| = ds.
Cartesian Coordinates
When working with fixed cartesian coordinates, vector differentiation takes a particularly simple form. Since
the vectors i, j, and k do not change, the derivative of a vector A(t) = Ax(t)i + Ay (t)j + Az (t)k, is simply
A˙^ (t) = A˙x(t)i + A˙y (t)j + A˙z (t)k. That is, the components of the derivative vector are simply the derivatives
of the components.
Thus, if we refer the position, velocity, and acceleration vectors to a fixed cartesian coordinate system, we
have,
r(t) = x(t)i + y(t)j + z(t)k (3)
v(t) = vx(t)i + vy (t)j + vz (t)k = x˙ (t)i + ˙y(t)j + ˙z(t)k = r˙ (t) (4)
a(t) = ax(t)i + ay (t)j + az (t)k = v˙x(t)i + ˙vy (t)j + ˙vz (t)k = v˙ (t) (5)
Here, the speed is given by v = vx^2 + vy^2 + vz^2 , and the magnitude of the acceleration is a = a^2 x + a^2 y + a^2 z.
The advantages of cartesian coordinate systems is that they are simple to use, and that if a is constant, or
a function of time only, we can integrate each component of the acceleration and velocity independently as
shown in the ballistic motion example.
Example Circular Motion
We consider motion of a particle along a circle of radius R at a constant speed v 0. The parametrization of
a circle in terms of the arc length is
s s r(s) = R cos( )i + R sin( )j. R R
Since we have a constant speed v 0 , we have s = v 0 t. Thus,
r(t) = R cos(
v 0 t )i + R sin(
v 0 t )j. R R
The velocity is
v(t) =
dr(t) = −v 0 sin(
v 0 t )i + v 0 cos(
v 0 t )j , dt R R
Example Ballistic Motion
Consider the free-flight motion of a projectile which is initially launched with a velocity v 0 = v 0 cos φi +
v 0 sin φj. If we neglect air resistance, the only force on the projectile is the weight, which causes the projectile
to have a constant acceleration a = −gj. In component form this equation can be written as dvx/dt = 0
and dvy /dt = −g. Integrating and imposing initial conditions, we get
vx = v 0 cos φ, vy = v 0 sin φ − gt ,
where we note that the horizontal velocity is constant. A further integration yields the trajectory
x = x 0 + (v 0 cos φ) t, y = y 0 + (v 0 sin φ) t − 2
gt 2 ,
which we recognize as the equation of a parabola.
The maximum height, ymh, occurs when vy (tmh) = 0, which gives tmh = (v 0 /g) sin φ, or,
v 0 2 sin 2 φ ymh = y 0 +. 2 g
The range, xr , can be obtained by setting y = y 0 , which gives tr = (2v 0 /g) sin φ, or,
2 v 0 2 sin φ cos φ v 0 2 sin(2φ) xr = x 0 + = x 0 +. g g
We see that if we want to maximize the range xr , for a given velocity v 0 , then sin(2φ) = 1, or φ = 45 o^.
Finally, we note that if we want to model a more realistic situation and include aerodynamic drag forces of
the form, say, −κv^2 , then we would not be able to solve for x and y independently, and this would make the
problem considerably more complicated (usually requiring numerical integration).
ADDITIONAL READING
J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition
MIT OpenCourseWare http://ocw.mit.edu
16.07 Dynamics
Fall 2009
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
