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TAM 212
Spring, 2004
1.68 Given:
The bar of length 2L leans up against a vertical wall.
It makes an angle
q
with the horizontal.
Required: The velocity and acceleration of the center of the bar as a function of
q
and its derivatives.
(Essentially, we are solving for the velocity and acceleration of C, assuming we might know
q
and its derivatives, as functions of time.)
Solution: We first need to express
j
i
r
OC
OC
OC
y
x
+
=
, where O is some fixed point (origin).
Let’s chose
the origin to be where the floor meets the wall.
If you can’t picture the expression for the
position for C immediately, it may be simpler to first think of point B where, B is where the bar
touches the floor.
( 29 ( 29 ( 29
( 29 ( 29 ( 29
( 29
j
i
i
j
i
j
i
r
r
r
q
q
q
sin
cos
0
cos
2
L
L
L
y
x
y
x
BC
BC
OB
OB
BC
OB
OC
+

+
+
=
+
+
+
=
+
=
( 29 ( 29
j
i
r
q
q
sin
cos
L
L
OC
+
=
, so
q
q
sin
;
cos
L
y
L
x
OC
OC
=
=
1.27, 1.28 give us:
j
i
v
OC
OC
OC
y
x
+
=
and
j
i
a
OC
OC
OC
y
x
+
=
So we simply take derivatives (w.r.t. time) of
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This note was uploaded on 04/14/2010 for the course TAM 212 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff

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