Lecture 2C.2
Newton-Euler Equations Most of us are familiar with Newton’s equations of motion for a single particle of mass m: F=ma
If we know the resultant force acting on a single particle of mass, acceleration is simply obtained from Newton's second law
m,
then the
!hat does this equation look like when we consider a system of particles or a rigid "body# $et’s start with a system of particles %efore we can define Newton's &econd $aw for a system of particles, we must first define the centre of mass he image below shows a set of particles (), (*, and so on
$et's assume that we can write position +ectors p1, p2, pi, for each of these particles $et's denote the mass of the ith particle by Mi !e can now compute the a+erage position +ector, by weighting each position +ector with the appropriate mass r c
)
mi
mi p i ), N
In this equation you see here, m p i i, is essentially the weighted sum of all the position +ectors !e di+ide that by the total mass, m, to obtain a new position +ector his position +ector defines the centre of mass It turns out that the centre of mass for a system of particles, &, beha+es in eactly the same way as a single particle, with mass m -equal to the total mass of the system. would ha+e beha+ed, if it had been located at the centre of mass
&o instead of writing / = ma for a system of particles, we write / = m acceleration of the centre of mass: A
N
F
F i i )
m
the
d Av C dt
!e sum the indi+idual forces to get /, and equate that to the total mass, m, times the deri+ati+e of the +elocity of the centre of mass he notation probably requires some eplanation ) he superscript 0 refers to the fact that we're computing these quantities in an inertial frame 0
* he superscript 1, refers to the fact that we're computing the +elocity of the centre of mass 2emember that the right"hand side is essentially the rate"of"change of linear momentum If $ the $inear Momentum is $, then the total force, /, is gi+en by: A
F
dL
dt
his is Newton's second law for a system of particles 0gain the deri+ati+e of each of these quantities, must be performed in an inertial frame in order for this equation to be +alid his is also true for a rigid body 3ne way to think of a rigid"body is to consider it as simply an infinite set of particles, which are all glued together rigidly &o if this is +alid for a set of particles, it must also be +alid for an infinite collection of particles &o that was Newton's second law !hat are the rotational laws of motion for a rigid" body# $et's deri+e the rotational equations of motion for a rigid body /or linear motion we considered the linear momentum of the rigid body and it's deri+ati+e /or rotational motion, the analogous quantity to consider is the angular momentum
he rate of change of angular momentum of a rigid"body %, relati+e to the centre"of" mass, 1, in an inertial frame 405 is equal to the resultant moment of all eternal forces acting on the body relati+e to 1: A
d A H B C dt
B M C
!here 6 is the angular momentum of a rigid body with origin 1 -the centre"of"mass., in an inertial frame 405 !e want to calculate the rate"of"change of angular momentum of the rigid"body % hat equals the net moment applied to the rigid"body 3nce again, the differentiation must be done in an inertial frame In order to actually perform the calculation on the left"hand side, we ha+e to replace 6 with something that we can easily measure It turns out that the angular momentum is nothing but the body’s inertia times its angular +elocity A
B H C
I C A B
hese computations ha+e to be done in three dimensions he angular momentum and the angular +elocity are 7"dimensional +ectors he quantity in between is the inertia tensor, whose components can be written as a 77 matri 6ere, the subscript 1 denotes the fact that we measured the components with the centre"of"mass, 1, as the origin he angular +elocity is also obtained in the inertial frame, and the leading superscript 0 captures this he trailing superscript %, tells us that this is the angular +elocity of the rigid body % that we are measuring /inally, M is the net moment ake all the eternal forces, compute their moments, and add these moments to all eternal couples or torques 3nce again, the trailing subscript 1 reminds us that we're computing moments with the centre"of"mass, 1, as the origin
