Transcript Physics 207: Lecture 2 Notes

Lecture 17
Goals:
• Chapter 12
 Define center of mass
 Analyze rolling motion
 Introduce and analyze torque
 Understand the equilibrium dynamics of an extended
object in response to forces
 Employ “conservation of angular momentum” concept
Assignment:
 HW7 due tomorrow
 Wednesday, Exam Review
Physics 207: Lecture 17, Pg 1
A special point for rotation
System of Particles: Center of Mass (CM)
 A supported object will rotate about its center of
mass.
 Center of mass: Where the system is balanced !
 Building a mobile is an exercise in finding
centers of mass.
m1
+
m2
m1
+
m2
mobile
Physics 207: Lecture 17, Pg 3
System of Particles: Center of Mass
 How do we describe the “position” of a system made
up of many parts ?
 Define the Center of Mass (average position):
 For a collection of N individual point like particles
whose masses and positions we know:

 mi ri
N

i 1
RCM 
M
RCM
m2
m1
r1
r2
y
x
(In this case, N = 2)
Physics 207: Lecture 17, Pg 4
Sample calculation:
 Consider the following mass distribution: m at ( 0, 0)

 mi ri
N

RCM  i 1
M
 XCM ˆi  YCM ˆj  ZCM kˆ
2m at (12,12)
m at (24, 0)
XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters
RCM = (12,6)
YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters
(12,12)
2m
XCM = 12 meters
YCM = 6 meters
m
(0,0)
m
(24,0)
Physics 207: Lecture 17, Pg 5
Connection with motion...
 So a rigid object that has rotation and translation
rotates about its center of mass!
 And Newton’s Laws apply to the center of mass

 mi ri
N
K TOTAL  K Rotation  K Translatio n
2
1
K TOTAL  K Rotation  2 MVCM

RCM  i 1
M
K

m
v

m
(

r
)
R
p
p
p
p
For a point p rotating:
p
p
p
p
2
1
2
1
2
VCM
p

p
p
p
Physics 207: Lecture 17, Pg 7
2
Work & Kinetic Energy:
 Recall the Work Kinetic-Energy Theorem: K = WNET
 This applies to both rotational as well as linear motion.
K  I      m(V
1
2
2
f
2
i
1
2
2
CM f
V
2
CM i
)  WNET
 What if there is rolling?
Physics 207: Lecture 17, Pg 8
Demo Example : A race rolling down an incline
 Two cylinders with identical radii and total masses roll down
an inclined plane.
 The 1st has more of the mass concentrated at the center
while the 2nd has more mass concentrated at the rim.
 Which gets down first?
M
Two cylinders with radius R and mass m
h
q
A) Mass 1
B) Mass 2
C) They both arrive at same time
M
who is 1st ?
Physics 207: Lecture 17, Pg 9
Same Example : Rolling, without slipping, Motion
 A solid disk is about to roll down an inclined plane.
 What is its speed at the bottom of the plane ?
M
h
q
M
v?
Physics 207: Lecture 17, Pg 10
Rolling without slipping motion
 Again consider a cylinder rolling at a constant speed.
2VCM
VCM
CM
Physics 207: Lecture 17, Pg 11
Motion
 Again consider a cylinder rolling at a constant speed.
Rotation only
VTang = R
CM
Both with
|VTang| = |VCM |
2VCM
VCM
CM
If acceleration acenter of mass = - aR
Sliding only
VCM
CM
Physics 207: Lecture 17, Pg 12
Example : Rolling Motion
 A solid cylinder is about to roll down an inclined plane.
What is its speed at the bottom of the plane ?
 Use Work-Energy theorem
Disk has radius R
M
h
q
Mgh = ½ Mv2 + ½ ICM 2
and
M
v?
v =R
Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2
v = 2(gh/3)½
Physics 207: Lecture 17, Pg 13
How do we reconcile force, angular
velocity and angular acceleration?
Physics 207: Lecture 17, Pg 16
Angular motion can be described by vectors
 With rotation the distribution of mass matters. Actual
result depends on the distance from the axis of
rotation.
 Hence, only the axis of rotation remains fixed in
reference to rotation.
We find that angular motions may be quantified by
defining a vector along the axis of rotation.
 We can employ the right hand rule to find the vector
direction
Physics 207: Lecture 17, Pg 17
The Angular Velocity Vector
• The magnitude of the angular velocity vector is ω.
• The angular velocity vector points along the axis
of rotation in the direction given by the right-hand
rule as illustrated above.
• As  increased the vector lengthens
Physics 207: Lecture 17, Pg 18
From force to spin (i.e., ) ?
A force applied at a distance from
the rotation axis gives a torque
q
FTangential
=|FTang| sin q



r
a
FTangential
F
Fradial
r
Fradial
If a force points at the axis of rotation the wheel won’t
turn
Thus, only the tangential component of the force matters
With torque the position & angle of the force matters
NET = |r| |FTang| ≡ |r| |F| sin q
Physics 207: Lecture 17, Pg 19
Rotational Dynamics: What makes it spin?
a
A force applied at a distance
FTangential
from the rotation axis
F
NET = |r| |FTang| ≡ |r| |F| sin q
Fradial
r

Torque is the rotational equivalent of force
Torque has units of kg m2/s2 = (kg m/s2) m = N m
NET =
r FTang = r m aTang
=rmra
= (m r2) a
For every little part of the wheel
Physics 207: Lecture 17, Pg 20
For a point mass
NET = m r2 a
The further a mass is away from
this axis the greater the inertia
(resistance) to rotation (as we
saw on Wednesday)
NET = I a

a
FTangential
F
Frandial
r
This is the rotational version of FNET = ma
Moment of inertia, I ≡ Si mi ri2 , is the rotational
equivalent of mass.
 If I is big, more torque is required to achieve a
given angular acceleration.

Physics 207: Lecture 17, Pg 21
Rotational Dynamics: What makes it spin?
A force applied at a distance from
the rotation axis gives a torque
a
FTangential
F
NET = |r| |FTang| ≡ |r| |F| sin q
Fradial
r

A constant torque gives constant angular acceleration
if and only if the mass distribution and the axis of
rotation remain constant.
Physics 207: Lecture 17, Pg 22
Torque, like , is a vector quantity
|r| |F| sin q
(2)
|Ftangential | |r|
(3)
|F| |rperpendicular to line of action |
 Direction is parallel to the axis of rotation with respect to the
“right hand rule”
r sin q
line
of
action
F cos(90°q) = FTang.
r
a
90°q
q
F
F
F
Fradial
 Magnitude is given by (1)
r
r
r
 And for a rigid object  = I a
Physics 207: Lecture 17, Pg 23
Example : Rolling Motion
 Newton’s Laws:
N
SFx  Max  Mg sin q  f
SFy  Ma y  0  N  Mg cos q
S CM  Ia CM  MR a CM   fR
1
2
f
Mg
x dir
q
2
 a CM R  ax
Notice rotation CW (i.e. negative) when ax is positive!
Combining 3rd and 4th expressions gives f = Max / 2
Top expression gives Max + f = 3/2 M ax = Mg sin q
So
ax =2/3 Mg sin q
Physics 207: Lecture 17, Pg 24
Statics
Equilibrium is established when
Translatio nal motion
Rotational motion

SFNet  0

S Net  0
In 3D this implies SIX expressions (x, y & z)
Physics 207: Lecture 17, Pg 29
Example
 Two children (60 kg and 30 kg) sit on a horizontal teeter-totter. The
larger child is 1.0 m from the pivot point while the smaller child is
trying to figure out where to sit so that the teeter-totter remains
motionless. The teeter-totter is a uniform bar of 30 kg its moment of
inertia about the support point is 30 kg m2.
 Assuming you can treat both children as point like particles, what is
the initial angular acceleration of the teeter-totter when the large
child lifts up their legs off the ground (the smaller child can’t reach)?
 For the static case:
Rotational motion

S Net  0
Physics 207: Lecture 17, Pg 30
Example: Soln.

Use S
Net
0
N
30 kg
30 kg
0.5 m
300 N 300 N
60 kg
1m
600 N
 Draw a Free Body diagram (assume g = 10 m/s2)
 0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0
0= 2d + 1 – 4
d = 1.5 m from pivot point
Physics 207: Lecture 17, Pg 31
Recap
Assignment:
 HW7 due tomorrow

Wednesday: review session
Physics 207: Lecture 17, Pg 32