Environmental Engineering Reference
In-Depth Information
P
q
R
P
R
C
W
=
m
g
C
(a)
(b)
Figure 4.6
A pendulum consisting of a rigid body free to rotate about an axis.
acts on the body (more specifically we need the component of the torque along the
axis of rotation) and this is provided by the weight. Now, the weight of the body
acts as if it is concentrated at the centre of mass
2
andsoweareabletoworkout
the torque about
P
. Our particular choice of the point
P
guarantees that the torque
is parallel to the axis of rotation, i.e.
τ
=−|
R
×
W
| =−
mg R
sin
θ,
(4.31)
where
W
mg
is the weight of the body. Note the minus sign: the way we defined
the angle
θ
means that
=
ω
points out of the plane of the page in Figure 4.6(b)
whereas
points into the page (from the definition of the vector product). We can
now substitute
τ
into Eq. (4.30):
τ
I
d
2
θ
−
=
=
mg R
sin
θ
Iα
d
t
2
.
(4.32)
As a double check that we got the sign right for the torque, you should note that
the torque clearly must act so as to try and pull the body back towards equilibrium,
i.e. the angular acceleration must be negative for positive
θ
. Note that the weight
is not the only force to act upon the body. There will also be a normal reaction
coming from the axis but that produces no torque about
P
.
We can keep things simple if we focus on the case where
θ
is small. Then
sin
θ
≈
θ
and
d
2
θ
d
t
2
mgR
I
≈−
θ.
(4.33)
Notice that this is none other than the equation for simple harmonic motion that
we met in Section 3.2.2. The solution is
θ
=
A
cos
(
2
πf t)
provided
mgR
I
1
2
π
f
=
.
(4.34)
2
Can you prove this? It is easiest to consider a collection of particles. Work out the total gravitational
torque about the origin and then show that this is the same torque that you would obtain if you had all
the mass concentrated at the centre of mass.
