Environmental Engineering Reference
In-Depth Information
We can use this expression to figure out the angular speed,
ω
, about a vertical
axis through
C
,i.e.
bp
=
I
C
ω,
(4.47)
where
I
C
is the moment of inertia of the bat about the vertical axis. Armed with
this information, we can now figure out what is happening at the handle,
H
.The
velocity of
H
, immediately after the impulse, is parallel to
p
and a superposition
of the velocity of the centre of mass (
V
c
) with the velocity relative to the centre
of mass, and the latter is the result of a pure rotation about the centre of mass.
Thus we can write
p
M
−
ωh
t,
x
=
(4.48)
where
x
is the displacement of the handle (in the direction of
p
) that occurs in
a short time
t
after the impact. The sweet spot is defined such that
x
=
0, i.e.
p
M
=
ωh.
(4.49)
Finally, substitute for
ω
using Eq. (4.47) to determine the position of the sweet spot:
p
M
bh
I
C
p
=
(4.50)
and so
I
C
hM
.
b
=
(4.51)
4.6 KINETIC ENERGY OF ROTATION
When an extended body rotates, its constituent particles will each have some
kinetic energy. We can figure out the total energy of a rigid body by summing up
all of these contributions. Referring to Figure 4.3, particle
j
has a kinetic energy
1
2
m
j
v
j
1
2
m
j
d
j
ω
2
=
(4.52)
and the total kinetic energy of a system of many particles rotating about a fixed
axis is therefore
2
j
ω
2
1
2
Iω
2
.
m
j
d
j
K
rot
=
=
(4.53)
If the body is only rotating about the fixed axis (i.e. the centre of mass is at rest)
then this is the sole contribution to the kinetic energy. However, if we allow the
body to undergo translation
3
as well as rotation then we must add together the
3
Pure rotational motion arises if each and every part of the body is undergoing circular motion about
some axis as determined in any inertial frame.
