Physics 207: Lecture 2 Notes
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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