ME 240: Introduction to Dynamics and Vibrations
Mechanical Engineering Department
The University of Michigan
Winter 2008
Computer Assignment #1
Solution
February 8, 2007
Prepared by Joosup Lim (
[email protected]
)
(i)
Free body diagram is shown in Fig.(
1
).
m
θ
R
e
^
n
e
^
t
θ
F
D
mg
N
Figure 1: Free Body Diagram
(ii)
Using tangentialnormal coordinates, and Newton’s second law, or
∑
F
=
m
a
, we can write the equations
as,
(
F

D

mg
sin
θ
)
e
t
+ (
N

mg
cos
θ
)
e
n
=
m
(
R
¨
θ
e
t
+
R
˙
θ
2
e
n
)
(1)
We can consider tangential components only because there are no friction between the ring and the mass.
Hence,
summationdisplay
F
t
=
F

D

mg
sin
θ
=
mR
¨
θ
(2)
As given in the handout, the drag force
D
can be expressed as,
D
=
bv
=
bv
t
=
bR
˙
θ
(3)
1
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By substituting
D
and rearranging Eq.(
2
), we can get the required equation of motion,
mR
¨
θ
+
bR
˙
θ
+
mg
sin
θ
=
F
(4)
After getting the answer, we can simplify Eq.(
4
) for numerical simulation,
¨
θ
+
b
m
˙
θ
+
g
R
sin
θ
=
F
mR
(5)
(iii)
Consider two cases of (a)
b
= 1
and (b)
b
= 10
for problem (iii) and (iv).
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