Dynamics
15. Kinematics of Rigid Bodies
Chanhwa Lee
Department of Intelligent Mechatronics Engineering
Sejong University
Contents
• Translation
• Rotation About a Fixed Axis
• General Plane Motion
• Velocity & Acceleration
• Instantaneous Center of Rotation
• Motion w.r.t a Rotating Frame
• Rate of Change w.r.t. a Rotating Frame
• Plane Motion Relative to a Rotating Frame
Rigid Body and its Motion
• Rigid Body
- A solid body in which deformation is zero or so small
- The distance between any two given points on a rigid body
remains constant in time regardless of external forces or
moments exerted on it
- Kinematics of rigid bodies: relations between time and the
positions, velocities, and accelerations of the particles forming
a rigid body.
Rigid Body and its Motion
• Classification of rigid body motions
- translation
 rectilinear translation
 curvilinear translation
- rotation about a fixed axis
- general plane motion
- motion about a fixed point
- general motion
15.1 Translation & Fixed-Axis Rotation
• Translation
15.1 Translation & Fixed-Axis Rotation
• Translation
- For any two particles in the body,
  
rB= rA + rB A
- Differentiating with respect to time,
  

rB =rA + rB A =rA


vB = v A
  


rB =
rA + rB A =
rA


aB = a A
- All particles have the same velocity
and acceleration.
15.1 Translation & Fixed-Axis Rotation
• Rotation About a Fixed Axis : Velocity
- Consider rotation of rigid body about a
fixed axis AA’


- Velocity vector v = dr dt of the particle P
is tangent to the path with magnitude
v = ds dt
∆
=
s
=
v
θ ( r sin ϕ ) ∆θ
( BP ) ∆=
ds
∆θ
= lim ( r sin ϕ ) = rθ sin ϕ
dt ∆t →0
∆t
- The same result is obtained from

 
dr

v=
= ω×r
dt
15.1 Translation & Fixed-Axis Rotation
• Rotation About a Fixed Axis : Velocity
- Consider rotation of rigid body about a
fixed axis AA’


- Velocity vector v = dr dt of the particle P
is tangent to the path with magnitude
v = ds dt
∆
=
s
=
v
θ ( r sin ϕ ) ∆θ
( BP ) ∆=
ds
∆θ
= lim ( r sin ϕ ) = rθ sin ϕ
dt ∆t →0
∆t
- The same result is obtained from

 dr  
v=
= ω×r
dt




=
ω ω=
k θ=
k angular velocity
15.1 Translation & Fixed-Axis Rotation
• Rotation About a Fixed Axis : Acceleration
- Differentiating the velocity,

 dv d  
a =
=
(ω × r )
dt dt


d ω   dr
=
×r +ω×
dt
dt

dω   
=
×r +ω×v
dt

- d ω= α= angular acceleration,
dt





= α=
k ω=
k θk
- Acceleration is combination of two vector
normal component
tangential component
15.1 Translation & Fixed-Axis Rotation
• Rotation About a Fixed Axis : Representative Slab
- Velocity of any point P of the slab,
 
  
v = ω × r = ωk × r
v = rω
- Acceleration of any point P of the slab,
     
a = α × r + ω ×(ω × r )
 

= α k × r − ω 2r
- Resolving the acceleration into tangential
and normal components,
 

at =
αk × r
a t=
rα


an =
an =
rω 2
−ω 2 r
15.1 Translation & Fixed-Axis Rotation
• Rotation About a Fixed Axis
- Uniform Rotation
θ= θ 0 + ω t
- Uniformly Accelerated Rotation
ω
= ω0 + α t
θ =θ 0 + ω0t + 12 α t 2
ω2 =
ω02 + 2α (θ − θ 0 )
15.1 Translation & Fixed-Axis Rotation
Ex 15.3)
(a) 𝜃𝜃 of the pulley in 2 s (when 𝜃𝜃0 = 0)
(b) 𝑣𝑣 and Δ𝑦𝑦 of B after 2 s
(c) 𝑎𝑎 of D at 𝑡𝑡=0
- Cable C has a constant
acceleration of 225mm/s2 and
an initial velocity of 300mm/s,
both directed to the right
Review
• Translation
Prob 15.CQ1) A rectangular plate swings from arms of equal length
as shown below. What is the magnitude of the angular velocity of the
plate?
a. 0 rad/s
b. 1 rad/s
c. 2 rad/s
d. 3 rad/s
e. Need to know the location of the center of gravity
15.2 General Plane Motion : Velocity
• Analyzing General Plane Motion
- General plane motion = translation + rotation
15.2 General Plane Motion : Velocity
• Absolute and Relative Velocity in Plane Motion
- General plane motion = translation + rotation

 
v=
v A + vB A
B
 

vB A =
ω k × rB A
vB A =
rω
 


vB =v A + ω k × rB A
15.2 General Plane Motion : Velocity
• Absolute and Relative Velocity in Plane Motion
- General plane motion = translation + rotation
15.2 General Plane Motion : Velocity
• Absolute and Relative Velocity in Plane Motion
- General plane motion = translation + rotation
15.3 Instantaneous Center of Rotation
• Instantaneous Center of Rotation
15.3 Instantaneous Center of Rotation
• Instantaneous Center of Rotation
15.3 Instantaneous Center of Rotation
• Instantaneous Center of Rotation
15.4 General Plane Motion : Acceleration
• Absolute and Relative Acceleration in Plane Motion
- General plane motion = translation + rotation



aB = a A + aB A
 

(a B A )t = α k × rB A
(aB A ) = −ω 2 rB A
n
(a B A )t = rα
(a B A )n = rω 2
15.4 General Plane Motion : Acceleration
• Absolute and Relative Acceleration in Plane Motion
- General plane motion = translation + rotation
15.4 General Plane Motion
Ex 15.7)
Use the absolute and relative velocity
(a) 𝜔𝜔 of BD
(b) 𝑣𝑣 of P
- AB has a constant clockwise
angular velocity of 2000rpm.
15.4 General Plane Motion
Ex 15.10)
Use the instantaneous center of rotation
(a) 𝜔𝜔 of BD
(b) 𝑣𝑣 of P
- AB has a constant clockwise
angular velocity of 2000rpm.
15.4 General Plane Motion
Ex 15.15)
Use the absolute and relative acceleration
(a) 𝛼𝛼 of BD
(b) 𝑎𝑎 of P
- AB has a constant clockwise
angular velocity of 2000rpm.
Review
• General Plane Motion : Velocity
Prob 15.CQ3) The ball rolls without slipping on the fixed surface as
shown. What is the direction of the velocity of Point A?
a. →
b. ↗
c. ↑
d. ↓
e. ↘
Review
• Instantaneous Center of Rotation
Prob 15.CQ5) The disk rolls without sliding on the fixed horizontal
surface. At the instant shown, the instantaneous center of zero velocity
for rod AB would be located in which region?
a. region 1
b. region 2
c. region 3
d. region 4
e. region 5
f. region 6
15.5 Analyzing Motion w.r.t a Rotating Frame
• Rate of Change with respect to a Rotating Frame
- With respect to the rotating 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 frame,




Q = Qx i + Q y j + Qz k







Q Oxyz = Qx i + Q y j + Qz k
()
- With respect to the fixed 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 frame,
()










Q OXYZ = Qx i + Q y j + Qz k + Qx i + Q y j + Qz k
- Frame 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 is fixed
- Frame 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 rotates
about fixed axis OA
with angular velocity Ω.
()







Qx i + Q y j + Qz k = Q Oxyz = rate of change
with respect to rotating frame.
15.5 Analyzing Motion w.r.t a Rotating Frame
• Rate of Change with respect to a Rotating Frame
()










Q OXYZ = Qx i + Q y j + Qz k + Qx i + Q y j + Qz k
()







Qx i + Q y j + Qz k = Q Oxyz
- Frame 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 is fixed
- Frame 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 rotates
about fixed axis OA
with angular velocity.



Qx i + Q y j + Qz k
- With respect to the rotating 𝑂𝑂𝑋𝑋𝑋𝑋𝑋𝑋 frame
()
()


 
Q OXYZ = Q Oxyz + Ω × Q
15.5 Analyzing Motion w.r.t a Rotating Frame
• Plane Motion Relative to a Rotating Frame : Velocity
- Frame 𝑂𝑂𝑂𝑂𝑂𝑂 is fixed
- Frame 𝑂𝑂𝑂𝑂𝑂𝑂 rotates about 𝑍𝑍 axis with angular
velocity Ω
- Imagine a rigid slab attached to the rotating
frame 𝑂𝑂𝑂𝑂𝑂𝑂 or F for short


v P F = (r )Oxy = velocity of 𝑃𝑃 along its path
on the slab
- Let 𝑃𝑃𝑃 be a point on the slab which
corresponds instantaneously to position of
particle 𝑃𝑃

v P ' = absolute velocity of point 𝑃𝑃𝑃 on the slab
15.5 Analyzing Motion w.r.t a Rotating Frame
• Plane Motion Relative to a Rotating Frame : Velocity
- Absolute velocity of the particle 𝑃𝑃 is,
 


v P = (r )OXY = Ω × r + (r )Oxy



v P = v P′ + v P F
15.5 Analyzing Motion w.r.t a Rotating Frame
• Plane Motion Relative to a Rotating Frame : Acceleration
- Absolute acceleration of the particle 𝑃𝑃 is,
P
[
   
d 
= Ω × r + Ω × (r )OXY + (r )Oxy
dt
]
  

(r )OXY = Ω × r + (r )Oxy
[
]

d 
(r )Oxy = (r)Oxy + Ω × (r )Oxy
dt
    
 


a P = Ω × r + Ω × (Ω × r ) + 2Ω × (r )Oxy + (r)Oxy
 

v P = Ω × r + (r )Oxy


= v P′ + v P F
15.5 Analyzing Motion w.r.t a Rotating Frame
• Plane Motion Relative to a Rotating Frame : Acceleration
- Utilizing the conceptual point 𝑃𝑃𝑃 on the slab,
    

a P′ = Ω × r + Ω × (Ω × r )


a P F = (r)
Oxy
- Absolute acceleration for 𝑃𝑃 becomes,
    
 


aP = Ω
× r + Ω × (Ω × r ) + 2Ω × (r )Oxy + (r)Oxy
 



a P = a P′ + a P F + 2Ω × (r )Oxy



= a P′ + a P F + ac
 

v P = Ω × r + (r )Oxy


= v P′ + v P F
()
 
 


- Coriolis acceleration ac = 2Ω × (r )Oxy = 2Ω × v P F
15.5 Analyzing Motion w.r.t a Rotating Frame
• Coriolis acceleration
- Absolute acceleration of the collar 𝑃𝑃 is,




a P = a A + a P F + ac
    

a A = Ω × r + Ω × (Ω × r )
a A = rω 2


a P F = (r)Oxy = 0
 

a c = 2Ω × v P F
ac = 2ωu
15.5 Analyzing Motion w.r.t a Rotating Frame
• Coriolis acceleration
- Change in velocity over Dt is represented by
the sum of three vectors

∆v = RR ′ + TT ′′ + T ′′T ′
-
at t ,
at t + ∆t ,
 

v = vA + u
 

v ′ = v A′ + u ′
TT ′′ is due to change in direction of the
velocity of point A on the rod,
TT ′′
∆θ
= lim v A
= rωω = rω 2 = a A
∆t
∆t →0 ∆t
∆t →0
lim
- RR ′ and T ′′T ′ result from combined effects of
relative motion of 𝑃𝑃 and rotation of the rod
 RR ′ T ′′T ′ 
∆r 
∆θ
 = lim  u
+ω 
lim 
+
∆t 
∆t  ∆t →0 ∆t
∆t →0 ∆t
= uω + ωu = 2ωu
15.5 Analyzing Motion w.r.t a Rotating Frame
• Radial and Transverse Components (in Sec.11.5)


r = re r




=
v r er + rθ eθ
(
)
(
)


2 



a = 
r − rθ er + rθ + 2rθ eθ
 


v P = (r )OXY = Ω × r + (r )Oxy
    
 


a P = Ω × r + Ω × (Ω × r ) + 2Ω × (r )Oxy + (r)Oxy
15.5 Analyzing Motion w.r.t a Rotating Frame
Ex 15.19)
(a) 𝜔𝜔𝑆𝑆 of the disk S
(b) 𝑣𝑣𝑃𝑃 relative to the disk S
- Disk D with constant
counterclockwise angular velocity
𝜔𝜔𝐷𝐷 = 10 rad/s.
- At the instant when φ = 150o,
15.5 Analyzing Motion w.r.t a Rotating Frame
Ex 15.19)
(a) 𝜔𝜔𝑆𝑆 of the disk S
(b) 𝑣𝑣𝑃𝑃 relative to the disk S
- Disk D with constant
counterclockwise angular velocity
𝜔𝜔𝐷𝐷 = 10 rad/s.
- At the instant when φ = 150o,
15.5 Analyzing Motion w.r.t a Rotating Frame
Ex 15.20)
(a) 𝛼𝛼𝑆𝑆 of the disk S
- Disk D with constant
counterclockwise angular velocity
𝜔𝜔𝐷𝐷 = 10 rad/s.
- At the instant when φ = 150o,
15.5 Analyzing Motion w.r.t a Rotating Frame
Ex 15.20)
(a) 𝛼𝛼𝑆𝑆 of the disk S
- Disk D with constant
counterclockwise angular velocity
𝜔𝜔𝐷𝐷 = 10 rad/s.
- At the instant when φ = 150o,
Review
• Plane Motion Relative to a Rotating Frame : Acceleration
Prob 15.CQ8) A person walks radially inward on a platform that is
rotating counterclockwise about its center. Knowing that the platform has
a constant angular velocity 𝜔𝜔 and the person walks with a constant
speed 𝑢𝑢 relative to the platform, what is the direction of the acceleration
of the person at the instant shown?
a. Negative x
b. Negative y
c. Negative x and positive y
d. Positive x and positive y
e. Negative x and negative y
THANK YOU!
QUESTIONS?
(chlee@sejong.ac.kr)