ME 352 - Machine Design I
Name of Student
___________________________
Fall Semester 2011
Lab Section Number
________________________
Homework No. 5 (30 points). Due at the beginning of lecture on Wednesday, September 28th.
Consider the inverted slider-crank linkage shown in Problem 3.15, see Figure P3.15, page 160. For
the given position of the linkage, the angular velocity and angular acceleration of the input link 2 are
specified as
2
10 rad s
ω
=
/
clockwise and
2
2
5 rad s
α
=
/
clockwise, respectively. Determine the
following:
(i) The first and second-order kinematic coefficients of the mechanism.
(ii) The velocity and the acceleration of point B fixed in link 3. Give the magnitude and the direction of
each vector.
(iii) The unit tangent vector and the unit normal vector to the path of point B. Show the directions of
these two unit vectors on a sketch of the mechanism.
(iv) The radius of curvature of the path of point B. Indicate your answer on a sketch of the mechanism.
(v) The X and Y coordinates of the center of curvature of the path of point B. Indicate the position of the
center of curvature of the path of point B on a sketch of the mechanism.

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
2
Solution to Homework Set 5.
The inverted slider-crank mechanism is shown in Figure 1.
Figure 1. The Inverted Slider-Crank Linkage.
Position Analysis.
The known position variables are
o
1
0 ,
θ =
1
R
125 mm,
=
o
2
150 ,
θ
=
and
2
R
75 mm.
=
The other position variables can be obtained from trigonometry. Consider the triangle
2
4
O O A,
+
using the law of cosines gives
(
)
(
)
(
)
(
)
(
)(
)
2
2
2
2
4
34
1
2
1
2
4
2
AO
R
R
R
2 R
R
cos
O O A
=
=
+
−
∠
(1)
Therefore
(
)
(
)
(
)
(
)(
)
2
2
2
34
R
125
75
2 125
75 cos150
=
+
−
°
(2a)
which gives
34
R
193.62 mm
=
(2b)
Also for the triangle
2
4
O O A,
+
using the law of sines gives
34
4
2
4
2
4
R
R
sin
O AO
sin
AO O
=
∠
∠
(3a)
Rearranging this equation gives
4
2
4
2
4
34
R
125
sin
O AO
sin
AO O
sin150
R
193.62
∠
=
∠
=
°
(3b)
that is
2
4
O AO
18.83
∠
=
°
(3c)
Therefore
3
34
(30
18.83 )
11.17
θ
= θ
= −
° −
° = −
°
(3d)
Velocity Analysis.
The vectors for the inverted slider-crank mechanism are shown in Figure 2. The
vector loop equation (VLE) can be written as
1
I
??
R
R
R
0
2
34
√
√√
+
−
=
(4)

3
The X and Y components of Equation (4) are
2
2
34
3
1
1
cos
cos
cos
0
R
R
R
+
−
=
θ
θ
θ
(5a)
and
2
2
34
3
1
1
sin
sin
sin
0
R
R
R
+
−
=
θ
θ
θ
(5b)
Figure 2. The Vectors for the Inverted Slider-Crank Mechanism.
Differentiating Equations (5a) and (5b) with respect to the input position
θ
2
gives
2
2
34
3
3
34
3
sin
sin
cos
0
R
R
R
′
′
−
−
+
=
θ
θ θ
θ
(6a)
and
2
2
34
3
3
34
3
cos
cos
sin
0
R
R
R
′
′
+
+
=
θ
θ θ
θ
(6b)
Then writing Equations (6) in matrix form gives
34
3
3
3
2
2
34
3
3
34
2
2
sin
cos
sin
cos
sin
cos
R
R
R
R
R
′
−
⎡
⎤ ⎡
⎤
⎡
⎤
=
⎢
⎥ ⎢
⎥
⎢
⎥
′
−
⎣
⎦
⎣
⎦ ⎣
⎦
θ
θ
θ
θ
θ
θ
θ
(7)
The determinant of the coefficient matrix in Equation (7) is
34
3
3
34
34
3
3
sin
cos
193.62 mm
cos
sin
−
=
= −
= −
R
DET
R
R
θ
θ
θ
θ
(8)

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This is the end of the preview.