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Physics 2211: Lecture 38. Rolling Motion Mass rolling down incline plane Energy solution Newton’s 2nd Law solution Angular Momentum. Translational & rotational motion combined. For a solid object which rotates about its center or mass and whose CM is moving:. V CM. . R. R. v cm. - PowerPoint PPT Presentation
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Lecture 38, Page 1 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Physics 2211: Lecture 38Physics 2211: Lecture 38
Rolling Motion Mass rolling down incline plane
Energy solution Newton’s 2nd Law solution
Angular Momentum
Lecture 38, Page 2 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
For a solid object which rotates about its center or mass and whose CM is moving:
VCM
Translational & rotational motion Translational & rotational motion combinedcombined
2 21 12 2NET CM CMK MV I
Lecture 38, Page 3 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Rolling MotionRolling Motion
Rolling without slipping
R
s
R
svcm
s Rds d
R Rdt dt
But CM
dsv
dt
CMv R
Lecture 38, Page 4 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Rolling MotionRolling Motion
Objects of different I rolling down an inclined plane: find their speed at the bottom of the plane and their acceleration on the plane.
h
R
v = R
M
2 21 12 2E Mv Mgy I
2 21 12 2finalE Mv I
v = 0
= 0
K = 0initalE Mgh
Lecture 38, Page 5 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
12
1
v gh
c
The rolling speed is always lower than in the case of simple sliding
since the kinetic energy is shared between CM motion and rotation.
2 21 12 2finalE Mv I
2
2 21 12 2
final
vE MR Mv
Rc 21
2 1 Mvc
final initialE E
212 1 Mvc Mgh
Use v = R and I = cMR2
1 hoop
1/2 disk
2/5 spherec =
Rolling MotionRolling Motion
Lecture 38, Page 6 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
2 2v a x
12
1
v gh
c
v
h
R
M
2sin
ha
sin
1
g
ac
1 hoop
1/2 disk
2/5 spherec =
Rolling MotionRolling Motion
x
Lecture 38, Page 7 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
An object with mass M, radius R, and moment of inertia I rolls without slipping down a plane inclined at an angle with respect to horizontal. What is its acceleration?
Consider CM motion and rotation about the CM separately when solving this problem (like we did with the lastproblem)...
R
I
M
Rolling MotionRolling Motion(alternate approach)(alternate approach)
Lecture 38, Page 8 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Static friction f causes rolling. It is an unknown, so we must solve for it.
First consider the free body diagram of the object and use FNET = MaCM . In the x direction
RM
y
x
Mg
f
sinMg
sin MMg af
Now consider rotation about the CMand use = I realizing that
= Rf and a = R
Rolling MotionRolling Motion(alternate approach)(alternate approach)
Ia
RfR
I2
af
R
Lecture 38, Page 9 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
We have two equations:
aR
I
M
I2
af
R
I
2
2
sinMRa g
MR
For a sphere, c = :
5
725
sinsin
1
ga g
25
2sin
1
I cc
ga for MR
We can combine these to eliminate f:
Rolling MotionRolling Motion(alternate approach)(alternate approach)
sin - Mg f ma
Lecture 38, Page 10 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
ExampleExampleRotationsRotations
Two uniform cylinders are machined out of solid aluminum. One has twice the radius of the other.If both are placed at the top of the same ramp and released, which is moving faster at the bottom?
(1) bigger one
(2) smaller one
(3) same
Lecture 38, Page 11 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
ExampleExample SolutionSolution
h
12
1
v gh
c
1 hoop
1/2 disk
2/5 spherec =
Answer:
(3) same
speed does not depend on size,
as long as the shape is the same!!
43v gh
Lecture 38, Page 12 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Direction of Rotational MotionDirection of Rotational Motion In general, the rotation variables (etc) are vectors
(have direction) If the plane of rotation is in the x-y plane, then the convention is
x
y
z
x
y
z
CCW rotation is in the + z direction
CW rotation is in the - z direction
Lecture 38, Page 13 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Direction of Rotation:Direction of Rotation:The Right Hand RuleThe Right Hand Rule
To figure out in which direction the rotation vector points, curl the fingers of your right hand the same way the object turns, and your thumb will point in the direction of the rotation vector!
We normally pick the z-axis to be the rotation axis as shown.= z
= z
= z
For simplicity we omit the subscripts unless explicitly needed.
x
y
z
x
y
z
Lecture 38, Page 14 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Angular momentum of a rigid bodyAngular momentum of a rigid bodyabout a fixed axis:about a fixed axis:
Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momenta of each particle:
i
j m2
m1
m3
1r
3r2r
3v
1v
2v
We see that is in the z direction.L
Using vi = ri , we get
2 ˆi i
i
L m r k
Analog of !IL
p mv
ˆi i i i i i i i
i i i
L r p m r v m r v k (since and are
perpendicular) iv
ir
Lecture 38, Page 15 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Angular momentum of a rigid bodyAngular momentum of a rigid bodyabout a fixed axis:about a fixed axis:
In general, for an object rotating about a fixed (z) axis we can write LZ = I
The direction of LZ is given by theright hand rule (same as ).
We will omit the Z subscript for simplicity,and write L = I
z
ZL I
Lecture 38, Page 16 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Conservation of Angular MomentumConservation of Angular Momentum
Total angular momentum is conservedTotal angular momentum is conserved
where EXT
dL
dt
L I
In the absence of external torquesIn the absence of external torques 0EXT
dL
dt
Lecture 38, Page 17 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Two DisksExample: Two Disks
A disk of mass M and radius R rotates around the z axis with angular velocity i. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity f.
i
z
f
z
Lecture 38, Page 18 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Two DisksExample: Two Disks
First realize that there are no external torques acting on the two-disk system.Angular momentum will be conserved!
Initially, the total angular momentum is due only to the disk on the bottom:
i
z
2
1I 211 2i i iL MR
Lecture 38, Page 19 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Two DisksExample: Two Disks
First realize that there are no external torques acting on the two-disk system.Angular momentum will be conserved!
Finally, the total angular momentum is dueto both disks spinning:
f
z
21
I I 2 2 21 11 2 2 2f f f f fL MR MR MR
or
I 2 212 2f f f f fL M R MR
Lecture 38, Page 20 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Two DisksExample: Two Disks
Since Li = Lf
z
f
z
Li Lf
if 2
1
An inelastic collision, since K is not conserved (friction)!
I 2 2 2 2 21 1 1 12 2 2 82f f f f iK M R MR
I 2 2 2 2 21 1 1 12 2 2 4i i i I iK MR MR
2 212 i fMR MR
Lecture 38, Page 21 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Rotating TableExample: Rotating Table
A student sits on a rotating stool with his arms extended and a weight in each hand. The total moment of inertia is Ii, and he is rotating with angular speed i. He then pulls his hands in toward his body so that the moment of inertia reduces to If. What is his final angular speed f?
i
Ii
f
If
Lecture 38, Page 22 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Rotating TableExample: Rotating Table
Again, there are no external torques acting on the student-stool system, so angular momentum will be conserved.Initially: Li = Iii
Finally: Lf = If f f
i
i
f
I
I
i
Ii
f
If
LiLf
Lecture 38, Page 23 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
ExampleExampleAngular MomentumAngular Momentum
A student sits on a freely turning stool and rotates with constant angular velocity 1. She pulls her arms in, and due to angular momentum conservation her angular velocity increases to 2. In doing this her kinetic energy:
(1) increases (2) decreases (3) stays the same
1 2
I2 I1
L L
Lecture 38, Page 24 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
ExampleExample SolutionSolution
I2
LI
21
K2
2 (using L = I)
L is conserved and
I2 < I1 K2 > K1 K increases!
1 2
I2 I1
L L
Lecture 38, Page 25 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
ExampleExample SolutionSolution
Since the student has to force her arms to move toward her body, she must be doing positive work! The work/kinetic energy theorem states that this will increase the kinetic energy of the system!
1
I1
2
I2
L L
Lecture 38, Page 26 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Angular Momentum of aAngular Momentum of aFreely Moving ParticleFreely Moving Particle
We have defined the angular momentum of a particle about the origin as
This does not demand that the particle is moving in a circle!We will show that this particle has a constant angular
momentum!
y
x
L r p
v
Lecture 38, Page 27 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Angular Momentum of aAngular Momentum of aFreely Moving ParticleFreely Moving Particle
Consider a particle of mass m moving with speed v along the line y = -d. What is its angular momentum as measured from the origin (0,0)?
x
md
y
v
Lecture 38, Page 28 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Angular Momentum of aAngular Momentum of aFreely Moving ParticleFreely Moving Particle
We need to figure out The magnitude of the angular momentum is:
Since and are both in the x-y plane, will be in the z direction (right hand rule):
y
xd
L pdZ
L r p
sin sinL r p rp p r pd
p x dis tance of closest approach
r
p
L
r
p mv
Lecture 38, Page 29 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Angular Momentum of aAngular Momentum of aFreely Moving ParticleFreely Moving Particle
So we see that the direction of is along the z axis, and its magnitude is given by LZ = pd = mvd. L is clearly conserved since d is constant (the distance of closest approach of the particle to the origin) and p is
constant (linear momentum conservation).
y
xd
L
p
Lecture 38, Page 30 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Bullet hitting stickExample: Bullet hitting stick
A uniform stick of mass M and length D is pivoted at the center. A bullet of mass m is shot through the stick at a point halfway between the pivot and the end. The initial speed of the bullet is v1, and the final speed is v2.
What is the angular speed F of the stick after the collision? (Ignore gravity)
v1 v2
M
F
initial final
mD
D/4
Lecture 38, Page 31 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Bullet hitting stickExample: Bullet hitting stick
Conserve angular momentum around the pivot (z) axis! The total angular momentum before the collision is due
only to the bullet (since the stick is not rotating yet).
v1
D
M
initial
D/4m
4
DmvapproachclosestofcedisxpL 1i )tan(
Lecture 38, Page 32 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Bullet hitting stickExample: Bullet hitting stick
Conserve angular momentum around the pivot (z) axis! The total angular momentum after the collision has
contributions from both the bullet and the stick.
where I is the moment of inertia of the stick about the
pivot.
v2
F
final
D/4
I2 4f F
DL mv
Lecture 38, Page 33 Physics 2211 Spring 2005© 2005 Dr. Bill Holm
Example: Bullet hitting stickExample: Bullet hitting stick
Set Li = Lf and using
v1 v2
M
F
initial final
mD
D/4
I 1
122MD
21 2
1
4 4 12 F
D Dmv mv MD 1 2
3F
mv v
MD