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Pendulum with springs
c
Paul
Stiverson
Problem
c
m, l
l/4
G
O
k
Consider the pendulum system above, it is comprised of a uniform slender rod of length l and mass m
that is pinned to the wall at l/4 below the center of gravity, the bottom point of the pendulum is connected
to the wall by a spring, and the top point is connected to the wall by a damper. Find the equation of motion
for small angular displacement of the system. Comment on the stability of the system.
Solution
As usual the first task is to assign coordinates to define our motion, in this case we only need a rotational
coordinate (because the rotation is about a fixed point). Once the coordinates are established we can displace
the body in the positive direction and draw a free body diagram. This is given in Figure 1.
Fc
θ
G
mg
Oθ
O
Or
Fk
Figure 1: Free Body Diagram of Our Pendulum
Since the spring is attached at a distance of 1/4l below the pin we know that the deflection of the spring
will be 1/4l sin(θ), thus the spring force is
1
Fk = kl sin(θ)
4
3
Likewise the damper is affixed at a distance of /4l so its deflection will be 3/4l sin(θ); since dampers respond
to velocity rather than displacement we will want to differentiate that term yielding 3/4l cos(θ)θ̇, this makes
our damper force
3
Fc = cl cos(θ)θ̇
4
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Pendulum with springs
c
Paul
Stiverson
Before proceeding to the summation of moments we need to determine the moment of inertia of our
pendulum about the fixed axis of rotation, centered at point O. We know the inertia of the slender rod
about the axis passing through the center of gravity is:
IG =
1
ml2
12
Using the parallel axis theorem we can transition to a general point, in this case point O is desired.
1
IO = IG + m|~bOG |2 =
ml2 + m
12
2
l
7
ml2
=
4
48
Now we can sum the moments on the body, since we will be summing about point O we will not need
to consider the reaction forces at the pin. Recall that the forces discussed above are only forces and we are
summing moments, so we will need to multiply by the moment-arm.
X
1
3
1
MO = IO θ̈ = −Fk
l cos(θ) − Fc
l cos(θ) + mg
l sin(θ)
4
4
4
In this summation the moments induced by the spring and damper are both negative, this is because
those forces will cause the assembly to rotate in the anti-clockwise direction while our coordinate states the
positive direction is clockwise. The force due to gravity is causing a clockwise rotation (same as +θ) so it
has a positive sign in the summation. Let’s make some substitutions in that summation equation.
IO θ̈
7ml2
θ̈
48
1
1
3
3
1
= − kl sin(θ)
l cos(θ) − cl cos(θ)θ̇
l cos(θ) + mg
l sin(θ)
4
4
4
4
4
kl2
9cl2
mgl
= −
sin(θ) cos(θ) −
cos2 (θ)θ̇ +
sin(θ)
16
16
4
The above is the equation of motion for the system, but since the problem asks us to analyze for small
angular motion we can make a further observation. Namely that for small angles (θ < π/12 = 15◦ ), sin(θ) ≈ θ
and cos(θ) ≈ 1. Making these substitutions gives us the linearized equations of motion.
kl2
9cl2
mgl
θ−
θ̇ +
θ
16
16
4
7ml2
9cl2
kl2
mgl
0 =
θ̈ +
θ̇ +
−
θ
48
16
16
4
kl2
48
mgl
48 9cl2
0 = θ̈ +
θ̇ +
−
θ
7ml2 16
7ml2 16
4
27c
3k
12g
0 = θ̈ +
θ̇ +
−
θ
7m
7m
7l
7ml2
θ̈
48
0
= −
= θ̈ + 2ζωn θ̇ + ωn2 θ
From this equation of motion we know that the natural frequency is
r
3k
12g
−
ωn =
7m
7l
That the natural frequency is a real number is a requirement for stability, and that requires the radicand1
1 Radicand
is the general name for a quantity under a radical (square root symbol).
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Pendulum with springs
c
Paul
Stiverson
to be strictly positive. Thus for stability:
12g
3k
−
7m
7l
3k
7m
> 0
>
k
>
k
>
12g
7l
12g 7m
7l 3
4mg
l
So long as our stiffness meets the above criteria our system will be stable.
And we’re done! I hope you found this useful, check my twitter often (@MEENTutorPaul) for updates
and more example problems.
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