School of Aerospace Engineering
Georgia Institute of Technology
AE3515A HW SET #3 Solutions Complete by 6/11/2004
______________________________________________________________________________
1) Consider the single link robotic arm shown in the figure with r and θ chosen as generalized
coordinates. A linear and rotary actuator apply input force f and torque τ in the direction of r
and θ, respectively.
m
r
θ
Ignoring the arm mass and moment of inertia (the payload mass m is the only significant inertia)
derive the equation of motion for the system.
Solution: The Kinetic and Potential (assuming the arm is in a vertical plane) energies in terms of
the genralized coordinates (r,θ) are given by
(
)
(
)
θ=
θ+==
sin
2
1
2
12
22
mgrU
rrmmvT m&
&
Thus the Lagrangian L=T-U is
(
)
(
)
θ−θ+= sin
2
12
2mgrrrmL &
&
The equations of motion according to Lagrangian formulation are
()
()
τ=θ+θ=
θ∂
∂
−
θ∂
∂
=θ+θ−=
∂
∂
−
∂
∂
cos
sin
2
2
mgrr
dt
dLL
dt
d
Fmgmrrm
dt
d
r
L
r
L
dt
d
&
&
&
&
&
Taking the time derivatives in the above equations using chair rule gives
mgrrrr
mFgrr
τ=θ+θ+θ
=θ+θ−
cos2
sin
2
2
&
&
&&
&
&&
b) Linearize the equations of motion of in (a) about the equilibrium points corresponding to θ=90
degrees and r=1m. Derive the transfer function Θ(s)/T(s) of the linarized system.
From the equations of motion, it can be seen that the equilibrium values of F and τ are mgF =
and 0=τ . Linearizing the nonlinear term sin(θ), cos(θ) , and 2
θ
&
r, and θ+θ &
&
&& rrr 2
2about
0,1,0,2/ ===θπ=θ rr &
& we get
