IEEE Robotics & Automation Magazine - September 2010 - 84

Solving a Two-Body Dynamics Problem Using 3-D Vectors

a1
a2
P
x_ 1
P
x_ 2
1
f2
P
n2
P

W

e are given a rigid-body system consisting of two bodies,
B1 and B2 , connected by a revolute joint [S1] The bodies
have masses of m1 and m2 , centers of mass located at the points
C1 and C2 , and rotational inertias of I1 and I2 about their respective centers of mass. Both bodies are initially at rest. The joint's
axis of rotation passes through the point P in the direction given
by s. A system of forces acts on B1 , causing both bodies to accelerate. This system is equivalent to a single force f acting on a line
passing through C1 together with couple n. These forces impart
an angular acceleration of x_ 1 to B1 and a linear acceleration of
a1 to its center of mass. The problem is to express a1 and x_ 1 in
terms of f and n (Figure S1).

P

¼ a1 À r 1 3 x_ 1 ,
¼ a2 À r 2 3 x_ 2 ,
¼ x_ 1 ,
¼ x_ 2 ,
¼ Pf 2 ¼ f 2 ,
¼ n2 À r 2 3 f 2 ,

(S5)
(S6)
(S7)
(S8)
(S9)
(S10)

and
n2 ¼ n2 þ (r 1 À r 2 ) 3 f 2 :

1

(S11)

If B1 exerts 1 f 2 and 1 n2 on B2 then B2 exerts À1 f 2 and
À n2 on B1 (Newton's third law expressed at C1 ); so, the net
force and couple acting on B1 are
1

P

f 1 ¼ f À1 f 2 ,
n1 ¼ n À1n2 ,

s
n

f

C1
.

ω1

r2

r1

from which we get [via (S9) and (S11)]

C2
(m2, I2)

a1

B1

(S12)

n ¼ n1 þ n2 þ (r 1 À r 2 ) 3 f 2 :

(S13)

and

B2

The joint allows B2 one degree of motion freedom relative
to B1 and imposes one constraint on the couple transmitted
from B1 to B2 . Expressed at P, the constraint equations are

Figure S1. Problem diagram using 3-D vectors.

Solution
Let f 1 , n1 , f 2 , and n2 be the net forces and couples acting on
B1 and B2 , respectively, where the lines of action of f 1 and f 2
pass through C1 and C2 , respectively; let x_ 2 and a2 be the
angular acceleration of B2 and the linear acceleration of its
center of mass; let P a1 , P x_ 1 , P a2 , and P x_ 2 be the linear and
angular accelerations of B1 and B2 expressed at P; and let P f 2 ,
P
n2 , 1 f 2 , and 1 n2 be the net force and couple acting on B2
expressed at P and C1 , respectively. As the system of applied
forces acts only on B1 , the net force and couple acting on B2
are also the net force and couple transmitted through the
!
!
joint. Let us also define r 1 ¼ C1 P and r 2 ¼ C2 P , and let a be
the joint acceleration variable.
The equations of motion of the two bodies, expressed at
their centers of mass, are
f 1 ¼ m 1 a1 ,
n1 ¼ I1 x_ 1 ,
f 2 ¼ m 2 a2 ,

(S1)
(S2)
(S3)

a2 ¼ Pa1 ,
x_ 2 ¼ Px_ 1 þ s a,

(S14)
(S15)

sT Pn2 ¼ 0,

(S16)

P
P

and

where a is the unknown joint acceleration variable. There is
no constraint on P f 2 : (S16) is sufficient to ensure that the
force and couple transmitted by the joint perform no work
in the direction of relative motion permitted by the joint.
We are now ready to solve the problem. Let us start by
calculating a2 and x_ 2 in terms of a1 , x_ 1 and a. From (S8),
(S15), and (S7), we have
x_ 2 ¼ Px_ 2
¼ Px_ 1 þ s a
¼ x_ 1 þ s a,

(S17)

and from (S6), (S14), (S17), and (S5) we have

and
n2 ¼ I2 x_ 2 :

(S4)

There are no velocity terms because the bodies are at rest.
The rules for transferring forces and accelerations (of bodies
at rest) from one point to another provide us with the following relationships between quantities referred to C1 , C2 ,
and P:

84

f ¼ f 1 þ f 2,

(m1, I1)

IEEE Robotics & Automation Magazine

a2 ¼ Pa2 þ r 2 3 x_ 2
¼ Pa1 þ r 2 3 (x_ 1 þ s a)
¼ a1 À r 1 3 x_ 1 þ r 2 3 (x_ 1 þ s a)
¼ a1 þ (r 2 À r 1 ) 3 x_ 1 þ r 2 3 s a:

(S18)

Now let us calculate a. From (S16), (S10), (S3), (S4), (S17),
and (S18), we get

SEPTEMBER 2010
Table of Contents for the Digital Edition of IEEE Robotics & Automation Magazine - September 2010
