### MECH 314 Instantaneous Points of Zero Velocity and Acceleration

```MECH 314
Dynamics of Mechanisms
March 2, 2011
Instantaneous Points of Zero Velocity and Acceleration
1
Instant Centre
The instant centre is the unique point C, on a moving planar rigid body, that is stationary at any given instant. As
we saw previously C is located at the intersection of two lines that are drawn respectively on two points, say, A and
B, on the body such that each line is normal to the velocity vector direction at A and B. In Fig. 1 the instantaneous
state of motion, of the rigid body shown there, is given by the velocity at A, vA and the angular velocity ω. Similarly,
A is considered to have an acceleration aA at this instant while the angular acceleration is α. To find C we can write
a
C
B
A [9:13.375:0]
=5s -2
(-2.37,5.3125)
v
A [10.625:4.75:0]
(-0.75305,2.40244)
A
(0,0)
=2s -1
ZAVBC0Ah
Figure 1: Instant Motion State of a Planar Rigid Body
vA + ω × rAC = 0
which produces the two simultaneous equations
vAx − ωrACy = 0, vAy + ωrACx = 0
or rACx = −2.375 and rACy = 5.3125dm/s.
2
Point of Instantaneous Zero Acceleration
The pair of corresponding conditions for a point B having no acceleration at this moment is



 
  
aAx
rABx
−αrABy
0
 aAy  − ω 2  rABy  +  αrABx  =  0 
0
0
0
0
Substituting given values for aA , ω and α and solving for rAB locates B at rABx = −0.75305dm and rABy =
2.40244dm.
2
2.1
Theorem
Arbitrary specification of instantaneous acceleration of two given points on a rigid body satisfies the compatibility
constraint that maintains the preservation of distance between points.
2.2
Proof by Example
Although this is not rigorous, a numerical example, involving the computation of the instantaneous zero acceleration
point C in the context of a numerical example, will be used to demonstrate the validity of the theorem statement
above. The example chosen is P4.6 on p.234 of . The problem statement is shown on the left of Fig. 2. Units have
been changed from feet and inches to proportionally equivalent dm. A pair of points A and B is separated by 20dm.
The acceleration vector at A, aA has a magnitude of 7200dm/s2 and points north east at 30◦ to the horizontal while
that at B, aB points north west at 60◦ to the horizontal. The task is to determine the angular velocity ω and the
angular acceleration α of the body at the instant specified. One proceeds by writing the relative acceleration of B
aA
aA
P4-613b
7200dm/s 2
A
30
1800dm/s 2
60
20dm
-a B
a A/B
aB
C
aB
-a A
A
B
B
a B/A
Figure 2: Instantaneous Acceleration State of a Planar Rigid Body
with respect to A or, conversely, that of A with respect to B.
aB/A = aB − aA = −ω 2 rAB + α × rAB →
aBx − aAx − ω 2 rABx + αrABy = 0
aBy − aAy − ω 2 rABy − αrABx = 0
√
−900 − 3600 3 + ω 2 = 0, 900 3 − 3600 − 20α = 0
√
or
2
aA/B = aA − aB = −ω rBA + α × rBA →
aAx − aBx − ω 2 rBAx + αrBAy = 0
aAy − aBy − ω 2 rBAy − αrBAx = 0
√
3600 3 + 900 − 20ω 2 = 0, 3600 − 900 3 + 20α = 0
√
The two lines with numerical values are obtained by substituting
√
√
aAx = 3600 3, aAy = 3600, aBx = −900, aBy = 900 3, rABx = 20, rABy = 0, rBAx = −20, rABy = 0
into the respective pairs of equations above. Both produce the following identical result.
√
√
ω 2 = 180 3 + 45(rdn/s)2 , α = 45 3 − 180rdn/s2
Note that in the problem statement ω is specified as CW, hence the negative square root of ω 2 is meant. There is
no way to determine the sense of ω in this context.
2.3
Zero Acceleration Point
Now the point C such that aC = 0 will be located. C is the instantaneous centre of zero acceleration so the
acceleration of A with respect to C and the acceleration of B with respect to C must define the original accelerations
3
of A and B as specified in order to validate the demonstration. Definitions aA and aB in terms of C are as follows.
aA − ω 2 rAC + α × rAC = 0
aB − ω 2 rBC + α × rBC = 0
2
aAx − ω rACx − αrACy = 0
aBx − ω 2 rBCx − αrBCy = 0
2
aBy − ω 2 rBCy√+ αrBCx = 0
√ aAy − ω rACy√+ αrACx = 0
(45 + 180 3)(xC − xA ) −900 −√
(45 + 180 3)(xC − xB )
3600 3 −√
−(45 3 − 180)(y
−
y
)
=
0
−(45
3 − 180)(y
A
√C
√
√C − yB ) = 0
3600 −√(45 + 180 3)(yC − yA ) 900 3 −√(45 + 180 3)(yC − yB )
+(45 3 − 180)(xC − xA ) = 0
+(45 3 − 180)(xC − xB ) = 0
Since aA and aB are given as are A(0, 0) and B(20, 0 one may compute C(xC , yC ) with either equation pair above
while noting that
rACx = xc − xA = xC , rACy = yC − yA = yC , rBCx = xC − xB = xC − 20, rBCy = yC − yB = yc
Using either equation pair that defines aA and the left, or that which defines aB on the right, and making the
appropriate substitutions, produces the same coordinates of C.
320 80
,
C
17 17
Reversing the procedure and substituting the newly establishesd values of xC and yC into the respective equation
pairs produces, canonically,
√
aAx − 3600 3 = 0
aBx + 900
√ =0
aAy − 3600 = 0 aBy − 900 3 = 0
thus completing the demonstration.
References
 J.J. Uicker, Jr., G.R. Pennock and J.E. Shigley (2011) Theory of Mechanisms and Machines, 4th ed., Oxford,
ISBN 9780-19-537123-9.
```