PHYSICS EXPERTS ONLY
aroojch
lecture_13__a_mech_393.pptxMECH 393
Lecture 13
Advanced and Coriolis Acceleration Analysis
Agenda
Questions
Chapter 7 Kepler?
Working Model simulation?
Review acceleration
Consider pure rotation RAFA motion for acceleration of a point not at a pin
Introduce relative acceleration equation for a point on a complex motion link
Introduce coriolis acceleration
RAFA acceleration review
Single link
Change in velocity VECTOR is acceleration VECTOR
Change in direction of V
Change in magnitude of V
At early time blue position
At later time red position
Review RAFA acceleration Normal component
an (Normal acceleration) is directed from point of interest toward RAFA axis
an (Normal acceleration) has a magnitude of r * w2
an = r * w2
Closed loop vector equation
Negative vector
Review RAFA acceleration Tangential component
Change in magnitude of V
If red V vector longer than blue, w is increasing
V= r * w
a = d/dt w or
(wlater – wearlier)/time change
a
w
w
RAFA motion
Change in magnitude of V
at (tangential acceleration)
has a direction tangential, along same line of action as velocity vector
at (tangential acceleration)
has a magnitude of r * a
at = r * a
a
w
w
Review RAFA motion acceleration of another point
For some point on a RAFA link that is not the pin, the physics are the same as for a pin located on the link
at (tangential acceleration)
has a direction tangential, along same line of action as velocity vector
at (tangential acceleration)
has a magnitude of r * a
at = r * a
an (Normal acceleration) is directed from point of interest toward RAFA axis
an (Normal acceleration) has a magnitude of r * w2
an = r * w2
a,w
at
an
r
Relative acceleration applied to center of gravity of a link
aB = aA + aB/A
Apply to center of gravity point G3 on complex motion link 3
aB = aA + aB/A replace B with G3
aG3 = aA + aG3/A
Example problem statement
4 bar all revolute joints
Given w2, a2
Find a all links
a of any point any link
Particular points of interest G2, G3, G4
O2
A
B
O4
G3
G2
G4
Why are we interested in acceleration of center of gravity?
Model of each link for dynamic force analysis
S F = ma
Model translation of link
S T = Ia
Model rotation of each link
Single model works for RAFA, pure translation and complex motions
O2
A
B
O4
G3
G2
G4
Example
Process
1. do velocity analysis to get w3 and w4 from known w2
2. consider RAFA link 2 and find aA
3. consider RAFA link 4 and evaluate aB
4.write and apply relative acceleration equation
Example
Process
4.write and apply relative acceleration equation
aB = aA + aB/A
Expanded version
aBn + aBt = aAn + aAt + aB/An+ aB/At
Each part has rotational motion
Each part has normal and tangential components
Example
After step 4 we know w2 w3 w4 and a2 a3 a4
Find a of G2, G3, G4
Link 2 is simple RAFA
aG2 is given as aG2n + aG2t
aG2n = r w22 points G2 to O2
aG2t = r a2 points perpendicular to G2 O2,
Following sense of a2
Distance r is between G2 and center O2
A
B
O4
G3
O2
G2
G4
O2
G2
A
Example
Similarly for RAFA link 4
aG4 is given as aG4n + aG4t
aG4n = r w42 points G4 to O4
aG4t = r a4 points perpendicular to G4 O4,
Following sense of a4
Distance r is between G4 and center O4
O2
A
B
O4
G3
G2
G4
B
O4
G4
Example
DIFFERENT for complex motion link 3
aG3 is given by relative acceleration equation
aG3 = aA + aG3/A
aA is known from earlier step as aAn + aAt
aG3/A = aG3/An+ aG3/At
aG3/An = r w32 points G3 to A
aG3/At = r a3 points perpendicular to G3A,
Following sense of a3
Distance r is between G3 and A
O2
O4
A
B
G3
G2
G4
A
B
G3
aG3/A is rotational aspect of link 3 motion, thus it has normal and tangential components
Summary
Given design parameters
Link lengths, driver speed, driver acceleration
Find linear V of any point, angular w all links
Find linear a of any point, angular a all links
Find linear aGn of the center of gravity of each link
Next F = Ma, T = Ia
S F = m * aG for each link in motion
Example: Slider crank mechanism
Given
w2, a2
OR VB, aB
Find
V any point
w all links
a any point
a all links
a of cg all links
O2
A
B
4
3
2
1
1
Example steps
1. do velocity analysis to get w’s
(verify correct answer by alternate method?)
2. solve relative acceleration equation for this mechanism
aB = aA + aB/A
aB direction along slide, no n and t components, w = 0, a = 0, pure translation, no rotation of link 4
aA as before with n and t components
aB/A as before with n and t components
O2
A
B
Example steps
3. knowing aB , aA as normal and tangential components, and aB/A as normal and tangential components, calculate
a3 = aB/At / rB/A
Determine aG2 as n and t components for RAFA motion
aG3 by using relative acceleration equation
because link 3 is complex motion
aG3 = aA + aG3/A
aG3/A = aG3/An+ aG3/At
Combined
aG3 = aAn+ aAt + aG3/An+ aG3/At
O2
A
B
Coriolis acceleration
When a slide joint is present on a rotating link there is an additional component of acceleration called Coriolis acceleration
Consider first constant velocity of slide in slide joint
Does velocity vector change?
Yes direction changes due to rotation, even if magnitude is constant
Coriolis acceleration
Direction of coriolis is perpendicular to V slide
Magnitude of coriolis is 2 V w
Also V slide may not be constant, leading to a slide along same line of action as V slide
Still we have normal and tangential accelerations due to RAFA link that contains the slide
Result: 4 components of acceleration!
