Environmental Engineering Reference
In-Depth Information
The angular momentum is then
10
−
33
kg m
2
s
−
1
.
L
=
Iω
∼
4
×
As you can see, SI units are not so convenient for the description of subatomic
objects. Physicists would normally express the angular momentum of a nucleus in
terms of the fundamental quantity
10
−
34
kg m
2
s
−
1
=
1
.
054589
×
in which case we can write the answer as
∼
L
40
.
In the above example we have used classical mechanics to obtain an
order-of-magnitude estimate for a calculation that should really be carried out
using quantum mechanics. While not strictly correct, such “ballpark” estimates
are frequently used as a quick first approach by professional physicists to get an
idea of an order of magnitude or to point the way forward to a more elaborate
(and more correct) calculation.
We have shown in this section that the calculation of the angular momentum of a
rigid body rotating at a known angular speed about a fixed axis boils down to being
able to calculate the appropriate moment of inertia. In the last section we stated
that it is an experimental fact that the sum of the internal torques is always zero.
We can now bring these two results together to establish the equation of motion
for the rotation of a rigid body about a fixed axis:
d
L
d
t
=
d
(I ω)
d
t
I
d
ω
τ
=
=
d
t
=
Iα,
(4.30)
where we have introduced the angular acceleration,
d
ω
d
t
.
In Eq. (4.30),
τ
is the component of net external torque
α
=
in the direction of
the rotation axis. Notice that this equation is very similar in structure to New-
ton's Second Law with torque replacing force, angular momentum replacing linear
momentum and angular acceleration replacing linear acceleration. This similarity
to Newton's Second Law is handy to remember when it comes to problem solving.
Finally, we are ready to go ahead and study the motion of a particular rigid
body. We shall consider the situation illustrated in Figure 4.6. The body is free
to rotate about a horizontal axis in the Earth's gravitational field and the centre
of mass of the object is at the point
C
. The perpendicular distance from the axis
of rotation to
C
is
R
and
P
is the point on the axis of rotation that lies directly
above
C
when the body is in equilibrium. The angle
θ
(see Figure 4.6) therefore
specifies the extent of deviations from equilibrium. Our task is to understand the
general motion of the body as it rotates about the axis and that is achieved if we
can figure out how
θ
varies with time. To do that, we must solve Eq. (4.30) above,
remembering that
ω
τ
d
2
θ/
d
t
2
. We need the torque about
P
that
=
d
θ/
d
t
and
α
=
