1.

DYNAMICS OF RIGID BODIES

Measuring angles in radian

Define the value of an angle 𝜃 in radian

!
as 𝜃= ,
"
or arc length 𝑠 = 𝑟𝜃

a pure number, without dimension

independent of radius r of the circle
one complete circle
#$"
𝜃= = 2𝜋 in radian ↔ 360°
"
𝜋 in radian ↔ 180°
𝜋 ⁄2 in radian ↔ 90°
Consider a rigid body rotating about a fixed axis

angular displacement: ∆𝜃 = 𝜃! − 𝜃"

angular velocity:
∆𝜃 ∆$→& 𝑑𝜃
𝜔=
∆𝑡 𝑑𝑡
(average) (instantaneous)

angular acceleration:
𝑑𝜔 𝑑! 𝜃
𝛼= = !
𝑑𝑡 𝑑𝑡

Convention: 𝜃 measured from x axis in counterclockwise direction

Convention: θ measured from x axis in counterclockwise direction
Angular velocity is a vector, direction defined by the right hand rule

direction of 𝝎
represents sense of
rotation
 𝒅𝝎
Angular acceleration is defined as 𝜶 = 𝒅𝒕

if rotation axis is fixed, 𝜶 along the direction of 𝝎

Rotation Rotation slowing

speeding up, down, 𝜶 and 𝝎
𝜶 and 𝝎 in the in the opposite
same direction direction
 Question
• The figure shows a graph of 𝜔 and 𝛼 versus
time. During which time intervals is the
rotation speeding up?
(i) 0 < t < 2 s; (ii) 2 s < t < 4 s; (iii) 4 s < t < 6 s.
Rotation with constant angular acceleration
Example
A Blu-ray disc is slowing down to a stop with constant angular acceleration
𝛼 = −10.0 rad/s2 . At 𝑡 = 0, 𝜔! = 27.5 rad/s, and a line PQ marked on the disc surface
is along the x axis.

angular velocity at 𝑡 = 0.300 s:

𝜔 = 𝜔% + 𝛼𝑡
= 27.5 rad/s + −10.0 rad⁄s 2 0.300 s
= 24.5 rad/s
Suppose 𝜃 is the angular position of PQ at
𝑡 = 0.300 s
𝜃 = 𝜔% 𝑡 + &#𝛼𝑡 # = 7.80 rad
360°
= 7.8 rad = 447° = 87°
2𝜋 rad

What are the directions of 𝝎 and 𝜶?

 Question
• In the above example, suppose the initial angular velocity is
doubled to 2𝜔& , and the angular acceleration (deceleration) is
also doubled to 2𝛼, it will take (more / less / the same
amount of) time for the disc to come to a stop compared to
the original problem.
Q9.2

A DVD is initially at rest so that the line PQ on the

disc’s surface is along the +x-axis. The disc begins to
turn with a constant az = 5.0 rad/s2.
At t = 0.40 s, what is the angle between the line PQ
and the +x-axis?

A. 0.40 rad
B. 0.80 rad
C. 1.0 rad
D. 2.0 rad
A9.2

A DVD is initially at rest so that the line PQ on the

disc’s surface is along the +x-axis. The disc begins
to turn with a constant az = 5.0 rad/s2.
At t = 0.40 s, what is the angle between the line PQ
and the +x-axis?

A. 0.40 rad
B. 0.80 rad
C. 1.0 rad
D. 2.0 rad
Rigid body rotation
In time ∆𝑡, angular displacement is ∆𝜃,
tangential displacement (arc length) is
∆𝑠 = 𝑟∆𝜃

∴ tangential speed

∆𝑠 ∆𝜃 𝑑𝜃
𝑣= =𝑟 →𝑟 = 𝑟𝜔
∆𝑡 ∆𝑡 𝑑𝑡

Velocity of point P, 𝒗, is tangential and

has magnitude 𝑣 = 𝑟𝜔

𝑑𝑣 𝑑𝜔
tangential acceleration 𝑎tan = =𝑟 = 𝑟𝛼
𝑑𝑡 𝑑𝑡
radial acceleration
𝑣#
(from circular motion 𝑎rad = = 𝜔# 𝑟
𝑟
of P)
Example

An athlete whirls a discus in a circle of radius 80.0 cm. At some instant 𝜔 =

10.0 rad/s, and 𝛼 = 50.0 rad/s2. Then
𝑎tan = 𝑟𝛼 = 0.800 m 50.0 rad⁄s" = 40.0 m/s2
𝑎rad = 𝜔" 𝑟 = 10.0 rad⁄s " 0.800 m = 80.0 m/s"
Magnitude of the linear acceleration is
" "
𝑎= 𝑎tan + 𝑎rad = 89.4 m/s"
2. Moment of inertia
Rotational kinetic energy of a rigid body

Consider a rigid body as a collection of particles, the kinetic energy due to rotation is
1
𝐾= I &#𝑚' 𝑣'# = I &#𝑚' 𝑟'# 𝜔 # = I 𝑚' 𝑟'# 𝜔 #
2

moment of inertia I, analogous

to mass in rectilinear motion

𝐾 = "!𝐼𝜔!
c.f. in rectilinear motion,
𝐼 = / 𝑚? 𝑟?! 𝐾 = !"𝑚𝑣 "

I depends on distribution of mass, and therefore on the location of the rotation axis.
 Question
• A pool cue is a wooden rod with a uniform composition and
tapered with a larger diameter at one end than at the other
end. Does it have a larger moment of inertia
 for an axis through the thicker end of the rod and
perpendicular to the length of the rod, or
‚ for an axis through the thinner end of the rod and
perpendicular to the length of the rod?
Q9.5

You want to double the radius of a rotating solid sphere while

keeping its kinetic energy constant. (The mass does not change.) To
do this, the final angular velocity of the sphere must be

A. 4 times its initial value.

B. twice its initial value.
C. the same as its initial value.
D. 1/2 of its initial value.
E. 1/4 of its initial value.
Q9.5

You want to double the radius of a rotating solid sphere while

keeping its kinetic energy constant. (The mass does not change.) To
do this, the final angular velocity of the sphere must be

A. 4 times its initial value.

B. twice its initial value.
C. the same as its initial value.
D. 1/2 of its initial value.
E. 1/4 of its initial value.
Q9.6

The three objects shown

here all have the same
mass M. Each object is
rotating about its axis of
symmetry (shown in blue).
All three objects have the
same rotational kinetic R
energy. Which one is
rotating with fastest
angular speed? B C
A
Q9.6

The three objects shown

here all have the same
mass M. Each object is
rotating about its axis of
symmetry (shown in blue).
All three objects have the
same rotational kinetic R
energy. Which one is
rotating with fastest
angular speed? B C
A
Gravitational potential energy of a rigid body

𝑈 = 𝑚" 𝑔𝑦" + 𝑚! 𝑔𝑦! + ⋯

= 𝑚" 𝑦" + 𝑚! 𝑦! + ⋯ 𝑔 = 𝑀𝑔𝑦cm

Gravitational PE is as if all the mass is

concentrated at the CM.
Example
Assumption: rotation of cylinder is frictionless
no slipping between cylinder and cable
At the moment the block hits the ground, speed of block is 𝑣, angular speed
of cylinder is 𝜔
𝑣 = 𝑅𝜔
0 + 𝑚𝑔ℎ = &#𝑚𝑣 # + &#𝐼𝜔 #
#
initial PE rotational KE, 𝐼 = " 𝑀𝑅 "
of block (we will tell you why later)

2𝑔ℎ
⇒ 𝑣=
1 + 𝑀/2𝑚

if 𝑀 = 0, 𝑣 = 2𝑔ℎ, same as free falling

 Question
• Suppose the cylinder and block have the same mass,
𝑚 = 𝑀. Just before the block hits the floor, it’s KE is
(larger than / less than / the same as) the KE of the
cylinder.
Parallel axis theorem
𝐼cm : moment of inertia about an axis through its CM
𝐼( : moment of inertia about another axis ∥ to the original one and at ⊥ distance 𝑑
𝐼C = 𝐼cm + 𝑀𝑑!

Proof: square of ⊥ distance of 𝑚# to rotation axis

𝐼cm = % 𝑚! 𝑥!" + 𝑦!"

" "
𝐼# = % 𝑚! 𝑥! − 𝑎 + 𝑦! − 𝑏

= % 𝑚! 𝑥!" + 𝑦!" − 2𝑎 % 𝑚! 𝑥! − 2𝑏 % 𝑚! 𝑦!

𝐼cm 𝑀𝑥cm = 0 𝑀𝑦!" = 0

+ 𝑎 " + 𝑏 " % 𝑚!

𝑀
Significance of the parallel axis theorem:
need formula for 𝐼cm only
Example A cylinder with uniform density
Before calculating moment of inertia, must specify
rotation axis
CM along axis of symmetry

𝐼 = D 𝑚$ 𝑟$" ⟶ F 𝑟 " 𝑑𝑚 = F 𝑟 " 𝜌𝑑𝑉

⊥ distance of 𝑚# to rotation axis uniform density

Key: choose 𝑑𝑉 (the volume element) wisely, as

symmetric as possible
𝑑𝑉 = 2𝜋𝑟 𝑑𝑟 𝐿
%" 𝜋𝜌𝐿 '
𝐼 = 2𝜋𝜌𝐿 F 𝑟 & 𝑑𝑟 = 𝑅" − 𝑅#'
𝑑𝑉 %! 2
𝜋𝜌𝐿 "
= 𝑅" − 𝑅#" 𝑅"" + 𝑅#"
2
But 𝑀 = 𝜌 𝜋𝑅"" 𝐿 − 𝜋𝑅#" 𝐿 = 𝜋𝜌𝐿 𝑅"" − 𝑅#"
𝐼 = #"𝑀 𝑅"" + 𝑅#"
independent of length
 Question
• Two hollow cylinders have the same inner and outer
radii and the same mass, but they have different
lengths. One is made of wood and the other of lead.
The wooden cylinder has (larger / smaller / the same)
moment of inertia about the symmetry axis than the
lead one.
Example A uniform sphere

Choose 𝑑𝑉 to be a disk of radius 𝑟 = 𝑅 " − 𝑥 " and thickness 𝑑𝑥

From Example 9.10, moment of inertia of this disk is
1 1 1
𝑑𝑚 𝑟 = 𝜌𝜋𝑟 𝑑𝑥 𝑟 = 𝜌𝜋 𝑅 " − 𝑥 " " 𝑑𝑥
" " "
2 2 2
Therefore
% )
𝜌𝜋 8𝜋𝜌𝑅
𝐼= F #" 𝑑𝑚 𝑟 " = F 𝑅 " − 𝑥 " " 𝑑𝑥 =
2 (% 15
* &*
Since 𝜌 = =
+ ',% $

𝐼 = ")𝑀𝑅 "
3. Torque
 Vector (Cross) Product

𝑪 = 𝑨×𝑩
Magnitude: 𝐶 = 𝐴𝐵 sin 𝜙
direction determined by Right Hand Rule

Important!
Special cases:
(i) if 𝑨 ∥ 𝑩, 𝑨×𝑩 = 0,
in particular, 𝚤×
̂ 𝚤̂ = 𝚥× @ 𝑘@ = 0
̂ 𝚥̂ = 𝑘×
(ii) if 𝑨 ⊥ 𝑩, 𝑨×𝑩 = 𝐴𝐵
in particular,
In analytical form (no need to memorize)
𝑨×𝑩
= 𝐴X 𝐵Y − 𝐴Y 𝐵X 𝚤̂ + 𝐴Y 𝐵Z − 𝐴Z 𝐵Y 𝚥̂
+ 𝐴Z 𝐵X − 𝐴X 𝐵Z 𝑘@ don’t worry if you
have not learnt
𝚤̂ 𝚥̂ 𝑘@ determinants in
= 𝐴Z 𝐴X 𝐴Y high school
𝐵Z 𝐵X 𝐵Y
Torque
Besides magnitude and direction, the line of action of a force is important because
it produces rotation effect.

𝑭+ and 𝑭, have the same

magnitudes and directions, but
different line of action: they
produce different physical effects
– which force would you apply if
you were to tighten/loosen the
screw?
 Define torque about a point O as
a vector
𝝉 = 𝒓×𝑭

• 𝝉 is ⊥ to both 𝒓 and 𝑭
• Magnitude:
𝜏 = 𝑟 𝐹 sin 𝜙 = 𝑟 sin 𝜙 𝐹

component of ⊥ distance from

𝑭 ⊥ to 𝒓 O to line of
actions of 𝑭

Direction gives the sense of rotation about O

through the right-hand-rule.
Notation: ⊙ out of the plane
⊗ into the plane

SI unit for torque: Nm (just like work done)

Q10.2
F1 F3
Which of the four forces shown here produces
a torque about O that is directed out of the
O
plane of the drawing?

F2
F4

A. F1
B. F2
C. F3
D. F4
E. more than one of these
A10.2
F1 F3
Which of the four forces shown here produces
a torque about O that is directed out of the
O
plane of the drawing?

F2
F4

A. F1
B. F2
C. F3
D. F4
E. more than one of these
 Question
A force P is applied to one end of a lever of length L. The magnitude of
the torque of this force about point A is (𝑃𝐿 sin 𝜃 / 𝑃𝐿 cos 𝜃 / 𝑃𝐿 tan 𝜃)
Suppose a rigid body is rotating about a fixed axis which we
arbitrarily call the z axis. 𝑚" is a small part of the total mass.

𝐹&,rad , 𝐹&,tan , and 𝐹&,z are the 3 components of

the total force acting on 𝑚&
Only 𝐹&,tan produces the desired rotation, 𝐹&,rad
and 𝐹&,z produce some other effects which are
irrelevant to the rotation about the z axis.
𝐹&,tan = 𝑚& 𝑎&,tan = 𝑚& 𝑟& 𝛼.
𝐹&,tan 𝑟& = 𝑚& 𝑟&# 𝛼.

torque on 𝑚$ about z, 𝜏$%

Sum over all mass in the body, since they all have the same 𝛼.

I 𝜏'. = I 𝑚" 𝑟'# 𝛼. = 𝐼𝛼.

Need to consider torque due to external forces
only. Internal forces (action and reaction pairs)
produce equal and opposite torques which have
no net rotational effect.
Conclusion: for rigid body rotation about a
fixed axis,

I 𝜏ext = 𝐼𝛼

c.f. Newton’s second law ∑ 𝑭ext = 𝑀𝒂

Example Pulley rotates about a fixed axis. What is the acceleration
a of the block?
For the cylinder
& #
𝑎
𝑇𝑅 = #
𝑀𝑅
𝑅
torque moment of angular
due to T inertia of acceleration
cylinder

i.e. 𝑇 = -.𝑀𝑎

For the block

𝑔
𝑚𝑔 − 𝑇 = 𝑚𝑎 → 𝑎 =
1 + 𝑀/2𝑚

Suppose the block is initially at rest at height h. At the moment it hits the floor:
𝑔 2𝑔ℎ
𝑣" =0+2 ℎ ⟹ 𝑣=
1 + 𝑀/2𝑚 1 + 𝑀/2𝑚
c.f. Previously we get the same result using energy conservation.
 Question
Mass 𝑚# slides on a frictionless track. The pulley has moment of inertia I about its
rotation axis, and the string does not slip nor stretch. When the hanging mass 𝑚" is
released, arrange the forces 𝑇# , 𝑇" , and 𝑚" 𝑔 in increasing order of magnitude.
4. Rolling without slipping
We know how to deal with:
translation of a point particle (or CM of a rigid body):
/ 𝑭ext = 𝑚𝒂

rotation of a rigid body about a fixed axis:

/ 𝜏ext = 𝐼𝛼

In general, a rigid body is rotating about a moving axis, i.e.,

has both types of motion simultaneously.

Every possible motion of a rigid body can be represented as

a combination of translational motion of the CM and
rotation about an axis through its CM.
e.g. tossing a baton

translation + rotation rotation translation of CM

about a fixed (considered as a
axis through particle)
CM
Energy consideration
𝑚$ is a small mass of the rigid body
𝒗-$ its velocity relative to the CM, its velocity relative to the
ground is 𝒗$ = 𝒗cm + 𝒗-$

𝐾$ = #"𝑚$ 𝒗$ = #"𝑚$ 𝒗cm + 𝒗-$ c 𝒗cm + 𝒗-$

"

= #"𝑚$ 𝒗cm c 𝒗cm + 2𝒗cm c 𝒗-$ + 𝒗-$ c 𝒗-$

= #"𝑚$ 𝑣cm
" + 2𝒗 - -"
cm c 𝒗$ + 𝑣$

Total KE of the rigid body

𝐾 = D 𝐾$

= #" D 𝑚$ 𝑣cm
"
+ 𝒗cm c D 𝑚$ 𝒗-$ + D #"𝑚$ 𝑣$-"

$ " "
𝑀 center of mass "𝑚! 𝑟! 𝜔 Therefore
velocity
relative to CM 𝐾 = "!𝑀𝑣cm
!
+ "!𝐼𝜔!
= zero
Rolling without slipping
No slipping at the point of contact ⟹ point of contact must be at rest (instantaneously), i.e.,
−𝑅𝜔 + 𝑣cm = 0 ⟹ 𝑣cm = 𝑅𝜔

translation + rotation rotation about instantaneous axis

𝐾 = #"𝑀𝑣cm
"
+ #"𝐼𝜔" of rotation ( a moving axis)
𝐾 = #" 𝐼 + 𝑀𝑅 " 𝜔"
= #"𝑀𝑣cm
" + #𝐼𝜔 "
"
parallel axis theorem
𝑣cm = 𝑅𝜔
Example

Rolling without slipping,

friction does no work

What determines which body rolls down the incline fastest?

Suppose a rigid body’s moment of inertia about its symmetry axis is 𝐼 = 𝑐𝑀𝑅 #
& # & #
𝑣cm #
0 + 𝑀𝑔ℎ = #
𝑀𝑣cm + #
𝑐𝑀𝑅 +0
𝑅

translation rotation KE about

initial KE initial PE final PE
KE of CM a fixed axis

Rigid body with smaller 𝑐 rolls faster :

2𝑔ℎ "
⇒ 𝑣56 = solid sphere (𝑐 = ))
1+𝑐 #
> solid cylinder (𝑐 = ")
"
depends on 𝑐 only, > thin walled hollow sphere (𝑐 = &)
independent of 𝑀 and 𝑅 > thin walled hollow cylinder (𝑐 = 1)
Role of friction: Example

Rolling without slipping is not possible without friction.

Consider a rigid sphere going freely down an inclined plane. If no friction, no
torque about the center and the sphere slides down the plane.

Assume rolling without slipping, friction must be (static / dynamics) and

must point (upward / downward) along the plane.

𝑣cm = 𝑅𝜔 ⇒ 𝑎cm = 𝑅𝛼
 𝑣cm = 𝑅𝜔 ⇒ 𝑎cm = 𝑅𝛼

Translation of CM: 𝑀𝑔 sin 𝛽 − 𝑓 = 𝑀𝑎cm

Rotation of sphere about its center: 𝑓𝑅 = 𝐼cm 𝛼 = "

%
𝑀𝑅 " 𝑎cm ⁄𝑅

) "
Get 𝑎./ = 𝑔 sin 𝛽 and 𝑓 = 𝑀𝑔 sin 𝛽
0 0

Rolling is slower than sliding because part of the PE is converted into rotation KE
If the sphere is rolling uphill with no slipping, the friction will point (upward /
downward) along the plane because its effect is to decelerate the rotation.
Puzzle: For rolling without slipping, friction does NO
work. Therefore a vehicle will go on forever if there
is no air resistance, just like a magnetic levitated train.
Too good to be true!
In reality energy is lost because the floor and/or the rolling body are deformed, e.g.
vehicle tyre.

Energy is lost because:

• due to deformation, normal reaction produces a torque opposing the rotation.
• sliding of the deformed surfaces causes energy lost.
These two effects give rise to rolling friction.
Consequence: trains, with metal wheels on metal tracks, are more fuel efficient
than vehicles with rubber tires.
A Yo-yo
To find 𝑣cm at point 2, need energy conservation
& # & & #
𝑣cm #
0 + 𝑀𝑔ℎ = #
𝑀𝑣cm + # #
𝑀𝑅 +0
𝑅

initial translation rotation KE about a final

initial
KE KE of CM fixed axis PE
PE

7
⇒ 𝑣cm = 8
𝑔ℎ c.f. for free falling 𝑣&' = 2𝑔ℎ

To find the downward acceleration of the yo-yo, need

dynamic equations
Translation of CM: 𝑀𝑔 − 𝑇 = 𝑀𝑎cm
Rotation of cylinder about its axis:
& #
𝑇𝑅 = 𝐼cm 𝛼 = #
𝑀𝑅 𝑎cm ⁄𝑅
Get
𝑎cm = #8𝑔
𝑇 = &8𝑀𝑔
Exercise: rotation of a dumbbell

A dumbbell consists of a weightless rod of length L and two masses (each with
mass M) on its two ends. Initially, the dumbbell sits on a frictionless table and
points north. A constant force F (towards east) is applied on one of the ball.
The dumbbell will accelerate and rotate due to the applied force.
Find the tension in the rod when the dumbbell rotation 90o

𝐹
𝐹
𝜑

𝑙
Due to the constant external force F, the CM of the dumbbell accelerates with constant
acceleration 𝑎 = 𝐹/2𝑀.
At the instance when the CM moves to the distance 𝑙, the CM velocity becomes 𝑣 = 2𝑎𝑙.
And the work-energy theorem gives
1
𝐹 𝑙 + 𝐿 sin 𝜑 = 𝐾1 + 𝐾2
2
where
1
𝐾1 = ×(2𝑀)𝑣 " = 𝐹𝑙
2
1 𝐿 " " 1
𝐾2 = ×(2𝑀) 𝜔 = 𝑀𝐿" 𝜔"
2 2 4
are the translational and rotational kinetic energies of the dumbbell respectively. Hence we
have
1
2𝐹 sin 𝜑 𝐹 𝑙 + 𝐿 sin 𝜑
𝜔=
𝑀𝐿
2
𝐹
𝜑

𝑙
Finally, focusing on the centripetal force acting on the mass 1.

T F sin ' = M acir M aCM sin '

2L F
= M! M sin '
2 2M
F ML 2 3
T = sin ' + ! = F sin '
2 2 2

𝐹
𝐹
𝜑
𝑇
𝑙
Work and power in rotational motion
A particle or rigid body, being pushed by an external force, is undergoing circular
motion about a fixed axis (such as a merry-go-round).

only the tangential component 𝐹tan does

work – no displacement along the radial
and z directions.
Work done after going through angle 𝑑𝜃
𝑑𝑊 = 𝐹tan 𝑅𝑑𝜃 = 𝜏𝑑𝜃

⇒ 𝑊 = h 𝜏 𝑑𝜃

c.f. in translation, 𝑊 = h 𝑭 i 𝑑𝒓
 𝑊 = h 𝜏 𝑑𝜃

By changing variable
𝑑𝜔
𝜏𝑑𝜃 = 𝐼𝛼 𝑑𝜃 = 𝐼 𝑑𝜃 = 𝐼 𝑑𝜔 𝜔
𝑑𝑡
m"
𝑊tot = N 𝐼𝜔𝑑𝜔 = "!𝐼𝜔!! − "!𝐼𝜔"!
m!

This is the work-energy theorem for

rotational motion.

How about power?

𝑑𝑊 𝑑𝜃
𝑃= =𝜏 = 𝜏𝜔
𝑑𝑡 𝑑𝑡
c.f. 𝑃 = 𝑭 P 𝒗 for translational motion.
 Question
• You apply equal torques to two different
cylinders, one of which has a moment of
inertial twice as large as the other. Each
cylinder is initially at rest. After one complete
rotation, the cylinder with larger moment of
inertia will have (larger / smaller / the same)
kinetic energy as the other one.
5. Angular momentum
For a point particle, define its angular momentum about the origin
O by
𝑳 = 𝒓×𝒑
the particle need not be rotating about any
axis, can be travelling in a straight line

trajectory of m 𝐿 = 𝑚𝑣𝑟 sin 𝜙 = 𝑚𝑣 sin 𝜙 𝑟

= 𝑚𝑣 𝑟 sin 𝜙

𝑑𝑳 𝑑𝒓 𝑑𝒑
= ×𝒑 + 𝒓× = 𝒓×𝑭 = 𝝉
𝑑𝑡 𝑑𝑡 𝑑𝑡

𝑑𝒓
𝑚 𝑭
𝑑𝑡

c.f.
i.e.
𝑑𝑳
=𝝉 𝑑𝑷
𝑑𝑡 =𝑭
𝑑𝑡
How to calculate the angular momentum 𝑳 of a rigid
body?

Answer:
In general, it is difficult.
However, for a symmetric rigid body….
For a rigid body Take the rotation axis as the z axis,
𝑚# is a small mass of the rigid body

𝐿# = 𝑚𝑣# 𝑟# = 𝑚 𝜔𝑟# sin 𝜃# 𝑟#

If rotation axis is a symmetry axis,

then there exist 𝑚" on the opposite
side whose x-y components of
angular momentum cancel those of
𝜃$
𝑚# .
Therefore only z component of any
𝑳$ is important.

q points along rotation axis with

Total angular momentum 𝑳 = ∑ 𝑳$ = ∑ 𝐿$ sin 𝜃$ 𝒌,
magnitude

#
𝐿 = I 𝑚' 𝜔𝑟' sin 𝜃' 𝑟' sin 𝜃' = I 𝑚' 𝑟' sin 𝜃' 𝜔

⊥ distance of 𝑚! to rotation axis

Conclusion: if rotation axis is a symmetry axis, then

𝑳 = 𝐼𝝎
c.f. 𝒑 = 𝑚𝒗

𝝎 and 𝑳 have the same direction

What if the rotation axis is not a symmetry axis?
The angular momentum is NOT parallel to the angular velocity, i.e.
𝑳 ≠ 𝐼𝝎

But the angular momentum about the rotation axis reads,

𝐼 is the moment of inertia with
𝐿Y = 𝐼𝜔
respect to the rotation axis

And we can apply,

𝑑𝐿Y 𝑑 𝐼𝜔
𝜏Y = =
𝑑𝑡 𝑑𝑡

For rotation about a fixed axis, “angular

momentum” often means the component of
𝑳 along the axis of rotation, but not 𝑳 itself.
Internal forces (action and reaction pairs) have the same line of action
→ no net torque.
Therefore for a system of particles or a rigid body

𝑑𝑷
𝑑𝑳 c.f. = % 𝑭ext
= /𝝉 𝑑𝑡
𝑑𝑡

Under no external torque ( not force)

𝑑𝑳
=0
𝑑𝑡

conservation of angular momentum

A spinning physics professor

Conservation of angular momentum

𝐼& 𝜔& = 𝐼# 𝜔#
If 𝐼# = 𝐼& /2, then 𝜔# = 2𝜔& , and
&
𝐾# = 𝐼# 𝜔## = _____𝐾& .
#

Where comes the extra energy?

And in the reverse process 𝐼# → 𝐼& ,
where goes the energy?
Example
A bullet hits a door in a perpendicular direction, embeds in it and swings it open.
Linear momentum is not conserved because _______________________________
Angular momentum along the rotation axis is conserved because ______________
initial angular 𝑀𝑑 #
momentum of 𝑚𝑣𝑙 = 𝜔 + 𝑚𝑙 # 𝜔
3
top view bullet about hinge
moment of
moment of inertia of bullet
inertia of about hinge
door about after embedded
hinge in door

𝑚𝑣𝑙
⇒ 𝜔=
1
𝑀𝑑 # + 𝑚𝑙 #
3
Question: If the polar ice caps were to completely melt due to global
warming, the melted ice would redistribute itself over the earth. This
change would cause the length of the day (the time needed for the earth to
rotate once on its axis) to (increase / decrease / remain the same).
Q10.11

A spinning figure skater pulls

his arms in as he rotates on
the ice. As he pulls his arms
in, what happens to his
angular momentum L and
kinetic energy K?

A. L and K both increase.

B. L stays the same; K increases.
C. L increases; K stays the same.
D. L and K both stay the same.
A10.11

A spinning figure skater pulls

his arms in as he rotates on
the ice. As he pulls his arms
in, what happens to his
angular momentum L and
kinetic energy K?

A. L and K both increase.

B. L stays the same; K increases.
C. L increases; K stays the same.
D. L and K both stay the same.
6. Gyroscope

https://www.youtube.com/wat
ch?v=cquvA_IpEsA&t=3s
Case 1: when the flywheel is not spinning – it falls down

torque 𝝉 due to weight of the

flywheel 𝒘 causes it to fall 𝑳 increases as flywheel falls
in the x-z plane
Case 2: when flywheel spinning with initial angular moment 𝑳' – it precesses

Since 𝑳 ⊥ 𝑑𝑳, flywheel axis execute circular motion

called precession, 𝑳 remains constant

faster spinning 𝜔 → larger 𝑳 → slower precession Ω

 If re
Rotational motion of the angular momentum rota
~ = ~⌧ dt ! dL = ⌧ dt = M grdt
vert
dL
ang
~ ?L
dL ~ ==> 𝐿 can only change its direction, but NOT its magnitude
dL M grdt Sim
d' = =
L I!

Precession rate
d' M gr
⌦= = L=
dt I!

Vec
Tor
Dire
animation of the vectors 𝒘, 𝝉, and 𝑳 at
http://phys23p.sl.psu.edu/phys_anim/mech/gyro_s1_p.avi
If 𝜔 ≫ Ω, can ignore angular momentum due to precession. Otherwise there
is nutation of the flywheel axis – it wobbles up and down
7. Kepler’s Laws of Planetary Motion

Purely phenomenological
– Kepler didn’t know why

Later derived by Newton using his laws of motion and

gravitation

– significance: heavenly objects obey the same physical laws

as terrestrial objects, don’t need, e.g., Greek myths!
First Law: Each planet moves in an elliptical orbit, with the sun at
one focus of the ellipse.
An ellipse is defined by the locus of a point P
such that 𝑃𝑆′ + 𝑆𝑃 = constant
S and S’ are the two foci of the ellipse

Semi-major axis a ( a length, not an axis)

𝑏
Semi-minor axis 𝑏

Eccentricity e (e = 0 for circle, 0 < e < 1 for

ellipse)

Aphelion – farthest [ 1 + 𝑒 𝑎] point from sun

The trajectory is:

𝑟% Perihelion – closest [ 1 − 𝑒 𝑎] point to sun
𝑟 𝜃 =
1 + 𝑒 cos 𝜃
(𝑟 is the distance measured from Note:
one focus) aphelion distance + perihelion distance = 2a
Second Law: A line from the sun to a given planet sweeps out equal areas in equal times.
See http://en.wikipedia.org/wiki/File:Kepler-second-law.gif

𝑑𝐴 ≈ area of blue triangle = #" 𝑟𝑑𝜃 𝑟

𝑑𝐴 1 " 𝑑𝜃
= 𝑟
𝑑𝑡 2 𝑑𝑡
𝑑𝜃
𝑣3 = 𝑣 sin 𝜙 = 𝑟
𝑑𝑡
𝑑𝐴 1 1 𝐿
∴ = 𝑟𝑣 sin 𝜙 = 𝒓×𝑚𝒗 =
𝑑𝑡 2 2𝑚 2𝑚

i.e., Kepler’s second law ⇔ conservation of

angular momentum

• Angular momentum is conserved because

gravitational force (a central force) produces no
torque
• Another consequence of conservation of
angular momentum – orbit lies in a plane
 & πµC ' 1
T= $
−E −2µE 0
Third Law: The periods of the planets are proportional to the .
powers of the
major axisorlengths of their orbits.
π2 µC 2
2
. !/#
T =
2𝜋𝑎
−2E 3
𝑇=
Finally, using Eq. (10.29),
𝐺𝑚
π2 µ $
T2 = A3 . (10.31)
2C
For the circular orbit, we have
Incidentally, we have just proved Kepler’s third law, T 2 = kA3 , where k is#
𝐺𝑚 𝑚 3𝑚𝑣2
essentially the same for all planets about the Sun. Table 10.1 !lists=A /T
#
𝑎 axis and 𝑎
for several planets. Despite variations of ≈ 100 in the major
≈ 1000 in the period, the value of A3 /T 2 is constant to within 2𝜋𝑎
0.05%.
𝑇=
𝑣
Table 10.1∗
Planet " A, km T, s A3 /T 2
Mercury 0.206 1.16 × 108 7.62 × 106 2.69 × 1010
Earth 0.017 2.99 × 108 3.16 × 107 2.68 × 1010
Mars 0.093 4.56 × 108 5.93 × 107 2.70 × 1010
Jupiter 0.048 1.557 × 109 3.743 × 108 2.69 × 1010
Neptune 0.007 9.05 × 109 5.25 × 109 2.69 × 1010
∗
Source: G. Woan, The Cambridge Handbook of Physics Formulas, Cambridge
University Press (2003).
Central force motion as a one-body problem

We consider an isolated system consisting of two particles (e.g. Sun and Earth) interacting
under the gravitational force. m2
The equations of motion (EOM) are rˆ

𝑚# 𝑟⃗#̈ = 𝑓 𝑟 𝑟̂
̈
𝑚" 𝑟⃗" = −𝑓 𝑟 𝑟̂
r2 r = r1 − r2

Where
r1 m1
𝐺𝑚# 𝑚"
𝑟⃗ = 𝑟⃗# − 𝑟⃗" , 𝑟 = 𝑟⃗ and 𝑓 𝑟 =
𝑟"
The EOM couple 𝑟⃗# and 𝑟⃗" . The problem is easier to handle if we replace 𝑟⃗# and 𝑟⃗" by 𝑟⃗ =
𝑟⃗# − 𝑟⃗" and the center of mass vector 𝑅,
𝑚# 𝑟⃗# + 𝑚" 𝑟⃗"
𝑅=
𝑚# + 𝑚"
If there is no external force, we have
𝑅̈ = 0 ⇒ 𝑅 𝑡 = 𝑅 + 𝑉𝑡
!
If we can take the origin at the center of mass and if the center of mass is stationary, we
have 𝑅 𝑡 = 0.
To get an equation of 𝑟,
⃗ we can subtract two EOM,
̈𝑟⃗ − 𝑟⃗̈ = 1 + 1 𝑓 𝑟 𝑟̂
# "
𝑚# 𝑚"
Or
𝑚# 𝑚"
𝑟⃗̈ = 𝑓 𝑟 𝑟̂
𝑚# + 𝑚"
/ /
Defining the reduced mass, 𝜇 = ! " , we have
/! 4/"
𝜇 𝑟⃗̈ = 𝑓 𝑟 𝑟̂
which is identical to the equation of motion for a particle of mass 𝜇 acted on by a force
𝑓 𝑟 𝑟.̂

• The two-particle problem has been transformed to a one-particle problem.

• If 𝑚" ≫ 𝑚# , we have
𝜇 ≈ 𝑚" , 𝑟⃗" ≈ 𝑅 = 0 𝑎𝑛𝑑 𝑟⃗# ≈ 𝑟⃗
i.e. The Sun (at 𝑟⃗" ) is at rest and the Earth (at 𝑟⃗# ) is moving relative to the Sun.

• The method cannot be generalized to three of more particles.

More on Kepler’s law
We assume the Sun (mass 𝑀) is located at the origin and the planet (mass 𝑚) is at the
point 𝑟⃗ = (𝑟, 𝜃) using the polar coordinate
The conservation of energy gives
1 𝐺𝑀𝑚 1 𝐺𝑀𝑚
𝐸 = 𝑚𝑣 − " " "
= 𝑚 𝑟̇ + 𝑟 𝜃 − ̇ "
2 𝑟 2 𝑟
and the conservation of angular momentum gives
𝐿 = 𝑚𝑟 " 𝜃̇
where 𝐸 and 𝐿 are some constants which depend on the initial condition.
5 #
Since 𝜃̇ = " , we have (let 𝜌 ≡ )
/2 2
1
𝐿 𝐿 1 𝜌 −
(# 𝑟!
𝜃=F 𝑑𝑡 = F 𝑑𝑟 = − F 𝑑𝜌 = cos 𝑒
𝑚𝑟 " 𝑚𝑟 " 𝑟̇ "
𝑒 " 1 𝑟!
− 𝜌−
𝑟! 𝑟!
𝑟!
→𝑟=
1 + 𝑒 cos(𝜃 − 𝜃! )
5" "72&
Here, 𝑟! = 𝑎𝑛𝑑 𝑒 " = 1 +
6*/" 6*/

" )*( " )*( ,!

𝑟̇ " + 𝑟 " 𝜃̇ " = 𝐸+ ⇒ 𝑟̇ " = 𝐸+ -
( + ( + (!+!
We have shown that a planet which has conserved mechanical energy 𝐸 and angular
momentum 𝐿 has the trajectory,
𝑟!
𝑟 𝜃 =
1 + 𝑒 cos(𝜃 − 𝜃! )
5" "72&
where 𝑟! = and 𝑒 " = 1 + . 𝐸 and 𝐿 are determined by the initial conditions.
6*/" 6*/

6*/
1. If 𝐸 < − (i.e. 𝑒 " < 0), the planet collide with the Sun.
"2&
6*/
2. If 𝐸=− (i.e. 𝑒 = 0), the trajectory is a circle.
"2&
6*/
3. If − <𝐸 < 0 (i.e. 0 < 𝑒 < 1), the trajectory is an ellipse.
"2&
4. If 𝐸 = 0 (i.e. 𝑒 = 1), the trajectory is a parabola.
5. if 𝐸 > 0 (i.e. 𝑒 > 1), the trajectory is a hyperbola.
1. The area swept out in a time 𝑑𝑡 is,
1
𝑑𝐴 = 𝑟𝑑𝑟 sin 𝛼
2
1
𝑑 𝐴⃗ = 𝑟×𝑑⃗ 𝑟⃗
2
𝑑 𝐴⃗ 1 ̇ 𝐿
→ = 𝑟× ⃗ 𝑟⃗ = is constant.
𝑑𝑡 2 2𝑚

8 :⃗
3. The total area of an ellipse is 𝐴 = 𝜋𝑎𝑏 = 𝜋𝑎" 1 − 𝑒 " . Since 81
is a
constant, the orbital period is
𝜋𝑎𝑏 𝑚 " "
2𝜋 &/"
𝑇= = 2𝜋𝑎 1 − 𝑒 = 𝑎
𝐿/2𝑚 𝐿 𝐺𝑀

after some simplification.

2&
𝑎= (next page) and 𝐿 = 𝐺𝑀𝑚" 𝑟!
#(< "
Please refer to the note: Proof of Kepler’s laws from Newtonian dynamics

Useful Formula for an ellipse,

𝑟%
𝑟 𝜃 =
1 + 𝑒 cos 𝜃
#
2𝐸𝑟%
𝑒 =1+
𝐺𝑀𝑚
and the semi-major axis
1 𝑟% 𝑟% 𝑟%
𝑎= + =
2 1−𝑒 1+𝑒 1 − 𝑒#
We have
𝐺𝑀𝑚 # 𝐺𝑀𝑚
𝐸= 𝑒 −1 =−
2𝑟% 2𝑎

𝑟! 𝑟!
1+𝑒 1−𝑒

𝑂
2𝑎