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Classical Mechanics: Rotational Motion, Lecture notes of Classical Mechanics

Rotational motion : Kinematics, Angular Momentum, Torque, Problems in Angular Momentum, Angular momenta of system of particles, Torque on system of particles

Typology: Lecture notes

2018/2019

Uploaded on 08/07/2019

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Rotational Motion
Rotational motion about a fixed-axis : non-zero ˆ
θ, ˆϕcomponent
IChange in orientation of extended object about a point / axis
ITranslational motion but observed from a fixed point / axis
IRotational motion about a point / axis
In 2-dimension, a particle moving with a velocity in a straight line,
~
v= ˙xˆ
i+ ˙yˆ
j, is equivalent to a rotational motion ~
v= ˙rˆr+r˙ϕˆϕabout the
origin or z-axis, where ˙ris the radial velocity and ˙ϕωis the angular
velocity. The term r˙ϕis the tangential velocity whose direction changes
with ˆϕbut its magnitude remains constant. In 2-dim ωis just a number.
In 3-dimension,~
v= ˙xˆ
i+ ˙yˆ
j+ ˙zˆ
k= ˙rˆr+r˙
θˆ
θ+r˙ϕsin θˆϕ. Tangential
velocity in ˆϕis rωsin θ, implying ~ω =ωˆ
k,ˆ
kbeing the axis of rotation.
~
v=~ω ×~
r=rωˆ
k׈r
=rωˆ
k×(sin θcos ϕˆ
i+ sin θsin ϕˆ
j+ cos θˆ
k)
=rωsin θ(sin ϕˆ
i+ cos ϕˆ
j) = rωsin θˆϕ
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Rotational Motion

Rotational motion about a fixed-axis : non-zero

θ, ϕˆ component

I Change in orientation of extended object about a point / axis

I Translational motion but observed from a fixed point / axis

I Rotational motion about a point / axis

In 2-dimension, a particle moving with a velocity in a straight line,

~v = ˙x

i + ˙y

j, is equivalent to a rotational motion ~v = ˙rˆr + r ϕ˙ ϕˆ about the

origin or z-axis, where ˙r is the radial velocity and ϕ˙ ≡ ω is the angular

velocity. The term r ϕ˙ is the tangential velocity whose direction changes

with ˆϕ but its magnitude remains constant. In 2-dim ω is just a number.

In 3-dimension, ~v = ˙x

i + ˙y

j + ˙z

k = ˙rˆr + r

θ

θ + r ϕ˙ sin θ ϕˆ. Tangential

velocity in ˆϕ is r ω sin θ, implying ~ω = ω

k, ˆk being the axis of rotation.

~v = ~ω × ~r = r ω

k × ˆr

= r ω

k × (sin θ cos ϕ

i + sin θ sin ϕ

j + cos θ

k)

= r ω sin θ(− sin ϕ

i + cos ϕ

j) = r ω sin θ ϕˆ

Rotational motion : Kinematics

In 3-dimension, we assign ˆk as axis of rotation, ϕ~ = ϕ

k. (N.B. angular

position cannot be represented by a vector.)

d ~ϕ

ω~ =

d ϕ~

dt

α ~ =

d~ω

dt

d

2 ϕ~

dt

2

d~r = d ϕ~ × ~r

v = ~ω ×

r

~a =

d~v

dt

d

dt

(~ω × ~r ) = α~ × ~r + ~ω × (~ω × ~r )

Since the relationship between ϕ, ~~ ω, ~α is the same as that between

~r , ~v , ~a, we have similar kinematical formulas,

a(t) =

a const.

~v (t) = ~v 0

  • ~at

r (t) =

r 0

v 0 t +

1

2

at

2

~α(t) = ~α const.

~ω(t) = ~ω 0

  • α~t

ϕ ~(t) = ϕ~ 0

  • ~ω 0 t +

1

2

~αt

2

Examine the R.H.S. of linear acceleration. Resolve ~r into components,

~r = ~r ⊥

  • ~r ‖

where ~r ⊥

is perpendicular to ~ω and |~r ⊥

| is just the distance of our point

from the axis of rotation. ~r ‖

is parallel to ~ω.

Angular Momentum

A disc is set in rotation about an axis from a state of rest. In order to

change its state of motion, some sort of force must have been applied i.e.

momentum has been changed. Can it be change of linear momentum?

Can’t be! Since linear momentum (tangential to the rotation) ~p = m~v of

two diametrically opposite points on the disc are opposite to each other,

the total linear momentum is zero!

Need a new momentum – angular momentum – defined as,

L = ~r × ~p = (~r ⊥

  • ~r ‖

) × ~p = ~r ⊥ × ~p OR

L = ~r × ~p ⊥

for a point particle having position vector

r , which is resolved into

components parallel ~r ‖

and perpendicular ~r ⊥

to its momentum ~p, which

in turn can be resolved into components parallel ~p ‖

and perpendicular ~p ⊥

to position vector.

| |

| p |

r

| |

r | |

p

x x

y y

p

r r

p

_Lz

p Lz = r_

Torque

Opening / closing a door requires different linear forces at different

distances from hinge to get the same effect i.e. same angular

displacement / velocity / acceleration.

Need a new force – torque – defined as,

~τ = ~r ×

F = ~r ⊥

×

F OR ~r ×

F

for a point particle at position vector ~r , with ~r ⊥

being component

perpendicular to the force

F acted upon by,

F

⊥ being the component

perpendicular to ~r. This is very similar to angular momentum.

Examine the time rate of change of angular momentum

d

L

dt

d

dt

(~r × ~p) =

d~r

dt

× ~p + ~r ×

d~p

dt

= ~r ×

F ≡ ~τ

The term

d~r

dt

= ~v and hence ~v × m~v = 0. From Newton’s law, it follows

that

d~p

dt

F. Hence, torque is time rate of change of angular momentum.

Important to note that torque and force are mutually perpendicular.

Problems in Angular Momentum

Example. Sliding block – a block of mass m is sliding on a surface having

coefficient of friction μ. At some instant its velocity is v

i. What would

be its angular momentum and torque on it about the (a) origin, (b) l

j?

At an instant, let position of the block be x

i and external force −μmg

i.

−μ mg

x

y

z

l

v

x

(a) Position vector ~r = x

i. Angular momentum

L and torque ~τ are

L = x

i × mv

i = 0, ~τ = x

i × −μmg

i = 0.

(b) Position vector ~r = x

i − l

j. Angular momentum

L and torque ~τ are

L = (x

i − l

j) × mv

i = mvl

k, ~τ = (x

i − l

j) × −μmg

i = −μmgl

k

Since mdv /dt = −μmg , we have ~τ = d

L/dt.

Example. Conical pendulum – instead of swinging in a vertical plane

about a pivot, it swings about an axis from a pivot tracing out circle on

horizontal plane containing the bob.

ω

α

p

r

La

A

B

z

r’

Lb

ω

α

A

B

z

r’

τ

l

M M

T

F

Origin A : Position vector ~r = rˆr of M is perpendicular to momentum

~p = Mr ω ϕˆ and force acting

F = T sin α(−ˆr ). Therefore,

L

a

= Mr

2

ω(ˆr × ϕˆ) = Mr

2

ω

k and ~τ a

= −Tr sin α(ˆr × ˆr ) = 0

~τ a = 0 is consistent with the fact d

L

a /dt = 0 i.e.

L

a is constant in both

magnitude and direction.

We used θ = π/2 in ˆr = sin θ cos ϕ

ˆ i + sin θ sin ϕ

ˆ j + cos θ

ˆ k, ϕˆ = − sin ϕ

ˆ i + cos ϕ

ˆ j.

Example. Kepler’s 2nd law of planetary motion – area swept out by

radius vector from Sun to a planet in a given time is the same for any

position of the planet in its orbit i.e. dA/dt is constant.

On a plane (θ = π/2), let two consecutive position vectors of the planet

in circular orbit be ~r 1

= r (cos ϕ 1

i + sin ϕ 1

j) and ~r 2

= r (cos ϕ 2

i + sin ϕ 2

j).

Area swept out for small change in angle ∆ϕ ≈ ϕ 1 ∼ ϕ 2 is,

~ A =

1

2

~r 1 × ~r 2 =

1

2

r

2 sin(ϕ 1 ∼ ϕ 2 ) ˆk =

1

2

r

2 sin ∆ϕ

ˆ k ≈

1

2

r

2 ∆ϕ

ˆ k.

Therefore, the rate at which area swept out is,

A = lim

∆→ 0

∆A/∆t = r

2

ϕ/˙ 2.

The velocity of the planet on its orbit is ~v = ˙rˆr + r ϕ˙ ϕˆ. Hence,

L =

r × m

v = mrˆr × (˙rˆr + r ϕ˙ ϕˆ) = mr

2

ϕ˙

k.

Which implies dA/dt = L z / 2 m. For central force i.e.

F = f (r )ˆr , torque

on the planet about the origin is ~τ = ~r × f (r )ˆr = 0 and in absence of any

external torque,

L is constant. So dA/dt is also constant.

Example. Getting captured by gravity – Aiming an unpowered

(hypothetical, of course) spacecraft to hit a far-off planet.

A planet of radius R presents us a disc of area πR

2

as a target. Without

its gravity, it would be quite a challenge to control the initial thrust to

keep the spacecraft within this area. However, situation is much more

favourable. Gravitational attraction of the planet tends to deflect the

spacecraft towards itself even if it is (not too much) outside the πR

2

area. Gravity essentially increases the effective area of hit.

b R

b’

b"

r

v

φ

planet

spacecraft

A

m

M

When the spacecraft is relatively far away from planet, take its trajectory

to be parallel to the axis A connecting some conveniently chosen origin

with the center of the planet. Define impact parameter b as distance

between the initial trajectory and the axis. Without gravity, b ≤ R for

assured hit. Impact parameter for grazing hit is b

and for misses b

′′

b

.

Angular momenta of system of particles

Rigid body : best way to view it as system of particles. The i-th particle

having mass m i at the position r i from a suitably chosen origin. Then

r ij = |~r i − ~r j | = constant. Typically origin is on the fixed axis of rotation.

Linear displacement, momentum and force(s) acting upon are the same

for each particles – point and extended bodies could be treated same way

  • motion of center of mass.

If rotation i.e. motion about an axis is involved – linear displacement,

momentum and force(s) acting upon are different for particles at different

positions from axis. For constant angular displacement ϕ and velocity ω,

d

r i = d ϕ~ ×

r i

v i = ~ω ×

r i

p i = m i

v i

Angular momentum of the system

L is the vector sum of the individual

angular momenta

L

i

L

i = ~r i × ~p i

L =

N ∑

i=

L

i

N ∑

i=

~r i × ~p i

Later we will connect

L with ~ω in a way analogous to ~p = m~v.

Torque on system of particles

Expression for torque ~τ follows from time rate of change of angular

momentum

L,

d

~ L

dt

=

d

dt

i

~ L i =

i

(

d~r i

dt

× ~p i

  • ~r i ×

d~p i

dt

)

Because d~r i

/dt = ~v i

, first term in parentheses vanishes and d~p i

/dt =

F

i

d

L

dt

i

~r i

×

F

i

i

~τ i

= ~τ

ext

int

We are going to assume all internal forces are directed along the line

connecting two interacting particles then the internal torque about the

origin ~τ

int

= 0, hence

N ∑

i=

(

~r i ×

~ F i

)

=

i

~τ i = ~τ

ext

=

d

~ L

dt

Connecting torque ~τ to ~α analogous to

F = m~a will follow later.

Moment of Inertia

Total rotational kinetic energy is

K −

N ∑

i=

K i =

1

2

N ∑

i=

m i r

2

i ⊥

ω

2 ≡

1

2

I ω

2

I is the moment of inertia of a rigid body. I serves the same purpose for

rotational kinetic energy as does mass for kinetic energy in linear motion.

Greater the I , harder is to change its rotational state.

Moment of inertia depends on geometry and distribution of mass in rigid

body. It depends on the position of axis about which it is calculated. For

continuous distribution of matter, with mass density ρ,

I =

i

m i r

2

i ⊥

→ I =

r

2

dm =

r

2

ρ dV

The angular momentum of a rigid body is thus,

~ L =

i

~ L i =

i

~r i ⊥ × ~p i =

i

m i ~r i ⊥ × (~ω × ~r i ⊥ )

=

i

m i [~ω (~r i ⊥ · ~r i ⊥ ) − ~r i ⊥ (~ω · ~r i ⊥ )]

=

i

m i

r

2

i ⊥

~ω = I ~ω

Rotational motion of rigid body

Hence, dynamical equations for pure rotation are

K =

1

2

I ω

2

,

~ L = I ~ω, ~τ = I ~α

An equivalent of center of mass motion (CM) for rotation can be

obtained by considering rotation about axis through CM and defining

radius of gyration k and mass M =

i

m i

I 0

=

i

m i

r

2

i ⊥

≡ M k

2 .

Let the position vector of CM be

R and the CM coordinate for i-th

particle be ~r

i

, then

M

~ R =

i

m i

~r i

and ~r i

=

~ R + ~r

i

, ⇒ ~v i

=

~ V + ~v

i

where

V is velocity of CM and v

i

is velocity of i-th particle in CM

frame. Total angular momentum of the body is,

~ L =

i

~r i × m i ~v i =

i

(

~ R + ~r

i

)

× m i

(

~ V + ~v

i

)

=

~ R ×

i

m i

~ V +

~ R ×

i

m i

~v

i

m i

~r

i

×

~ V +

i

m i

~r

i

× ~v

i

Example. Moment of inertia of a disk of mass M and radius R.

























R

ρ Iz

Ix

Iy

d

I Iz = Io

parallel

theorem

perpendicular

axis theorem axis

Axis through center & perpendicular to the plane of the disk. Let the

axis be in z-direction. Consider a concentric circular strip of radius ρ and

thickness dρ. Therefore, the area and mass of the strip is,

dA = 2πρ dρ and dm =

M

πR

2

dA.

Therefore, the moment of inertia of the disk about z-axis I z

is

I z =

ρ

2 dm =

∫ R

0

ρ

2

M

πR

2

2 πρ dρ =

2 M

R

2

∫ R

0

ρ

3 dρ =

1

2

M R

2

Axis through center & parallel to the plane of the disk. Let the axis be

in x-direction. Then you might be tempted to use

I x =

x

y

M

πR

2

x

2 dx dy

The above integration is horribly complicated! Best way is to invoke

Perpendicular axis theorem : For a planar object, the moment of inertia

about an axis perpendicular to the plane is the sum of the moments of

inertia of two perpendicular axes through the same point in the plane of

the object i.e. I z

= I

x

+ I

y

Hence, we straight away get I x

= I

y

= MR

2 /4.

Axis through the edge Once again, easiest would be to invoke

Parallel axis theorem : I = I 0

  • Md

2 , where d is the distance of the axis

from CM axis. The I 0

can either be I z

or I x

or I y

Axis through edge & parallel to I z

I = I z

  • MR

2

1

2

MR

2

  • MR

2

3

2

MR

2

Axis through edge & parallel to I x

I = I x

  • MR

2

1

4

MR

2

  • MR

2

5

4

MR

2