Note: Grading Breakdown and Comments Shown in Blue.
+5
+20
+5
+5
+5
+10
+5
+5
+5
Simulink Block Diagram of Ex5.10: Blue Boxed Elements Not Required For Hand Solution.
Time Clock To Workspace3 Omega2
S
0 8.11
(a) Summing torques on the disk and summing forces on the mass, we f-
have {t ( c
.. M 41 X
J6+K9Rf.2 = 0 8,2...) M..
M56 + Bac + K21 + fa = fa(t) x
where fc is the contact force between the m
1am
(a) We dene
1 1 1
=mq Wzmg %[email protected]
and transform Equation (11.19) with zero initial conditions, getting
a1
saw): 4N§®+a$+MmQ®)
SQ): waoywm+mmgg
Then we take the right sides of these tr
o 4.2
We rearrange the given equations to solve for if and 2':
.. 1 . .
.L = (31 + 33)::3 - K12: + B22]
M1
2 = imam M255 K225 322' K22]
M2
A chain of two integrators is used to produce :5 fro
________________________.___--
g 3.2
In the given equations, we replace in, :2, and by v, a, and ('1, respectively. Then the state-variable and output
equations are
1)
II
a
-Yfv aa+fa(t)
v+a
@ a
EE251: Dynamical Systems, Spring 2002
2001 Catalog Data:
R-3, C-3. Corequisite: MA232. System classification, mathematical
modeling of dynamic systems, state space representation, equilibrium points a