Lecture 6b rolling thithout sliding

Rolling

Lecturer: Tran Thi Ngoc Dung

HCMUT


What does it mean “Without sliding”?

The sphere is not sliding relative to the bar
if the velocities of the two adjacent points A and B are the same.
where A is on the sphere and B is on the bar


Without sliding : v A / ground  v B / ground
or



vA  vB

The velocity of A relative to B is :



v A / B  v A / ground  v B / ground  0

A
B
Ground

If there is ' sliding':



v A  v B and v A / B  0


Rotation about a moving axis
= Rolling
A cylinder is rolling on the
ground.
CM


v cm

A
Ground

Two types:
- Rolling with sliding
- Rolling without sliding

B




CM
A

B

Rolling

Rolling

Rolling


v cm

=

Translation
of a point of the object

=

Translation
of the center of mass

=

Translation
of point A

+

Rotation
about the axis

through this point

+

Rotation
about the axis
through CM

+

Rotation
about the axis
through A


Velocity in rolling motion
Velocity of a point M of the rolling object.



v M / ground  v M / cm  v cm / ground
Velocity of a point M in the rotation about the axis through CM


v M / cm    CM

v E / cm
E



v A / cm
Ground

CM


v cm

A
B

D

v B / cm


| v A / cm | R

| v D / cm | R

R
| v E / cm | 
2





v M / ground  v M / cm  v cm / ground

y

x

CM
R


v A / cm
Ground


v cm



Av

CM

Rolling without sliding,


D v
CM

 v D / gr.
v D / cm

B

Projecting (1)on the x axis :


vA  0

because the ground is not
moving, vB=0




v A  v A / cm  v cm  0
 R  v cm  0

R  v cm
: angular velocity in rotational motion
: angular acceleration
Vcm,acm: speed and acceleration of the CM

R  a cm

(1)


How to Solve Problems ‘Rolling without Sliding’
Method 1

Rolling

=

Translation
of the center of mass



ma cm  Fnet ,ext

+

Rotation
about the axis
through CM

Icm  net,ext / cm

Step 1. Write Eq. of Motion of the CM
Step 2. Write Eq. of Rotation about the axis through CM
Step 3. Find relation between acm and 


How to Solve Problems ‘Rolling without Sliding’
Method 2

Rolling

=

Translation

of point A
having vA=0

+

Rotation
about the axis
through A

I A  net,ext / A
Step 1. Write Eq. of Rotation about the axis through A =>
Step 2. Find relation between acm and  =>acm





y
x

N

CM
f

A
B

W


Method 1.
Example
A cylinder of mass m, radius R rolls
without sliding on an inclined plane.
Find
a) acceleration of CM
b) angular acceleration
c) friction force
d) condition for having ‘rolling without
sliding’


Step1. Eq. of motion of CM :


  
ma cm  mg  N  f friction

Pr ojecting on x and y axis
(1) / x : ma cm  mg sin   f friction

(2)

(1) / y : 0   mg cos   N

(3)  N  mg cos (3' )

Step2. Eq. of Rotation about the axis though CM :
Icm   f friction R

(4)




y

Step3. Re lation between a cm and  :
a cm  R

(1)

(5)

I cm
From (4), (5) : 2 a cm  f friction (6)
R
I
mg sin 
(3)  (6) : (m  cm2 )a cm  mg sin   a cm 
Icm
R
m 2
R
from(2) : f friction  mg sin   ma cm

x

N

CM
f

A
B

W



mg sin 
Icm
m 2
R
  a cm / R

a cm 

x

f

f friction  mg sin   ma cm
a) solid cylinder : Icm 

b) solid sphere : Icm

a cm

1

mR 2
2

1
f friction  mg sin 
3
2
 mR 2
5
2
f friction  mg sin 
7

f friction

A
B

W


5
a cm  g sin ;
7
b) Hoop : Icm  mR 2
1
 g sin ;
2

N

CM

N  mg cos 

2
a cm  g sin ;
3




y

1
 mg sin 
2

• condition for having
' rolling without sliding '
f fric  f s max
f fric  s N
Cylinder
1
mg sin   s mg cos 
3
tan   3s






y
x

Method 2.
Example
A cylinder of mass m, radius R rolls
without sliding on an inclined plane.
Find
a) acceleration of CM
b) angular acceleration
c) friction force
d) condition for having ‘rolling without
sliding’

N

CM
f

A
B

W


Eq. of Rotation about the axis though CM :
I A   mgR sin 





mgR sin 
IA

3
Parallel Axis Theory : I A  Icm  mR  mR 2
2
2
g sin 


3R
2
a cm  R  g sin 

3
2



2
g sin 
3R

a cm

2

 g sin 
3

Find f fric.from
a ) ma cm  mg sin   f friction
or b)Icm  f friction .R


Rolling without and with sliding
For rolling without sliding :
f friction  f static  f s. max
v cm  R
a cm  R
Work of static friction force is zero

For rolling with sliding :
f friction   k .N
v cm  R
a cm  R
Wfric   k N.s


Kinetic Energy in Rolling
K

1
2
mvcm
2






K.E of Translational motion
of CM

1
Icm2
2



K .E of Rotational Motion
about the axis throughCM

Homework
1) A cylinder rolls down a plane inclined at  = 50°.
What is the minimum value of the coefficient
of static friction for which the cylinder will roll
without slipping? (Answer 0.40)

2) For a hoop rolling down an incline,
(a) what is the force of friction,
(b) what is the maximum value of tan for which the
hoop will roll without slipping?
1
(Answers (a) f  mgsin, (b) tan = 2µs )
2