ROTATIONAL KINEMATICS

1. Any rotation is specified by giving two pieces of information:
a) The point about which the rotation occurs, i.e. the origin.
b) The angle of rotation is denoted by in the diagram.
2. The a
φ
rc length = radius angular displacement, or .
Here is measured in radians. The maximum value of is
2π radians, which corresponds to 360 degrees or one full revolution. From this it
s rφ
φ φ
× =
0
follows
that 1 radian 57.3 or 1 radian 0.159 revolution. Obviously, if = 2π, then 2 ,
which is the total circumference.
3. Suppose that there is a particle located at the tip of the radi
= = φ s= πr
1 2
us vector. Now we wish to
describe the rotational of this particle, i.e. describe its motion as goes around
the circle. So, suppose that the particle moves from angle to in t
kinematics
φ φ 2 1
2 1
2 1
0
ime . Then, the
is defined as, . Suppose that we look at
over a very short time. Then, lim is called the instantaneous angular
t
t t
average angular speed
t t t
d
t dt
φ φ φ
ω ω ω
φ φ
ω
Δ →
−
− Δ
= =
− Δ
Δ
= =
Δ
second
speed.
4. To familarize ourselves with the notion of angular speed, let us compute for a clock
second, minute and hour hands:
2 0.105 / ,
60
rad s
ω
π
ω = =
3
minute
4
hour
2 1.75 10 / ,
60 60
2 1.45 10 / .
60 60 12
5. Just as we defined accelaration for linear motion, we also define acceleration fo
rad s
rad s
π
ω
π
ω
−
−
= = ×
×
= = ×
× ×
2 1
2 1
2
0 2
r
circular motion:
(average angular speed)
Hence, lim becomes (angular acceleration). Let us
see wha
t
t t t
d dd d
t dt dt dt dt
ω ω ω
α
ω ω φ φ
α α
Δ →
− Δ
≡ =
− Δ
Δ
= = = =
Δ
t this means for the speed with which a particle goes around. Now use .
Differentiate with respect to time : . The rate of change of arc length is clearly
s r
tds rd s
dt dt
φ
φ
=
=
A B C
A r C r
what we should call the circular speed, v. So v . Since r is held fixed, it follows that
v . Now define v. Obviously, . Here stands for tangential, i.e.
in the direc
T T
r
d rd a d a r T
dt dt dt
ω
ω
α
=
= = =
tion of increasing .
6. Compare the formulae for constant linear and angular accelerations:
LINEAR ANGULAR
s
( )
0 0
2 2
0 0 0 0
2 2
0 0
v v a
v1a 1
2 2
v v 2a
t t
x x t t t t
x x
ω ω α
φ φ ω α
= + = +
= + + = + +
= + − 2 2 ( )
0 0 2
Why are they almost identical even though they describe two totally different physical
situations. Answer: because the mathematics is identic
ω =ω + α φ −φ
2
al!
7. The angular speed of a car engine is increased from 1170 rev/min to 2880 rev/min in
12.6 s. a)Find the average angular acceleration in rev/min . (b) How many revolutions
does t
2 2
he engine make during this time?
SOLUTION: this is a straightforward application of the formulae in point 5 above.
8140rev/min , 1 425rev.
2
8. W
f i
it t
t
ω ω
α φ ω α
−
= = = + =
heel A of radius 10.0 cm is coupled by a chain B to wheel C of radius 25.0 cm.
Wheel A increases its angular speed from rest at a uniform rate of 1.60 rad/s2. Determine
the time for
A C r= r=
wheel C to reach a rotational speed of 100 rev/min.
SOLUTION: Obviously every part of the chain moves with the same speed and so
v v . Hence . From the definition of acceleration,
0 . From this, 16.4
C C
A C A A C C A
A
A A C C
A
r r r
r
t r
t r
ω
ω ω ω
ω ω ω
α
α α
= = ⇒ =
−
= = = = .
9. Imagine a disc going around. All particles on the disc will have same ' ' and ' '
but different 'v' and 'a' . Clearly' ' and ' ' are simpler choices !!
s
ω α
ω α
2 v
1 v
10. Now consider a particle going around a circle at constant speed. You might think that
constant speed means no acceleration. Bu this is wrong! It is changing its direction and
accelerating. This is called "centripetal acceleration", meaning acceleration directed
towards the centre of the circle. Look at the figure below:
1 2
2 2
0
2
Note that the distance between points and is v . Similarly,
v v v v v . More generally, lim v v . In vector form,
/ v
a v . The negative sign in
t
R
P P r t r
a a
t r r t r
r
r
θ
θ
θ
θ Δ →
Δ = Δ ≈
Δ Δ
Δ ≈ ⇒ = ≈ = = =
Δ Δ
= −
􀁇 􀀃 dicates that the acceleration is towards the centre.
12. Vector Cross Products: The vector crossproduct of two vectors is defined as:
A×B=ABsinθ nˆ where nˆ is a unit vector that is perpendicul
􀁇 􀁇
ar to both and .
Apply this definition to unit vectors in 3-dimensions: , , .
A B
i j k k i j j k i
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧
× = × = × =
􀁇 􀁇
( ) ( ) ( )
13. Some key properties of the crossproduct:
0
ˆ ˆ ˆ
( ) ˆ ( x y z y z z y
x y z
A B B A
A A
A B C A B A C
i j k
A B A A A AB AB i
B B B
• × = − ×
• × =
• + × = × + ×
• × = = − +
􀁇 􀁇 􀁇 􀁇
􀁇 􀁇
􀁇 􀁇 􀁇 􀁇 􀁇 􀁇 􀁇
􀁇 􀁇 )ˆ ( )ˆ
14. The cross product is only definable in 3 dimensions and has no meaning in 2-d. This is
unlike the dot product which as a meaning in any number of dimensions.
z x x z x y y x A B −A B j+ A B −A B k
1 P
2 P
2 v
C r
r
θ
1 v
2 1 v −v
1 m
2 m
1 r􀁇
2 r􀁇