7.5. THREE-DIMENSIONAL MOTION OF A POINT IN A RIGID BODY 127
at is, in general, the relative velocity at P or A
4
with respect to A
2
is not zero as A
4
and
A
2
are associated with different links (rigid bodies). One may wonder why the last equation
holds true. e implicit assumption in this problem is that the roller of the follower does
not separate from the surface of the cam and it does not penetrate into the cam. In other
words, the only relative motion possible is along the surface of the cam. However, the last
term on the rhs of Equation (7.14a),
*
*
r
A
4
=A
2
D 0; since
*
r
A
4
=A
2
D 0:
Substituting all the terms into Equation (7.14a), one has
v
P
0:4310
*
{ C 0:9024
*
|
D 63:50
*
{ C 30:33
*
| C v
*
{ : (7.14b)
Equating coefficients of terms associated with
*
| results
v
P
.
0:9024
/
D 30:33;
giving v
P
D 33:61 mm/s.
Substituting into Equation (7.14b) and equating coefficients of terms associated with
*
{ ,
one finds v D 77:986 mm/s. erefore, the required velocity is
*
v
P
D
*
v
A
4
D 14:486
*
{ C 30:33
*
| mm/s:
(c) Acceleration analysis
e acceleration from Equation (7.11) is
*
a
A
4
D
*
a
A
2
C
d
2
*
r
A
4
=A
2
dt
2
Axyz
C
*
˛
*
r
A
4
=A
2
C
*
*
*
r
A
4
=A
2
C 2
*
d
*
r
A
4
=A
2
dt
Axyz
:
Since
*
r
P =A
D
*
r
A
4
=A
2
D 0; therefore it reduces to
*
a
A
4
D
*
a
A
2
C
d
2
*
r
A
4
=A
2
dt
2
Axyz
C 2
*
d
*
r
A
4
=A
2
dt
Axyz
: (7.14c)
Consider first the lhs term,
*
a
A
4
D
*
a
A
4
n
C
*
a
A
4
t
in which the normal component
*
a
A
4
n
D 0 because the follower is supposed to move
along the axis of the shaft only. Otherwise, the follower would not move because the
128 7. KINEMATICS OF RIGID BODIES
non-zero normal component (perpendicular to the axis of the shaft of the follower) of the
acceleration at A
4
will result in a couple that will jam the follower. us, the tangential
component (along the axis of the shaft of the follower; also parallel to the velocity at A
4
)
is
*
a
A
4
t
D
a
A
4
t
sin 25:53
ı
*
{ C cos 25:53
ı
*
|
D
a
A
4
t
0:4310
*
{ C 0:9024
*
|
:
e first term on the rhs of Equation (7.14c) is
*
a
A
2
D !
2
2
.O
2
A/
sin 25:53
ı
*
{ cos 25:53
ı
*
|
D 60:66
*
{ 127:0
*
| mm/s
2
:
is is the normal component and it is along the axis of the shaft of the follower. e
direction is from A
2
toward O
2
(recall, the normal component of the acceleration at a
point is always directing toward the center of curvature in a circular motion). e tangential
component associated with the angular acceleration is zero since the given angular velocity
*
!
2
of the RFR is constant. Physically, if this tangential component is not zero it would
means a force along this direction will result a moment jamming the shaft of the follower.
In order to avoid jamming the shaft of the follower the tangential component of this
acceleration should be zero and therefore the angular velocity
*
!
2
has to be constant.
e second term on the rhs of Equation (7.14c) is
d
2
*
r
A
4
=A
2
dt
2
Axyz
D
d
2
*
r
A
4
=A
2
dt
2
t
Axyz
C
d
2
*
r
A
4
=A
2
dt
2
n
Axyz
D
d
2
*
r
P =A
dt
2
t
Axyz
D a
*
{ ;
where the superscript t and n denote, respectively, the tangential and normal components
of the acceleration. e magnitude of the acceleration a is unknown. Since only the relative
acceleration parallel (that is, tangential) to the direction of the flat surface of the cam is
possible (that is, the normal component is zero; otherwise, the roller of the follower will
leave or lose contact with the surface of the cam or penetrate into the surface which is
impossible as all the links are assumed to be rigid bodies).
e third term on the rhs of Equation (7.14c) is
2
*
d
*
r
A
4
=A
2
dt
Axyz
D 2
!
2
*
k
v
*
{ D 2
!
2
*
k
.77:986/
*
{
D 311:944
*
| mm/s
2
:
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