EGR 232 Engineering Dynamics
Lecture 26: Rotating Reference Frames
Fall 2013
Today:
Homework Questions
Read Chap 15 Sections 10 and 11
Homework: Chap. 15, Problems 150, 156, 158, 177
Lecture Outcomes:
Following today's lecture you should be able to
-- understand what a rotating reference frame is.
-- be able to describe the relative and absolute velocities of a body moving on a
rotating reference frame.
-- be able to describe the relative and absolute accelerations of a body moving on a
rotating reference frame.
-- be familiar with the term Coriolis acceleration
Rotating Reference Frames:
So far we have been describing velocities and accelerations of points firmly attached to
the rigid bodies that are translating and rotation through space. The particles we have
been describing are fixed particles in the rigid body.
Another model exists where you may have a point given with respect to a rigid body, but
the point or particle is not fixed on the rigid body. As the rigid body undergoes rotation,
the point may move along some prescribed path on the rigid body and can be described as
if the rigid body did not rotate. The particle will have its own frame of reference attached
to the rigid body. However, because the frame of reference changes as the rigid body
rotates, there will be an effect on the velocity and acceleration of the particle.
The absolute velocity of a particle P moving along a reference frame which is also
rotating my be given as
v P  ( r )OXY  Ω  r  ( r )Oxy
or
v P  v P'  v P / Frame
where vP is the absolute velocity of Particle P
vP' is the velocity of point P' of moving Frame coinciding with point P
vP/Frame is the velocity of point P relative to the moving frame F.
The acceleration of this same particle moving along a reference frame which is also
rotating is given by
aP  Ω  r  Ω  ( Ω  r )  ( r )Oxy  2Ω  ( r )Oxy
or
aP  aP'  aP / Frame  aCoriolis
where
where aP is the absolute acceleration of Particle P
aP' is the acceleration of a point on the moving frame at the same
instantaneous point as point P.
aP/Frame is the acceleration of point P relative to the moving frame F.
aCoriolis is the Coriolis acceleration of point P.
Example 1:
Pin P slides in a circular slot cut in the plate shown at a constant relative speed u = 180 mm/s.
Knowing that t the instant shown the plate rotates clockwise about A at the constant rate
ω = 6 rad/s, determine the acceleration of pin if it is located a) at point A b) point B c) point C.
Solution:
Plate:
ˆ rad / s
Ω  6 k
vA  0
Pin at A:
r  rA / A  0
v P / Frame  u ˆi  018
. ˆi m / s
Velocity:
v P  v P'  v P / Frame  Ω  r  ( r )Oxy
ˆ  0 )  ( 018
v  ( 6 k
. ˆi )  018
. ˆi m / s
P
Acceleration:
aP  aP'  aP / Frame  aCoriolis  Ω  r  Ω  ( Ω  r )  ( r )Oxy  2Ω  ( r )Oxy
aP'  Ω  r  Ω  ( Ω  r )
ˆ  ( 6 k
ˆ  0 ))  0
 0  0  ( 6 k
aP / Frame  ( r )Oxy
2
018
.
u
ˆj  01620
ˆj m / s2
.
 at ˆi  an ˆj  0 ˆi  ˆj  0 ˆi 

02
.
2
aCoriolis  2Ω  ( r )Oxy
ˆ )  ( 018
 2( 6 k
. ˆi )  216
. ˆj m / s2
so
aP  aP'  aP / Frame  aCoriolis
ˆj  216
ˆj m / s2
 0  0162
.
. ˆj  2322
.
Pin at B:
r  rB / A  02
. ˆi  02
. ˆj m
v P / Frame  ( r )Oxy  u ˆj  018
. ˆj m / s
Velocity:
v P  v P'  v P / Frame  Ω  r  ( r )Oxy
ˆ )  ( 02
v P  ( 6 k
. ˆi  02
. ˆj )  018
. ˆj
 12
. ˆj  12
. ˆi  018
. ˆj  12
. ˆi  138
. ˆj m / s
Acceleration:
aP  aP'  aP / Frame  aCoriolis  Ω  r  Ω  ( Ω  r )  ( r )Oxy  2Ω  ( r )Oxy
aP'  Ω  r  Ω  ( Ω  r )
ˆ  ( 6 k
ˆ  ( 02
 0  ( 02
. ˆi  02
. ˆj )  ( 6 k
. ˆi  02
. ˆj ))
 72
. ˆi  72
. ˆj m / s
aP / Frame  ( r )Oxy
2
2
u
018
.
ˆi  0162
ˆi m / s2
 at ˆj  anˆi  0ˆj  ˆi 
.

02
.
aCoriolis  2Ω  ( r )Oxy
ˆ )  ( 018
 2( 6 k
. ˆj )  216
. ˆi m / s2
so
aP  aP'  aP / Frame  aCoriolis
ˆi  216
ˆi  72
 ( 72
. ˆi  72
. ˆj )  0162
.
. ˆi  9522
.
. ˆj m / s2
Pin at C:
r  rC / A  0.4 ˆj m
v P / Frame  ( r )Oxy  u ˆi  018
. ˆi m / s
Velocity:
v P  v P'  v P / Frame  Ω  r  ( r )Oxy
ˆ )  ( 0.4 ˆj )  018
v P  ( 6 k
. ˆi
 2.4 ˆi  018
. ˆi  258
. ˆi m / s
Acceleration:
aP  aP'  aP / Frame  aCoriolis  Ω  r  Ω  ( Ω  r )  ( r )Oxy  2Ω  ( r )Oxy
aP'  Ω  r  Ω  ( Ω  r )
ˆ  ( 6 k
ˆ  0.4 ˆj )
 0  ( 0.4 ˆj )  ( 6 k
 14.4 ˆj m / s
aP / Frame  ( r )Oxy
2
u
 at ˆi  anˆj  0ˆi  ˆj

018
. 2ˆ
ˆj m / s2

j   0162
.
02
.
aCoriolis  2Ω  ( r )Oxy
ˆ )  ( 018
 2( 6 k
. ˆi )  216
. ˆj m / s2
so
aP  aP'  aP / Frame  aCoriolis
ˆj  216
ˆj m / s2
 ( 14.4ˆj )  0162
.
. ˆj  16722
.
Example 2:
Pin P is attached to the collar shown; the motion of the pin is guided by a slot cut in bar
BD and by the collar that slides on rod AE. Bar BD rotates with a constant angular
velocity of 5 rad/s counterclockwise and the distance from B to P decreases at a constant
rate of 3.5 m/s. Determine at the instant shown a) the angular acceleration of rod AE, b)
the relative acceleration of pin P with respect to rod AE.
-----------------------------------------------------------------------------------------------------------Solution:
Note: Pin P moves with respect to the rotating
frame BD and with respect to the rotating frame AE.
Given:
BD  5 rad / s
constant
vP / B  35
. m / s constant
and
Displacement:
rP / A  0 ˆi  0.4 ˆj m
r  0.4 ˆi  0.4 ˆj m
P/ B
Velocities:
ˆ rad / s
ΩBD  5 k
ˆ
Ω  k
AE
AE
v P / FrameBD  vP / B uˆ BD
v P / FrameAE
ˆi  0707
ˆj )  2.475 ˆi  2.475ˆj m / s
 35
. ( 0707
.
.
v
uˆ
 v ˆj
P/ A
AE
P/ A
then:
v P  v P' onBD  v P / FrameBD  Ω BE  rP / B  ( r )Oxy
ˆ )  ( 0.4ˆi  0.4 ˆj )  ( 2.475 ˆi  2.475ˆj )
 (5 k
 2 ˆj  2ˆi  2.475 ˆi  2.475ˆj   4.475ˆi  0.475ˆj m / s
v P  v P' onAE  v P / FrameAE  Ω AE  rP / A  ( r )Oxy
ˆ )  ( 0.4 ˆj )  v ˆj
 (  AE k
P/ A
 0.4  AE ˆi  vP / A ˆj
Let the two equations equal each other:
( 4.475ˆi  0.475ˆj ) m / s  ( 0.4m )  AE ˆi  vP / A ˆj
Separating out the i and j components:
i:
4.475m / s  ( 0.4m )  AE
j:
 AE  1119
. rad / s
0.475 m / s  vP / A
vP / A  0.475 m / s
therefore:
v P  0.4  AE ˆi  vP / A ˆj  0.4( 1119
. )ˆi  0.475 ˆj
 4.476ˆi  0.475 ˆj m / s
Acceleration for Frame BD:
 BD  0
aP / FrameBD  0
aP  aP' onBD  aP / FrameBD  aCoriolis  Ω  r  Ω  ( Ω  r )  ( r )Oxy  2Ω  ( r )Oxy
aP'_ BD  ΩBD  rP / B  ΩBD  ( ΩBD  rP / B )
ˆ ( 5 k
ˆ  ( 0.4 ˆi  0.4ˆj ))
 0  ( 0.4 ˆi  0.4ˆj )  ( 5 k
 10ˆi  10ˆj m / s2
aP / FrameBD  ( r )P / FrameBD  0
aCoriolis  2ΩBD  ( r )P / FrameBD
ˆ )  ( 2.475 ˆi  2.475ˆj )
 2( 5 k
 2475
. ˆj  2475
. ˆi m / s2
so
aP  aP' onBD  aP / FrameBD  aCoriolis
 ( 10ˆi 10ˆj )  0  ( 2475
. ˆj  2475
. ˆi )  1475
. ˆi  3475
. ˆj m / s2
Acceleration of Frame AE
ˆ
 AE   AEk
ˆ AE  aP / FrameAE ˆj
aP / FrameAE  aP / FrameAE u
aP  aP' onAE  aP / FrameAE  aCoriolis  Ω  r  Ω  ( Ω  r )  ( r )Oxy  2Ω  ( r )Oxy
aP' onAE  Ω AE  rP / A  Ω AE  ( Ω AE  rP / A )
ˆ  ( 0.4 ˆj )  ( 1119
ˆ  ( 1119
ˆ  0.4ˆj )
  AEk
. k
. k
 0.4 AEˆi  50ˆj m / s2
aP / FrameAE  ( r )P / FrameAE  aP / FrameAE ˆj
aCoriolis  2Ω AE  ( r )P / FrameAE
ˆ )  ( 0.475 ˆj )  1063
 2( 1119
. k
. ˆi m / s2
so
aP  aP' onAE  aP / FrameAE  aCoriolis
 ( 0.4 AEˆi  50ˆj )  aP / FrameAE ˆj  1063
. ˆi
 ( 0.4  1063
. ) ˆi  ( a
 50 ) ˆj
AE
P / FrameAE
Next equate the two acceleration equations:
( 0.4 AE  1063
. ) ˆi  ( aP / FrameAE  50 ) ˆj  1475
. ˆi  3475
. ˆj
Extract the i and j component equations:
i:
0.4 AE  1063
.  1475
.
1475
.  10.63
 AE 
0.4
 AE  103
. rad / s2
j:
aP / FrameAE  50   3475
.
aP / FrameAE  50  3475
.
aP / FrameAE  1525
. m / s2
EGR 232 Dynamics
HW 26
Fall 2013
Chap. 15, Problem 150,
Two rotating rods are connected by slider block P. The rod attached at A rotates with a
constant angular velocity ωA. For the given data, determine for the position shown a) the
angular velocity of the rod attached at B, b) the relative velocity of slider block P with
respect to the rod on which it slides. Let b = 8 in and ωA= 6 rad/s
EGR 232 Dynamics
HW 26
Fall 2013
Chap. 15, Problem 156
Two rods AE and BD pass through holes drilled into a hexagonal block. The holes are
drilled in different planes so that the rods will not touch each other. Knowing that at the
instant considered rod AE rotates counterclockwise with a constant angular velocity, ω,
determine, for the given data, the relative velocity of the block with respect to each rod
for a) θ = 90 deg. b) θ = 60 deg.
EGR 232 Dynamics
HW 26
Fall 2013
Chap. 15, Problem 158
Four pins slide in four separate slots cut in a circular plate as shown. When the plate is at
rest each pin has a velocity directed as shown and of the same constant magnitude u. If
each pin maintains the same velocity relative to the plate when the plate rotates about O
with a constant counterclockwise angular velocity ω determine the acceleration of each
pin.
EGR 232 Dynamics
HW 26
Fall 2013
Chap. 15, Problem 177
At the instant shown bar BC has an angular velocity of 3 rad/s and an angular
acceleration of 2 rad/s2, both counter clockwise, determine the angular acceleration of
the plate.