Physics Beyond 2000 Chapter 6 Angular Momentum. Rotational Motion The body is rigid. (i.e. It does not suffer deformation by external forces.) The forces

Physics Beyond 2000

Chapter 6

Angular Momentum

Rotational Motion

• The body is rigid. (i.e. It does not suffer deformation by external forces.)

• The forces on the body may act at different points.

The Kinematics of Rotation

• Axis of rotation – the body is rotating about a fixed axis.

axis of rotation

side view


• Axis of rotation – the body is rotating about a fixed axis.

top view

axis ofrotation

ω


• Angular displacement – The reference line moves an angle Δθ about the axis of rotation.

top view axis ofrotation

ωΔθ


• Average angular speed - tav


ωΔθ


• Instantaneous angular speed - dt

d


ωΔθ


• Example 1 – Find the angular speed.


• Average angular acceleration -

tav


• Instantaneous angular acceleration -

dt

d


• Constant angular acceleration αωo = initial angular velocityω = final angular velocityθ = angular displacement t = time taken

tt o

o .

The Kinematics of Rotation• Constant angular acceleration α

ωo = initial angular velocityω = final angular velocityθ = angular displacement t = time taken

2.2

1. tto

.222 o

to .

The Kinematics of Rotation• Constant angular acceleration α

ωo = initial angular velocityω = final angular velocityθ = angular displacement t = time taken

to .2


• Constant angular acceleration α

2.2

1. tto

.222 o

to . to .

2

The Kinematics of Rotation• Note that the following quantities, except

time t, are vectors.

2.2

1. tto

.222 o

to . to .

2

The Kinematics of Rotation• We may use + and – signs to indicate the

direction of the vectors.

2.2

1. tto

.222 o

to . to .

2


• Example 2 – to find the angular acceleration.

• The negative sign of α indicates that it is in opposite direction to the positive angular velocity.

ω αO

Linear Acceleration

• When the object is rotating, it has two components of linear acceleration.

• Tangential acceleration at

– It is the linear acceleration along the tangent.

• Radial acceleration ar

– It is the centripetal acceleration pointing radially inwards.

Tangential acceleration• at = r. α• It changes the angular velocity.

at

r

O

Radial acceleration

22

rr

var

at

rarO

Linear velocity and

angular velocity

rAO A BrB

Points A and B have the same angular velocity but different linear velocities.

vA

vB

ω

Linear acceleration and

angular acceleration

rAO A BrB

Points A and B have the same angular acceleration but different linear tangential accelerations.

vA

vB

ω

aA aB

Example 3

• Find the tangential acceleration.

equatorAr

Kinetic energy of a rotating object

• A rigid body of mass M is rotating about a fixed axis at angular speed ω.

• Treat the body as a composition of N particles.

axis of rotation

ω


• The ith particle has mass mi and speed vi

• The distance of the ith particle from the axis of rotation is ri

ωmiri

viNote that

N

iimM

1


ωmiri

vi

The kinetic energy of the ith particle is 2

2

1iivm

22

2

2

1

).(2

1

ii

ii

rm

rm


ωmiri

vi

The kinetic energy of all N particles is 2

12

1i

N

iivm

22

1

2

1

2

1

).(2

1

i

N

ii

i

N

ii

rm

rm


ωmiri

vi

The rotational kinetic energy Kr of the rigid body is

2

12

1i

N

iivm

22

1

2

1

2

1

).(2

1

i

N

ii

i

N

ii

rm

rm


ωmiri

vi

2

1

2 )(2

1

N

iiir rmK Define

N

iiirmI

1

2

2

2

1 IK r

I is called the moment ofinertia of the body about this axis of rotation.

Moment of inertia

• The value of I depends on– the mass of the body– the way the mass is

distributed– the axis of rotation

N

iiirmI

1

2

ωmiri

axis ofrotation

Example 4

• Find the moment of inertia and thus the rotational kinetic energy.

• Change the axis of rotation and find the moment of inertia.

axis of rotation

Radius of gyration

• For a rotating body, its I M.

• So we cab write I = M.k2.

• The k is known as the radius of gyration of the body about the given axis.

M

Ik

Example

• Find the radius of gyration k.

Experiment to determine the moment of inertia of a flywheel

• Supplement Ch.6

The gravitational potential energy of the weight is converted into the rotational kinetic energy of the flywheel and the kinetic energy of the weight.However there is loss of energy due to friction.

• Hoop about cylindrical axisI = MR2

• Hoop about any diameterI = MR2

Table for Moment of Inertia

2

1

M = mass of the hoopR = radius of the hoop

• Solid Cylinder about cylindrical axisI = MR2

• Solid Cylinder about central diameterI = MR2 + ML2


2

1

M = mass of the cylinderR = radius of the cylinderL = length of the cylinder

4

1

12

1

• Thin Rod about axis through centre perpendicular to its lengthI = ML2

• Thin Rod about axis through one end and perpendicular to its lengthI = ML2


12

1

M = mass of the rodL = length of the rod

3

1

• Solid sphere about any diameterI = MR2

• Hollow sphere about any diameterI = MR2


5

2

M = mass of the sphereR = radius of the sphere

3

2

Parallel Axes Theorem2mhII GP

G

G is the centre of gravityof the object

IG is the moment of inertiaabout the axis through thecentre of gravity

G

New axis of rotation

P

h IP is the moment of inertiaabout the axis through thepoint P.

m is the mass of the object

Parallel Axes Theorem2mhII GP

G

G


P

h

Note that the two axesare parallel.

Parallel Axes Theorem

G

G


P

h

IG

IP

Example 7

Perpendicular Axes Theorem

IZ = IX + IY

For a lamina lying in the x-y plane, the momentof inertia IX , IY and IZ, about three mutuallyperpendicular axes which meets at the samepoint are related by

Perpendicular Axes Theorem

IZ = IX + IYExample 8

Moment of force Γ

• Moment of force (torque) It is the product of a force and its perpendicular distance from a point about which an object rotates.

• Unit: NmF

axis of rotation

O

Moment of force ΓF

axis of rotation

O

Γ = F d

Top view

O

F

d

Moment of force Γ

Γ = F d

Top view

O

F

d

•The force F acts at point P of the object.• The distance vector from O, the point of rotation, to P is r.• θis the angle between the force F and the distance vector. Γ = F.r.sinθ

rP

θ

Moment of force Γ

Γ = F d

Moment of force Γis a vector.In the following diagram, the moment of force is an anticlockwise moment.It produces an angular acceleration α in clockwise direction.

Top view

O

F

dr

P

θ α

Moment of force on a flywheel

F

r

Γ = F × r

A force F acts tangentially on the rim of a flywheel.

Work done by a torqueSuppose a force F acts at right angle to thedistance vector r.

Fr

O

Work done by a torqueWhat is the moment of force about O?

Γ= F × r

Fr

O

Work done by a torque

F

The moment of force turns the object throughan angle θ with a displacement s.

Fr

r

θ

O

Γ= F × r

s

Work done by a torqueWhat is the work done by the force?

W= F × s

F

Fr

r

θ

O

s

Work done by a torqueExpress the work done by the force in terms of Γ and θ.

F

Fr

r

θ

O

s

Use F =and s = r. θ

r

W = F × s = Γ× θ

Example 9

• Work done against the moment of friction is equal to the loss of rotational kinetic energy

of the flywheel.

Torque and Angular acceleration

= . is the torque– I is the moment of inertia is the angular acceleration

• Compare to F = m.a in linear motion.

Torque and Angular acceleration

In an angular motion with uniform angular acceleration :

..2

...2

1.

.

22

2

o

o

o

tt

t

Example 10

• Torque and angular acceleration

Conditions for equilibrium

• A body will be in static equilibrium, if– 1. net force is zero

– 2. net moment of force about any point is zero

N

iiF

1

0

N

ii

1

0

Angular momentum L

• The angular momentum L of an object about an axis is the product of the angular velocity and its moment of inertia.

• L = I.• Unit of L: kg m2 s-1 or Nms.

• L is a vector. Its direction is determined by the direction of the angular velocity .

Angular momentum of a rotating point mass

• A point mass m is rotating tangentially at speed v at a distance r from an axis.

• From I = mr2 , L = I and v = r mvrL

v

rmaxis of

rotation

Example 11

• Find the angular momentum of a solid sphere.

Newton’s 2nd law for rotation

• The torque acting on a rotating body is equal to the time rate of change of the angular momentum.

dt

dLI .

II .

Newton’s 2nd law for rotation

• If the net torque is zero, the angular momentum is a constant. The angular acceleration is zero.

dt

dLI .

II .

Example 12

• Find the change in angular momentum.

Torsional pendulum

• A disk is suspended by a wire.

• The wire is twisted through an angle θ

• The restoring torque is

Γ= c. θ where c is the torsional constant.

wire

disk

Torsional pendulum

• The restoring torque is

Γ= c. θ where c is the torsional constant.

• Prove that the torsional oscillation is a SHM with the equation

wire

disk

.I

c

Torsional pendulum

wire

disk

.I

c

and

c

IT 2

Typical examples of second law

• Flywheel with moment of inertia I.

r

mass m

axis

α

Find the angular acceleration αin terms of I, m and r.

2

2

mrI

gmr


• Flywheel with moment of inertia I.

raxis

α

T

mass m

T

mg

a

T.r = I. α mg – T = maa = r. α


• Smooth pulley with moment of inertia I and radius r.

r

m1

m2

α

a

a

Find the linear accelerationa of the two masses in terms of m1, m2, I and r.

221

221

)(

)(

rmmI

grmma

Typical examples of second law• Smooth pulley with moment of inertia I and

radius r.

αm2

a

T2

m2g

r

T2T1

m1

aT1

m1g

T2.r-T1.r = I.α T1-m1g = m1a m2g-T2 = m2a

a = rα

The law of conservation ofangular momentum

• If external net torque = 0, the sum of angular momentum of the system is zero.

N

iiiIL

1

constantIf Γ=0,

The law of conservation ofangular momentum

• For a system with initial moment of inertia I1 and initial angular velocity ω1, its initial angular momentum is I1ω1.

• If the system changes its moment of inertia to I2 and angular velocity ω2, its final angular momentum is I2ω2.

• If there is not any external net torque, then

I1ω1 = I2ω2

Documents

Physics Beyond 2000 Chapter 6 Angular Momentum. Rotational Motion The body is rigid. (i.e. It does not suffer deformation by external forces.) The forces