Transcript Document

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
PE  MgH
KE  0
Energy Conservation
PE  0
KE  Ktrans  Krot
H
K trans 
1
1
Mv 2 ; K rot  I 2
2
2
Rolling without slipping 
  v/R
2
1 2 12
 v  1
I   MR 2    Mv 2
2
25
 R  5
7
 K trans  K rot  Mv 2
10
K rot 
K tot
U  Mgh
U  K tot  0  Mgh 
v
7
Mv 2
10
10
gh
7
Mechanics Lecture 16, Slide 5
Acceleration of Rolling Ball
Newton’s Second Law
mgsin   f  max
a
f
Newton’s 2nd Law for rotations
a
 net,CM  ICM aCM
 net,CM  fR
Mg
 net,CM  fR  ICM aCM
fR
f 
2
Ma x
5
2
mg sin   ma x  ma x
5
7
mg sin   ma x
5
I CM
Rolling without slipping 
5
a x  g sin 
7
 a CM
a CM 
ax
R
fR
ax
R
2
MR 2
5

Mechanics Lecture 16, Slide 6
Rolling Motion
Objects of different I rolling down an inclined
plane:
R
M
h
v0
0
K0
 K   U  Mgh
1 2 1 2
K  I  Mv
2
2
v = R
Mechanics Lecture 16, Slide 7
Rolling
If there is no slipping:
v
v

v
Where v  R
In the lab reference frame
In the CM reference frame
Mechanics Lecture 16, Slide 8
Rolling
v
1 2 1 2
K  I  Mv Use v  R and I  cMR2 . Hoop: c  1
2
2
K
1
1 2 1
2 2
MR


Mv   c  1Mv2
c
2
2
2
Disk: c  1/2
Sphere: c  2/5
etc...
So:
1
 c  1Mv2  Mgh
2
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.
B.
C.
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
K rot
1
A)
1
K trans
 Mv
K rot 
K rot
K trans
2
2
K rot
3

B)
Ktrans 4
C)
K rot
1

Ktrans 2
1 2 1
v
1
I  MR 2 ( ) 2  Mv 2
2
2
R
2
1
Mv 2
2
1
1
2
Mv
2
Recall that I  MR2 for a hoop about
33%
33%
33%
an axis through its CM:
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?
v
A) Block
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
Sliding Block
K tot  K trans  K rot
K trans 
1
1
Mv 2 ; K rot  I 2
2
2
Rolling without slipping 
K trans 
K rot  0
  v/R
K tot  K trans  K rot 
2
1
12
 v  1
K rot  I 2   MR 2    Mv 2
2
25
 R  5
7
K tot  K trans  K rot  Mv 2
10
U  Mgh
U  K tot  0  Mgh 
hball
1
Mv 2
2
1
Mv 2
2
U  Mgh
U  K tot  0  Mgh 
hblock
1
Mv 2
2
1 v2

2 g
7
Mv 2
10
7 v2

10 g
hball
hblock
7 v2
10 g 7


5 v2 5
10 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
a
f
v
M R
  Ia
a

I
M g R 5  g
fR



2
I
2R
2
5
MR
Mechanics Lecture 16, Slide 18
a
f
v
M R
F  Ma
F  Mg
a
 g

M
M
Mechanics Lecture 16, Slide 19
a  g

a
5 g
a
2R
v
M R
v  R
v
v0
v  v0  at
t

  at
Once vR it rolls
without slipping
v0
R  a Rt
v  v0  at
t
t
v0  at  a tR
5 g
v0  gt 
t
2
7 g
v0 
t
2
2
7 g
t
v0
Mechanics Lecture 16, Slide 20

a
M R
v
1 2
x  v0t  at
2
t
2
v0
7 g
a  g
Plug in a and t found in parts 2) & 3)
v  v0  at
Mechanics Lecture 16, Slide 21

a
M R
v
Interesting aside: how v is related to v0 :
v  v0  at
 2v0 

v  v0  (  g )
 7 g 
2
v  v0  v0
7
5
v  v0 Doesn’t depend on 
7
v  0.714v0
a  g
2
t
v0
7 g
We can try this…
Mechanics Lecture 16, Slide 22
a
f
M R
K tran
v
1
 Mv 2
2
2
K rot
1
2
1 2 12
2  v 
 I    MR    Mv
25
2
 R  5
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.

y
Find the acceleration of the blocks.
x
For the hanging masses use F  ma
m1g  T1  m1a
m2g  T2  m2a
a
For the pulley use   Ia  I
R
a 1
T1R  T2R  I  MRa
R 2
1
(Since I  MR 2 for a disk)
2
M
a
R
T2
T1
a
m2
m1
a
m1g
m2g
Mechanics Lecture 16, Slide 24
Atwood's Machine with Massive Pulley:
We have three equations and three
unknowns (T1, T2, a).
y
Solve for a.
x
m1g  T1  m1a
m2g  T2  m2a
(1)
M
a
R
(2)
1
T1  T2  Ma (3)
2


m1  m2
 g
a  
 m1  m2  M 2 
T2
T1
a
m2
m1
a
m1g
m2g
Mechanics Lecture 16, Slide 25
Three Masses
Tsphere
 disk ,net  Thoop  Tsphere Rdisk  I disk a disk  I disk
Tsphere
Thoop
 Thoop  Tsphere  
Thoop
 sphere ,net   f sphere Rsphere  I sphere a sphere  I sphere
f sphere
a
1
a
1
2
 mdisk Rdisk
 mdisk Rdisk a
Rdisk 2
Rdisk 2
1
mdisk a
2
a
Rsphere
2
 msphere Rsphere a
5
2
 f sphere   msphere a
5
Tsphere  f sphere  msphere a  Tsphere  msphere a  f sphere 
mhoop g  Thoop  mhoop a  Thoop  mhoop ( g  a)
7
msphere a
5
7
1
mhoop ( g  a )  msphere a  mdisk a
5
2
7
1

a mdisk  mhoop  msphere   mhoop g
5
2

mhoop g
a 
7
1

 mdisk  mhoop  msphere 
5
2

Mechanics Lecture 16, Slide 26
Three Masses
asphere  a 
a disk 
a sphere 
mhoop g
7
1

 mdisk  mhoop  msphere 
5
2

mhoop g
adisk
a


7
Rdisk Rdisk  1

 mdisk  mhoop  msphere  Rdisk
5
2

asphere
Rsphere

a
Rsphere

1
 mdisk
2
mhoop g
7

 mhoop  msphere  Rsphere
5

Mechanics Lecture 16, Slide 27
Three Masses
Tsphere  f sphere  msphere a  Tsphere  msphere a  f sphere 
7
msphere a
5
mhoop g  Thoop  mhoop a  Thoop  mhoop ( g  a)
Thoop  mhoop ( g 
mhoop g
7
1

 mdisk  mhoop  msphere 
5
2

)
1 2
at
2
2y
t 
a
y 
Mechanics Lecture 16, Slide 28
Three Masses
1 2
at
2
2y
t 
a
y 
v  at  a
2y
 2ay
a
sphere  a sphere t 
asphere
Rsphere
t
a
Rsphere
t
Mechanics Lecture 16, Slide 29
Three Masses 2
T1  T2   1 m3a
2
 m1 g  T1   m1a
T2  m2 g  m2 a


m1  m2
 g
 a  
m

m

m
2
2
3
 1

Mechanics Lecture 16, Slide 30
Three Masses 2
a pulley 
a
R pulley
T1  m1 g  a 
T2  m2 g  a 
y 
1 2
at
2
v  at
 pulley  a pulleyt
Mechanics Lecture 16, Slide 31