t. k. ng, hkust lecture iv. mechanics of rigid bodies

T. K. Ng, HKUSTT. K. Ng, HKUST

Lecture IV. Mechanics of rigid bodies

Mechanics of Rigid Bodies

• 1) Mathematics description of rigid bodies

• 2)Rigid Body in static equilibrium

• 3)Dynamics of rigid bodies

1). Mathematical Description of Rigid Bodies

Before discussing the laws of motion for rigid bodies, first we have to understand how many ways can a rigid object move?

Example: how many ways a dumbbell can move (assuming the solids at the two ends of the dumbbell can be regarded as point particles) and how can we describe its motion?

Dumbbell can

(i) move as a whole, and

(ii) rotate upon its center

To describe motion as a whole, need

(i) x (t) (x = position of center for example), and

(ii) angles (t) and (t), describing the angular orientation of the dumbbell with respect to a chosen coordinate system (rotation).

Exercise!! How can we describe mathematically the motion of the following objects:

(i) bowling ball rolling on floor,

(ii) rigid triangle moving in space,

(iii) A piece of rock

(iv) A piece of rubber

Question: how many ways can a general rigid body move? (or how many mathematical

variables do we need to describe it’s motion)

Answer: Translation (3-degrees of freedom)

+ Rotation (3-degrees of freedom)

(why?)

Therefore, to understand motion of rigid bodies, we just need to know how Newton’s

Law governs Translation & Rotation!

To describe translation and rotation, we introduce the notion of center of mass

Recall: center of mass

• The center of mass is a special point in a rigid body with position defined by

i

ii

ii masstotalmMxmM

X )(,1

This point stays at rest or with uniform motion when there is no net force acting on the body

Center of mass (CM)

• For rigid bodies the rest of the body may be rotating upon this point which acts as the center of rotation.

• i.e., the motion of a rigid body can be described as motion of CM (translation) + rotation of the rest of the body upon CM.

Proof of the above statements

Assumption: We imagine a rigid body as composed of many small point masses.

mi

ri

Proof of the above statements

Therefore, A rigid body is characterized by the set {mi,ri}.

mi

ri

Newton’s Law for translation of rigid body

ext

toti

ii

ii

iii

ii

iii

FFamAM

vmVM

mMxmXM

)(,

We consider motion of center of mass

Newton’s Law for CM motion

Newton’s Law for rotation

(1)Angular momentum & rotation

The angular momentum of a rigid body defined by a set {mi,ri} is

i

iii vrmL



''

', )()(

ii

ii

iiii

vrmVRM

vVrRmL

It is convenient to separate the CM motion by writing ri = R + ri’

(where R = CM coordinate)

since i

iirm 0'



''

', )()(

ii

ii

iiii

vrmVRM

vVrRmL

It is convenient to separate the CM motion by writing ri = R + ri’

(where R = CM coordinate)

CM motion Rotation about CM


(1)Angular momentum & torque

'

'

''

totCMtot

iii

tot

ii

ii

frFR

armARMt

L

The rate of change of angular momentum is given by




0)(

)(

:

''

'''

ijji

j i

extiiij

exti

i

itot

exttot

i

exti

i

iCMtot

j

ijext

ii

frrif

frffr

FRfRfR

fffNotice

Force directed along the line joining the particles

(central force)



'ext

CMextt

L

i.e., the rate of change of angular momentum is govern by external force only.

Notice: internal force is needed to maintain rigidity of the body

Necessary conditions for a rigid body in static equilibrium

(a)net force acting on the body=0

(b)net torque acting on the body=0

Q. Are these conditions sufficient?

Necessary conditions for a rigid body in static equilibrium

Notice: A rigid body can

translate with a uniform velocity +

rotate with a uniform speed about CM

even if external force = external torque = 0

Rigid body in static equilibrium:stability problem

• Question: Which configuration shown below is stable? And why?

a b


• a is unstable because the orange block will fall if displaced a little bit away from equilibrium.

a b


• Question: A piece of wood with uniform density is put on the table in two ways (a) and (b). Which way is more stable? And why?

a b


• Ans: (b) is more stable because its center of mass is lower, and is more difficult to be turned over by a force. (or its potential energy is lower)

a b

Sufficient conditions for a rigid body in static equilibrium :

When the rigid body is displaced a little bit away from its equilibrium position, the force it felt pushes the body

back to its equilibrium position

Sufficient conditions for a rigid body in static equilibrium :

A rigid body is more stable if it has a configuration with lower

potential energy.

Examples:

Questions (2) & (3) in assignment I. Questions (6) & (7) in last year assignment II + ….

Dynamics of rigid bodies

• a) Rotation about a single axis passing through CM.

• b) Rotation ….. + translational motion of CM.

Rotation about a single axis

Example: rotation of an object about a given axis (z) passing through CM.

Rotation about a single axis

Let the angular speed of rotation be.

Therefore

)0),cos(),sin((

)),sin(),cos((),,(

tirtiriviztirtiriziyixir

vrmLi

iii


Show that

i

ziiz IrmL 2

Notice:

222

sin,cos

),,,(

iii

iiiiii

iiii

yxr

ryrx

zyxx


With external torque

,extzz

zt

It

L

i.e. torque change in rate of rotation


With external torque

,extzz

zt

It

L

Notice, IZ acts as “mass” (inertia) in Newton’s Law

of translation

Examples

What is the moment of inertia for the following objects?

(a) a sphere with uniform density?

(b)a cylinder with uniform density?

i

iiz rmI 2

Examples


(a) a sphere with uniform density

223

2

0

2

0

22

5

2

3

4

5

2

)(sinsin)(2

MRRR

ddrrrIR

z

Examples


(b)a cylinder with uniform density

About z-axis

z-axis

length = l

radius=R

Examples

(b)a cylinder with uniform density

About z-axis

222

00

2

2

1

2

1

)(2

MRRlR

dzdrrrIlR

z

Question:

For a cylinder with uniform density under a constant torque. Will it’s angular

acceleration larger if the torque is acting upon x- or z- directions?

Rotational kinetic energy

2

22

2

1

)(2

1

2

1..

z

i iiiii

I

rmvmEK

Example (Physical pendulum):

What determines the oscillation frequency of a rigid body suspended and free to swing under it’s own weight about a fixed axis of rotation?

I

CM


Moment of inertial about fixed point I

)0(

)(

'2

2'2

i

iiCMCM

i

i

CMi

i

ii

rmIMR

rRmrmI

CM

RCM

I



sin||)(

2

2

CMCM RMgzMgRdt

dI

CM

I


Small angle of oscillation:

CMCM MgRRMg

dt

dI ||

2

2

CM

IOscillation frequency

CMCM

CMCM

IMR

MgR

I

MgR

2

Example 1: A person moving with velocity v and mass m jumps on a merry-go-around (MGA) which is initially at rest. The mass of the MGA is M and its radius is R. The person lands on the MGA at a point x with distance r from the origin. The velocity v is perpendicular to the line joining the origin and x. What is the final angular velocity of the MGA.

Example 1: A person moving with velocity v and mass m jumps on a merry-go-around (MGA) which is initially at rest. The mass of the MGA is M and its radius is R. The person lands on the MGA at a point x with distance r from the origin. The velocity v is perpendicular to the line joining the origin and x. What is the final angular velocity of the MGA. How long will it take for the MGA to stop if the friction coefficient between ground and MGA is ?

Solution:

What are the conservation laws that hold in example (1)?

Energy? Momentum?Angular Momentum?

(N) (N) (Y)

Solution (1):

Conservation of angular momentum

22

22

2

1

2

1

mrMR

mvr

I

mvr

mrMRmvr

Solution (1): Let’s check energy change

2

22

22

222

2

1

212

1

2

1

2

1

mvmrMR

mrmv

mrMRE f

Solution (1): the torque due to friction is

gmr

gMRR

rdrr

P

R

MGA

PMGAtot

3

2

3

2)2(

3

0

Newton’s Law implies.

Idt

d tot

Therefore, time taken for MGA to slow down is

.)

3

2(

)0(

mrMRg

mvrIt

tot

Example 2: A person moving with velocity v and mass m jumps on a merry-go-around (MGA) which is initially at rest. The mass of the MGA is M and its radius is R. The person lands on the MGA at a point x with distance r from the origin. The velocity v is perpendicular to the line joining the origin and x. What is the final angular velocity of the MGA. What happens if the MGA is replaced by a disc of same mass and is resting on a frictionless ground?

This is an example of translational motion of CM + rotation about CM (one axis).

Solution:

We decompose the motion into

(i) Translational motion of CM

(ii)Rotation of the body above CM.

This is an example of translational motion of CM + rotation about CM (one axis).

Conservation Laws:

Translational motion of CM – conservation of momentum

Rotation about CM – conservation of angular momentum

CM position:

.

)0(

rMm

mMm

MmrRCM

Solution (2): conservation of momentum (translational motion of CM)

vMm

mv

vmMmv

f

f

)(

Final velocity

Solution (2): conservation of angular momentum (rotation about CM)

22

222

2

1

2

1)(

)(

rmM

MmMR

MRMRmM

MrmI

ImM

MrmvRrmv

CMCM

CMCM

Final frequency of

rotation about CM


mM

Mm

rMR

v

ImM

mMv

CM

22

2

1)(


Notice that the system is rotating about CM, NOT about center of

MGA!

CM

Center of MGA

Example: A dumbbell is formed by 2 equal masses m joint by a light rod of length l. Discuss the system’s motion if a force F acts on the first mass for a very short time t. The angle between the force and the rod joining the masses is .

t. k. ng, hkust lecture iv. mechanics of rigid bodies

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