Classical Mechanics
Lecture 16
Today’s Concepts: a) Rolling Kinetic Energy b) Angular Acceleration
Physics 211 Lecture 16, Slide 1

Schedule

One unit per lecture!
 I will rely on you watching and understanding pre-lecture videos!!!!

Lectures will only contain summary, homework problems, clicker questions,
Example exam problems….
Midterm 3 Wed Dec 11
Mechanics Lecture 14, Slide 2

Main Points
Mechanics Lecture 16, Slide 3

Main Points
Rolling without slipping 
  v / R
Mechanics Lecture 16, Slide 4

Rotational Kinetic Energy
Total Kinetic Energy = Translational Kinetic Energy + Rotational Kinetic Energy
H
PE

KE
MgH

0
Energy Conservation
PE
KE


0
K trans

K rot
K trans

1
2
Mv
2
; K rot

1
2
I

2
Rolling without slipping 
  v / R
K rot

1
2
I

2 
1
2
2
5
MR
2

  v
R
K tot

K trans

K rot

7
10
Mv
2
2

1
5
Mv
2

U

Mg
 h

U
 
K tot

0

Mg
 h

7
10
Mv
2 v

10 g
 h
7
Mechanics Lecture 16, Slide 5

Acceleration of Rolling Ball f

2
5
Ma x f mg sin
 
2
5 ma x
 ma x mg sin
 
7
5 ma x a
Mg a
Newton’s Second Law mg sin
  f
 ma x
Newton’s 2 nd Law for rotations
 net , CM

I
CM a
CM
 net , CM
 net , CM

 fR
 fR
I
CM a
CM
Rolling without slipping  fR
 a
CM
I
CM a
CM
 a x
R fR
2
5
MR
2
 a x
R a x

5
7 g sin

Mechanics Lecture 16, Slide 6

Rolling Motion
Objects of different I rolling down an inclined plane: h
R
M v

0
 
0
K

0

K
  
U

Mgh
K

1
2
I

2 
1
2
Mv
2 v =

R
Mechanics Lecture 16, Slide 7

If there is no slipping:
Rolling v
In the lab reference frame v
 v
Where v
 
R
In the CM reference frame
Mechanics Lecture 16, Slide 8

Rolling v
K

1
2
I

2 
1
2
Mv
2
K

1
2
Use v

So: 1
2
 c 
1

Mv
2 
Mgh
 c MR
2

2 
1
2
Mv
2
R
 and I
 cMR
1
2
 c

1

Mv
2
2 .
Hoop: c

1
Disk: c

1/2
Sphere: c

2/5 etc...
v

2 gh
1 c 
1
Doesn’t depend on M or R , just on c (the shape)
Mechanics Lecture 16, Slide 9

Clicker Question
A hula-hoop rolls along the floor without slipping.
What is the ratio of its rotational kinetic energy to its translational kinetic energy?
K trans

1
2
Mv
2
K
A)
K rot trans

1
K
B)
K rot trans

3
4
K rot

1
2
I

2 
1
2
MR
2
( v
R
)
2 
1
2
Mv
2
K
1
2
K rot
Mv
 2 
1 trans
1
2
Mv
2
Recall that I

MR 2 for a hoop about
C)
K rot
K trans

1
2
33% 33% 33% an axis through its CM :
A.
B.
C.
Mechanics Lecture 16, Slide 10

CheckPoint
A block and a ball have the same mass and move with the same initial velocity across a floor and then encounter identical ramps. The block slides without friction and the ball rolls without slipping . Which one makes it furthest up the ramp?
A) Block
B) Ball
C) Both reach the same height.
v
 v
Mechanics Lecture 16, Slide 11

CheckPoint
The block slides without friction and the ball rolls without slipping .
Which one makes it furthest up the ramp?
A) Block v
B) Ball
C) Same
 v
B) The ball has more total kinetic energy since it also has rotational kinetic energy.
Therefore, it makes it higher up the ramp.
Mechanics Lecture 16, Slide 12

Rolling vs Sliding
Rolling Ball
K tot

K trans

K rot
K trans

1
2
Mv
2
; K rot
Rolling without slipping

1

2

I


2 v / R
K rot

1
2
I

2 
1
2
2
5
MR
2

  v
R
K tot

K trans

K rot

7
10
Mv
2

U

Mg
 h
2

1
5
Mv
2

U
 
K tot

0

Mg
 h

7
10
Mv
2
 h ball

7
10 v
2 g
7 v
2
 h ball
 h block

10
5 g v
2
10 g

7
5
Sliding Block
K trans
K rot

1

0
2
Mv
2
K tot

K trans

K rot

1
2
Mv
2

U

Mg
 h

U
 
K tot

0

Mg
 h

1
2
Mv
2
 h block

1
2 v
2 g
Ball goes 40% higher!
Mechanics Lecture 16, Slide 13

CheckPoint
A cylinder and a hoop have the same mass and radius. They are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first?
A) Cylinder
B) Hoop
C) Both reach the bottom at the same time
Mechanics Lecture 16, Slide 14

Which one reaches the bottom first?
A) Cylinder
B) Hoop
C) Both reach the bottom at the same time
A) same PE but the hoop has a larger rotational inertia so more energy will turn into rotational kinetic energy, thus cylinder reaches it first.
Mechanics Lecture 16, Slide 15

CheckPoint
A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first?
A) Small cylinder
B) Large cylinder
C) Both reach the bottom at the same time
Mechanics Lecture 16, Slide 16

CheckPoint
A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one reaches the bottom first?
A) Small cylinder
B) Large cylinder
C) Both reach the bottom at the same time
C) The mass is canceled out in the velocity equation and they are the same shape so they move at the same speed.
Therefore, they reach the bottom at the same time.
Mechanics Lecture 16, Slide 17

f
M
R a v
 
I a a 

I
 f R
I


M g R
2
5
M R
2

5
 g
2 R
Mechanics Lecture 16, Slide 18

f
M
R a v
F

Ma a

F
M


Mg
M
  g
Mechanics Lecture 16, Slide 19

a
M
R
 v v
0 v
 v

0 at
   a t t t v v
0
Once v
 
R it rolls without slipping

R
 a
Rt v
 
0 at t a
  g v a 

5

2

R
R g v
0 v
0
 at
 a t R
  gt v
0


5
 g
7
2
 g t t
2 t

2
7
 g v
0
Mechanics Lecture 16, Slide 20

a
M
R
 v t
 a
7

2
 g
 g v
0 x
 v
0 t

1
2 at
2 v
 v

0 at
Plug in a and t found in parts 2) & 3)
Mechanics Lecture 16, Slide 21

a
M
R
 v
Interesting aside: how v is related to v
0 v v

 v
0 v
0


( at
 g )


7
2 v
0
 g

 v
 v
0

2 v
0
7
: v

5 v
0
7
Doesn’t depend on
 v


0 .
714
 v
0
We can try this… a
  g t

2
7
 g v
0
Mechanics Lecture 16, Slide 22

f
M
R a v
K tran

1
2
Mv
2
K rot

1
2
I

2 
1
2
2
5
MR
2

  v
R
2

1
5
Mv
2
Mechanics Lecture 16, Slide 23

Atwood's Machine with Massive Pulley:
A pair of masses are hung over a massive disk-shaped pulley as shown.

Find the acceleration of the blocks.
y x
For the hanging masses use F
 ma
 m
1 g

T
1
 m
2 g

T
2
 
 m m
2
1 a a
For the pulley use
 
I a
T
1
R

T
2
R

I a
R


I a
1
2
MRa
R
(Since I

1
2
MR
2 for a disk) a m
1 m
1 g
T
1
M a
R
T
2 m
2 a m
2 g
Mechanics Lecture 16, Slide 24

Atwood's Machine with Massive Pulley:
We have three equations and three unknowns ( T
1
, T
2
, a ).
Solve for a .
 m
1 g

T
1
  m
1 a (1)
 m
2 g

T
2
 m
2 a (2)
T
1

T
2

1
2
Ma (3) a


 m
1
 m
1 m

2 m
2

M 2

 g y a
M
R x m
1
T
1 a m
1 g
T
2 m m
2
2 g a
Mechanics Lecture 16, Slide 25

f sphere
T sphere
T sphere
Three Masses
T hoop
T hoop
 disk , net


T hoop

T sphere

R disk

I disk a disk

I disk a
R disk

1
2 m disk
2
R disk a
R disk


T hoop

T sphere


1
2 m disk a

1
2 m disk
R disk a
 sphere , net
  f sphere
R sphere

I sphere a sphere

I sphere a
R sphere

2 m sphere
R sphere a
5
 f sphere
 
2
5 m sphere a
T sphere
 f sphere
 m sphere a

T sphere
 m sphere a
 f sphere

7
5 m sphere a m hoop g

T hoop
 m hoop a

T hoop
 m hoop
( g
 a ) m hoop
( g
 a )

7
5 m sphere a

1
2 m disk a a
1
2 m disk
 m hoop

7
5 m sphere
 m hoop g
 a
 m hoop g
1
2 m disk
 m hoop

7
5 m sphere
Mechanics Lecture 16, Slide 26

Three Masses a sphere
 a
 m hoop g
1
2 m disk
 m hoop

7
5 m sphere a disk
 a disk
R disk
 a
R disk
 m hoop g
1
2 m disk
 m hoop

7
5 m sphere
R disk a sphere
 a sphere
R sphere
 a
R sphere
 m hoop g
1
2 m disk
 m hoop

7
5 m sphere
R sphere
Mechanics Lecture 16, Slide 27

Three Masses
T sphere
 f sphere
 m sphere a

T sphere
 m sphere a
 f sphere

7
5 m sphere a m hoop g

T hoop
 m hoop a

T hoop
 m hoop
( g
 a )
T hoop
 m hoop
( g
 m hoop g
1
2 m disk
 m hoop

7
5 m sphere
)
 y

 t

1
2 at
2
2
 y a
Mechanics Lecture 16, Slide 28

Three Masses
 y
 t


1
2 at
2
2
 y a
2
 y v
 at
 a a

2 a
 y
 sphere
 a sphere t
 a sphere
R sphere t
 a
R sphere t
Mechanics Lecture 16, Slide 29

Three Masses 2

T
1

T
2


1 m
3 a

T
2 m

1 g m
2
 g
T
1

2
  m m
2 a
1 a
 a


 m
1
 m
1 m
2

 m
2 m
3
2

 g
Mechanics Lecture 16, Slide 30

Three Masses 2 a pulley

 a
R pulley
T
1
 m
1
 g
 a

T
2
 m
2
 g
 a

 y

1
2 at
2 v
 at
 pulley
 a pulley t
Mechanics Lecture 16, Slide 31