Biomedical Engineering Reference
In-Depth Information
relation to this axis as illustrated in Fig.
4.9
. Determine the moment of inertia of this
person about the main longitudinal axis. Compare the degree of difficulty to rotate
about this axis when she or he is with opened or closed arms.
4.5 Angular Momentum and Its Conservation
Consider an ice skater rotating with a certain angular velocity about her or his
longitudinal axis with horizontally outstretched arms. What happens if she or he
closes their arms, hugging the axis of rotation? Why does an athlete change the
shape of his body to perform an ornamental jump? To understand such questions,
let us introduce a physical quantity called angular momentum
L
.
Every body rotating about an axis has an angular momentum
L
(
4.10
) given by
the product of the moment of inertia
I
of the body and the angular velocity,
ω
:
L
¼
I
ω:
(4.10)
Hence, the angular momentum depends not only on the mass but on its distribu-
tion about a rotational axis and the angular velocity. Its unit in the SI is kg m
2
/s,
with the radian (rad) of angular velocity dropped.
The angular momentum is conserved (the value does not change) as the net total
torque due to the external actions on the body is zero. Then, if no external torques
act on a rotating body, it will maintain its rotation indefinitely, conserving its angular
momentum. If no external torques act on the body that is not in rotation, on the
other hand, the body will not rotate and its angular momentum will continue zero.
Equation (
4.10
) shows that, as the value of
L
must be maintained constant, if
I
increases,
ω
must diminish and vice versa.
Example 4.3
Consider the wheel of a bicycle with radius
R
of 30 cm and mass
M
of
2.0 kg that is supposed to be concentrated at its rim. The linear velocity of the wheel
is 5.0 m/s. For objects with this geometric shape, the moment of inertia about the
axis passing through its C.G. can be calculated by
I
MR
2
. Find for the wheel
(a) its moment of inertia, (b) the radius of gyration, (c) the angular velocity, and
(d) the angular momentum.
¼
(2.0 kg)(0.30 m)
2
0.18 kg m
2
.
(a)
I
¼
¼
(b)
k
¼
R
¼
30 cm.
(c)
ω ¼
v
/
R
¼
(5.0 m/s)/0.30 m
¼
16.7 rad/s.
(0.18 kg m
2
)(16.7 s
1
)
3.0 kg m
2
/s.
(d)
L
¼
I
ω ¼
¼
Exercise 4.3
A car with 1,500 kg mass is moving with a velocity of 144 km/h
¼
40 m/s in a circular racecourse with 50 m radius. Obtain the angular velocity and
the angular momentum about the center of the racecourse. In this case use the equation
I
mR
2
, since the auto can be approximated as a particle with mass
m
, rotating about
the axis of rotation as shown in Fig.
4.2
, as the radius is very large.
¼
