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Problem 5: Consider the motion of a thin flat disk rolling on a plane shown in Figure: Problem 3. The configuration space of the system is parameterized by the xy location of the contact point of the...

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Problem 5:
Consider the motion of a thin flat disk rolling on a plane shown in Figure: Problem 3. The
configuration space of the system is parameterized by the xy location of the contact point of the disk
with the plane, the angle 0 that the disk makes with the horizontal line, and the angle ¢ of a fixed line on
the disk with respect to the vertical axis.

Figure: Problem 3: Disc Rolling on a Plane
‘We assume that the disk rolls without slipping. Writing these equations with q = (6, x, y, ©) we have the
co
esponding kinematic equations of motion
z— pcosf =0
§— psinfé =0,
where, p > 0 is the radius of the disk.
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Page 1 of 2
‘Writing the above equations in the co
esponding control system format as
0
ee ba]. ul] _ _
a0=|3|=lo1 921] = 6@u=g@u + gu:
¢ :
Here, g1 and g; are 4x1 column vectors; control inputs u; = ¢ = the rate of rolling, and
uz = 6 = rate of turning about the vertical axis.
1) Find the column vectors gi and ga.
Using Lie alge
a,
2) Evaluate g3 = [g1, g2] and ga = [g2, g3].
3) Check if the complete span {g1, g2, g3, g4} is of full rank fo
the controllability of the system.
Problem1:
For trajectory generation of an industrial robot, evaluate the coefficients of the fifth-order polynomial
s(t) = ast’ + ast* + ast’ + at? + ast + ao with time scaling that satisfies s(T) = 1 and s(0) = $(0) =5(0) =
$(T) =8(T) = 0, where the initial time t = 0 and the final time T =1.
Problem 2:
For a one degree-of-freedom (DOF) robotic system of the form mi + bx + kx =f, where fis the control
force and m = 4 kg, b=2 Ns/m, and £ = 0.1 N/m, obtain a proportional — derivative (PD) controller that
yields critical damping and a 2% settling time of 0.01 sec.
Answered Same Day Dec 16, 2022

Solution

Karthi answered on Dec 17 2022
28 Votes
1.
To evaluate the coefficients of a fifth-order polynomial for trajectory generation of an industrial
obot, we need to satisfy the given conditions at the initial and final times: s(0) = ṡ(0) = s(̈0) =
ṡ(T) = s(̈T) = 0.
The polynomial s(t) can be written in the following form:
s(t) = a5t^5 + a4t^4 + a3t^3 + a2t^2 + a1*t + a0
To satisfy the condition s(0) = 0, we set t = 0 and solve for a0:
s(0) = a50^5 + a40^4 + a30^3 + a20^2 + a1*0 + a0 = 0
Therefore, a0 = 0.
To satisfy the condition ṡ(0) = 0, we take the first derivative of s(t) and set t = 0:
ṡ(t) = 5a5t^4 + 4a4t^3 + 3a3t^2 + 2a2t + a1
ṡ(0) = 5a50^4 + 4a40^3 + 3a30^2 + 2a20 + a1 = 0
Therefore, a1 = 0.
To satisfy the condition s(̈0) = 0, we take the second derivative of s(t) and set t = 0:
s(̈t) = 20a5t^3 + 12a4t^2 + 6a3t + 2*a2
s(̈0) = 20a50^3 + 12a40^2 + 6a30 + 2*a2 = 0
Therefore, a2 = 0.
To satisfy the condition ṡ(T) = 0, we take the first derivative of s(t) and set t = T:
ṡ(T) = 5a5T^4 + 4a4T^3 + 3a3T^2 + 2a2T + a1
ṡ(T) = 5a51^4 + 4a41^3 + 3a31^2 + 2a21 + a1 = 0
Therefore, a4 = 0.
To satisfy the condition s(̈T) = 0, we take the second derivative of s(t) and set t = T:
s(̈T) = 20a5T^3 + 12a4T^2 + 6a3T + 2*a2
s(̈T) = 20a51^3 + 12a41^2 + 6a31 + 2*a2 = 0
Therefore, a3 = 0.
Finally, we need to satisfy the...
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