Date post: | 01-Jan-2016 |
Category: |
Documents |
Upload: | channing-vega |
View: | 17 times |
Download: | 4 times |
Physics 207: Lecture 17, Pg 1
Lecture 17Goals:Goals:
• Chapter 12Chapter 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 3
A special point for rotationSystem 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.
m1m2
+m1 m2
+
mobile
Physics 207: Lecture 17, Pg 4
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:
M
mN
iii
1CM
rR
(In this case, N = 2)
y
x
r2r1
m1m2
RCM
Physics 207: Lecture 17, Pg 5
Sample calculation:
Consider the following mass distribution:
(24,0)(0,0)
(12,12)
m
2m
m
RCM = (12,6)
kji CM CM CM1
CM ZYXM
mN
iii
r
R
XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters
YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters
XCM = 12 meters
YCM = 6 meters
m at ( 0, 0)
2m at (12,12)
m at (24, 0)
Physics 207: Lecture 17, Pg 7
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
For a point p rotating:
nTranslatioRotationTOTALK KK M
mN
iii
1CM
rR
2212
21 )(K ppppR rmvm
p
2CM2
1RotationTOTAL VK MK
VCM
p p p
p p p p
Physics 207: Lecture 17, Pg 8
Work & Kinetic Energy:
Recall the Work Kinetic-Energy Theorem: K = WNET
This applies to both rotational as well as linear motion.
What if there is rolling?
NET2
CM2
CM2122
21 )( I WVVmK ifif
Physics 207: Lecture 17, Pg 9
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?
Two cylinders with radius R and mass mM
h
M who is 1st ?
M M
M
M
M
A) Mass 1
B) Mass 2
C) They both arrive at same time
Physics 207: Lecture 17, Pg 10
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
Mv ?
M M
M
M
M
Physics 207: Lecture 17, Pg 11
Rolling without slipping motion
Again consider a cylinder rolling at a constant speed.
VCMCM
2VCM
Physics 207: Lecture 17, Pg 12
Motion Again consider a cylinder rolling at a constant speed.
VCM
CM
2VCM
CMVCM
Sliding only
CM
Rotation only
VTang = R
Both with
|VTang| = |VCM |
If acceleration acenter of mass = - R
Physics 207: Lecture 17, Pg 13
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
M
h
Mv ?
Disk has radius R
M M
M
M
M
Mgh = ½ Mv2 + ½ ICM 2 and v =R
Mgh = ½ Mv2 + ½ (½ M R2 )(v/R)2 = ¾ Mv2
v = 2(gh/3)½
Physics 207: Lecture 17, Pg 16
How do we reconcile force, angular velocity and angular acceleration?
Physics 207: Lecture 17, Pg 17
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 18
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 19
From force to spin (i.e., ) ?
NET = |r| |FTang| ≡ |r| |F| sin
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
FTangential
a
r
Fradial
F
A force applied at a distance from the rotation axis gives a torque
Fradial
FTangential
r
=|FTang| sin
Physics 207: Lecture 17, Pg 20
Rotational Dynamics: What makes it spin?
NET = |r| |FTang| ≡ |r| |F| sin
Torque is the rotational equivalent of forceTorque is the rotational equivalent of forceTorque has units of kg m2/s2 = (kg m/s2) m = N m
FTangential aa
r
Fradial
FF
A force applied at a distance from the rotation axis
NET = r FTang = r m aTang
= r m r = (m r2)
For every little part of the wheel
Physics 207: Lecture 17, Pg 21
For a point mass NET = m r2
This is the rotational version of FNET = ma
Moment of inertia, Moment of inertia, I ≡ imi ri2 , is the rotational is the rotational
equivalent of mass.equivalent of mass. If I is big, more torque is required to achieve a
given angular acceleration.
FTangential aa
r
Frandial
FF
The further a mass is away from this axis the greater the inertia (resistance) to rotation (as wesaw on Wednesday)
NET = I
Physics 207: Lecture 17, Pg 22
Rotational Dynamics: What makes it spin?
NET = |r| |FTang| ≡ |r| |F| sin
A constant torque gives constant angular acceleration if and only if the mass distribution and the axis of rotation remain constant.
FTangential a
r
Fradial
F
A force applied at a distance from the rotation axis gives a torque
Physics 207: Lecture 17, Pg 23
Torque, like , is a vector quantity
Magnitude is given by (1) |r| |F| sin
(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”
And for a rigid object = I
r
F
F cos(90°) = FTang.
r
FFradial
F
a
r
F
r 90°
r sin line of action
Physics 207: Lecture 17, Pg 24
Example : Rolling Motion
Notice rotation CW (i.e. negative) when ax is positive!Combining 3rd and 4th expressions gives f = Max / 2Top expression gives Max + f = 3/2 M ax = Mg sin So ax =2/3 Mg sin
M
Newton’s Laws:
cos0y MgNMay F
fMgMax sinxF
fRMRI CMCM 221
CM
xCM aR Mg
f
N
x dir
Physics 207: Lecture 17, Pg 29
Statics
Equilibrium is established when
0 motion nalTranslatio Net F
0 motion RotationalNet
In 3D this implies SIX expressions (x, y & z)
Physics 207: Lecture 17, Pg 30
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:
0 motion RotationalNet
Physics 207: Lecture 17, Pg 31
Example: Soln.
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
0 UseNet
30 kg
1 m
60 kg30 kg
0.5 m
300 N 300 N 600 N
N
Physics 207: Lecture 17, Pg 32
Recap
Assignment: HW7 due tomorrow
Wednesday: review session