Lecture 6 rotation sự quay

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Lecture 6

ROTATION

Lecturer Tran Thi Ngoc Dung

dungttn@gmail.com

Rotation in Reality

• Rotational motion is all around us from

molecules to galaxies

- The earth rotates about its axis

- Wheels, gears (sự truyền động bánh răng), propellers [( chân vịt (tàu), cánh quạt (máy bay) ], motors, a CD in its player, a

pirouetting (múa xoay tròn) ice skater, all rotate

Outline

- Angular Velocity-Angular Acceleration

- Torque

- Equation of Rotation

- Moment of Inertia

O



d

d

M1(t)

Mi(t+dt)

Mi(t)

M1(t+dt)

Rigid object rotates about a fixed axis 

The trajectory of each point (except those on the axis) is a circle

a rigid body = system of particles, that the distance between any two

particles that make up the object remains fixed

During a time dt, all points move the same angle d

) /

( rad s dt

d 

 

) /

( rad s 2 dt

d 

 

Angular velocity

Angular Acceleration

r i : distance from ith particle Mi to the rotation axis

During a time dt, i th particle move

a distance ds i



 r d

dsi i

Angular Velocity-Angular Acceleration

O



d

ri

i

v  i 

a 

in

a 

The speed of the i th particle:





i i

i

dt

d r dt

ds

v   

The tangential acceleration of the ith particle :

i i

i

dt

d dt

dv

a     

The centripetal acceleration

of the ith particle :

i i

r

v

2







The acceleration of the ith particle

4 2

2

 i ni i

a

O



 

i

v 



i

a 

in

a 

 

The rotation is accelerated

i















i in

i i

i i

r a

r a

r v

















2





















O



 

i

v 



i

a 

in

a 

 

The object rotates

counter-clockwise

i















The rotation is decelerated

O



  v  i ain a i



 

The rotation is accelerated

i















i in

i i

i i

r a

r a

r v

















2





















i















The rotation is decelerated

O



  v  i 

i

a 

in

a 

 

The object rotates

clockwise

Torque

O

i

F 

i

r 

ir

F  F i

//

F 

Consider a force F i exerting on the object, that can rotale about a

fixed axis

t

F F





Only the tangential force can rotate the object

The torque of the Force F about a fixed point O:

F r

O F





 /  







F











/ (  / ).



The torque of the Force F about the 

axis going through the point O





e 

evector unit

e  : _











































 / /

/

0

//

/

//

/

) (

) (

) (

) (

) (

//

t

F t

O F F

e rF

t r

e rF

O

F

t r

O

F

M e

rF e

e M

F r

F r

F r

F F

F r

F r



























 









 









 



































Only the tangential force can rotate the object

0

/

//  

F 





The torque of a parallel force F // is zero

The torque of the force that has the line of

action go through the axis is zero



  /

/ Ft

F   



0

/  

r

F 





Dynamics of Rotation



  I

) 1 (



i i it

i

F  

Let Force F i exerts on the ith particle

Force F it causes a tangential acceleration a ti

of the ith particle:

Multiply the eq(1) by r i

Torque of F i about the rotation axis



i

ti r

a 

) 2 (

2

i i it

iF m r

) ' 2 (

2

i  miri

) 3 ( )

( 2 

 

i

i i i

 : the net torque acting on the object:

I: Moment of Inertia





i

i i i

i I m r2





Summing over all the particles in the object

(Angular acceleration is the same for all the

particles of the object and can therefore be

taken out of the sum )

Equation of Rotation

Moment of Inertia

object) continuous

(

particles) of

system (

2

2









dm

r

I

r m

I

i

i

from the rotation axis 

r is the distance of the mass element dm from the rotation axis



m1

r1



mi

ri

mN

r1

dm

r



Moment of Inertia

of Homogenous Rigid Objects about the axis through the Center of Mass



R

2

2

1

mR

I CM 

Solid cylinder/disk



R

Hoop or thin cylindrical shell

2

mR

I CM 

Solid sphere

2

5

2

mR

I CM 



R



R1

R2

) (

2

2

2

1 R R m

Hollow cylinder

Thin spherical shell

2

3

2

mR

I CM 



R

CM

L

Thin rod

2

12

1

mL

I CM 

L

Thin rod

2

3

1

mL

I 



a

b

Sheet

) (

12

1 2 2

b a m

The Parallel-Axis Theorem

2

I

The moment of inertia about an axis is equal to the sum of moment

of inertia about an axis through the center of mass I cm and Md 2

d : distance between through parallel axes  and  cm

CM



2 2

2 2 2

2

3 2

1

, 2

1

mR mR

mR I

R d

mR I

md I

I

CM

CM



















Kinetic Energy of Rotation

2

2

1

i i

2 2

2 2 2

2

1 2

1 2

1





















i

i i i

i i i

i i i

i

K

The kinetic energy of a rotating object is the sum of the kinetic energies of the individual particle s in t he object

The kinetic energy of a mass element mi is

Summing over all the elements and using vi = riω gives

2

2 1

Example 2

A 4-kg block resting on a frictionless horizontal ledge is attached to a

string that passes over a pulley and is attached to a hanging 2-kg block (Figure 9-45 ) The pulley is a uniform disk of radius 8 cm and mass 0.6

kg (a) Find the speed of the 2-kg block after it falls from rest a distance of 2.5 m (b) What is the angular velocity of the pulley at this time?

Example 1

Two blocks are connected by a string that passes over a pulley of radius

R and moment of inertia I The block of mass m1 slides on a frictionless, horizontal surface; the block of mass m2 is suspended from the string Find the acceleration a of the blocks and the tensions T 1 and

T 2 assuming that the string does not slip on the pulley

Example 2

A uniform thin stick of le ngth L and mass M is pivoted at one end It is held horizontal and released ( Figure 9-24) Assume the pivot is

frictionless Find (a) the angular acceleration of the stick immediately

after it is released, and (b) the force F0 exerted on the stick by the pivot

at this time.

Example1

Two blocks are connected by a string that passes over a pulley of radius

R and moment of inertia I The block of mass m1 slides on a frictionless, horizontal surface; the block of mass m2 is suspended from the string Find the acceleration a of the blocks and the tensions T 1 and

T 2 assuming that the string does not slip on the pulley

1

2

1

2

P 2

P 1

N T 1 T’ 1

T’ 2

T 2

+ m1 moves right + m2 moves down + Pulley rotates clockwise + The magnitude of acceleration :a1 = a2 = a + The mqgnitude of the tensions:T1 = T’1 , T2 = T’2

1

2

P 2

P 1

N T 1 T’ 1

T’ 2

T 2

) 4 (

) 3 (

) 2 (

) 1 (

2 2

2 2

1 1

1

2 2

2 2

1 1

1 1

T g m a

m

T a

m

T P

a m

T N

P a

m































Applying Newton’s 2nd law and projecting (1) and (2) on the direction of motion :

Equation of rotation of pulley I   T2'R  T1'R ( 5 )

Retionship between a and  a   R ( 6 )

) 7 (

' 1

' 2

R

a

R

I m

2 2









  

2 2

1

2

/ R

I m m

g m a







2 2

1

2 1 1

1 1

/ R

I m m

g m m a

m T









(3)

R I m

m

R I m a

m g m

2 2

1

2 1

2 2

2

/

/













Example 2

A uniform thin stick of le ngth L and mass M is pivoted at one end It is

held horizontal and released ( Figure 9-24) Assume the pivot is

frictionless Find (a) the angular acceleration of the stick immediately

after it is released, and (b) the force F o exerted on the stick by the pivot

at this time.



P

CM

F o

Applying the Newton’s 2nd law for the CM gives:

) 1 (

o

cm mg F

2

L mg

I 

2 3

1

mL

I 

2 3

mg

2

s

rad L

g





mg ma

mg

4

1







g

L

acm

4

3

2 

 

(1)

Remark Just after the

stick is released, the

pivot exerts an upward

force equal to

one-fourth the weight of t he

stick

Ex 3 A 4-kg block resting on a frictionless horizontal ledge is attached to

a string that passes over a pulley and is attached to a hanging 2-kg block The pulley is a uniform disk of radius 8 cm and mass 0.6 kg (a) Find the speed of the 2-kg block after it falls from rest a distance of 2.5 m (b) What

is the angular velocity of the pulley at this time?

1

2

P 2

P 1

T’ 2

T 2

2 2

1

2

/ R

I m

m

g m a







v

v o 2

0

2

2  



From Ex 9-7

The motion of m 2 is linear with constant acceleration

as

The speed