Physics 121,Lecture 18,Pg 1
Physics 121,Sections 9,10,11,and 12
Lecture 18
Today’s Topics:
? Homework 7,due Friday Nov,4 @ 6:00 PM.
? Chap,7,# 3,11,20,21,25,27,30,40,
46,47,52,and 68.
? Chapter 8:
? Rotation
? Moment of inertia
? Rolling motion
Physics 121,Lecture 18,Pg 2
Summary
(with comparison to 1-D kinematics)
Angular Linear
c o n s ta n t??
??
? ? ? 0 ??t
? ? ? ?? ? ?0 0 212t t
co n st a n ta ?
v v at? ?0
x x v t at? ? ?0 0 212
And for a point at a distance R from the rotation axis:
x = R????????????v = ?R ??????????a = ?R
Physics 121,Lecture 18,Pg 3
Rotation & Kinetic Energy...
? The kinetic energy of a rotating system looks similar to that of a
point particle:
Point Particle Rotating System
2I
2
1 ??KE
I ? ? m ri i
i
2
2
2
1 mvKE ?
v is,linear” velocity
m is the mass.
? is angular velocity
I is the moment of inertia
about the rotation axis.
Physics 121,Lecture 18,Pg 4
Moment of Inertia
? Notice that the moment of inertia I depends on the
distribution of mass in the system.
?The further the mass is from the rotation axis,the bigger
the moment of inertia.
? For a given object,the moment of inertia will depend on
where we choose the rotation axis (unlike the center of
mass).
? We will see that in rotational dynamics,the moment of inertia
I appears in the same way that mass m does when we study
linear dynamics !
2I
2
1 ??KE
I ? ? m ri i
i
2
? So where
Physics 121,Lecture 18,Pg 5
Calculating Moment of Inertia
? We have shown that for N discrete point masses distributed
about a fixed axis,the moment of inertia is:
I ?
?
? m ri i
i
N
2
1where r is the distance from the mass to the axis of rotation.
Example,Calculate the moment of inertia of four point masses
(m) on the corners of a square whose sides have length L,
about a perpendicular axis through the center of the square:
mm
mm
L
Physics 121,Lecture 18,Pg 6
Calculating Moment of Inertia...
? The squared distance from each point mass to the axis is:
mm
mm
L
r
L/2
2
L
2
L2r 222 ??
?
??
?
??
2
Lm4
2
Lm
2
Lm
2
Lm
2
LmrmI 22222N
1i
2ii ???????
?
so
I = 2mL2
Using the Pythagorean Theorem
Physics 121,Lecture 18,Pg 7
Calculating Moment of Inertia...
? Now calculate I for the same object about an axis through
the center,parallel to the plane (as shown):
mm
mm
L
r
4
Lm4
4
Lm
4
Lm
4
Lm
4
LmrmI 22222N
1i
2ii ???????
?
I = mL2
Physics 121,Lecture 18,Pg 8
Calculating Moment of Inertia...
? Finally,calculate I for the same object about an axis along
one side (as shown):
mm
mm
L
r
2222N
1i
2ii 0m0mmLmLrmI ??????
?
I = 2mL2
Physics 121,Lecture 18,Pg 9
Calculating Moment of Inertia...
? For a single object,I clearly depends on the rotation axis !!
L
I = 2mL2I = mL2
mm
mm
I = 2mL2
Physics 121,Lecture 18,Pg 10
Lecture 18,Act 1
Moment of Inertia
? A triangular shape is made from identical balls and identical
rigid,massless rods as shown,The moment of inertia about
the a,b,and c axes is Ia,Ib,and Ic respectively.
?Which of the following is correct:
(a) Ia > Ib > Ic
(b) Ia > Ic > Ib
(c) Ib > Ia > Ic
a
b
c
Physics 121,Lecture 18,Pg 11
Calculating Moment of Inertia...
? For a discrete collection of point
masses we found:
? For a continuous solid object we have to add up the mr2
contribution for every infinitesimal mass element dm.
?We have to do an
integral to find I,
I ?
?
? m ri i
i
N
2
1
r
dm
I ? ? r dm2
Physics 121,Lecture 18,Pg 12
Moments of Inertia
? Some examples of I for solid objects:
Thin hoop (or cylinder) of mass M and
radius R,about an axis through its center,
perpendicular to the plane of the hoop.
I ? MR 2
R
I ? 12 2MR
Thin hoop of mass M and radius R,
about an axis through a diameter.
R
Physics 121,Lecture 18,Pg 13
Moments of Inertia...
? Some examples of I for solid objects:
Solid sphere of mass M and radius R,
about an axis through its center.
I ? 25 2MR
R
Thin spherical shell of mass M
and radius R,about an axis
through its center.
R??
I ? 23 MR 2
I ? 12 2MRR Solid disk or cylinder of mass M and radius R,about a perpendicular axis
through its center.
Physics 121,Lecture 18,Pg 14
Lecture 18,Act 2
Moment of Inertia
? Two spheres have the same radius and equal masses,One
is made of solid aluminum,and the other is made from a
hollow shell of gold.
?Which one has the biggest moment of inertia about an
axis through its center?
same mass & radius
solid hollow
(a) solid aluminum (b) hollow gold (c) same
Physics 121,Lecture 18,Pg 15
Moments of Inertia..,Other examples
? Some examples of I for solid objects
Thin rod of mass M and length L,about
a perpendicular axis through its center.
I ? 112 2ML
L
Thin rod of mass M and length L,about
a perpendicular axis through its end.
I ? 13 2ML
L
Physics 121,Lecture 18,Pg 16
Parallel Axis Theorem
? Suppose the moment of inertia of a solid object of
mass M about an axis through the center of mass
is known,= ICM
? The moment of inertia about an axis parallel to this
axis but a distance R away is given by:
IPARALLEL = ICM + MR2
? So if we know ICM,it is easy to calculate the
moment of inertia about a parallel axis.
Physics 121,Lecture 18,Pg 17
Parallel Axis Theorem,Example
? Consider a thin uniform rod of mass M and length D,Figure
out the moment of inertia about an axis through the end of
the rod.
IPARALLEL = ICM + MD2
L
D=L/2 M
xCM
ICMIEND
I EN D ML M L ML? ? ??? ??? ?112 2 132
2
2
So
I CM ML? 112 2
We know
We get the same result if evaluating as in the previous slide.
Physics 121,Lecture 18,Pg 18
Lecture 18,Example 1
? A rope is wrapped around the circumference of a solid disk
(R=0.2m) of mass M=10kg and an object of mass m=10 kg is
attached to the end of the rope 10m above the ground,as shown
in the figure.
a) How long will it take for the object to hit the ground?
b) What will be the velocity of the object when it hits the ground?
c) What is the tension on the cord? ?
M
m
h =10 m
T
Physics 121,Lecture 18,Pg 19
Lecture 18,Example 1
Solution?
M
m
h =10 m
T
v2 - vo2 = 2 a h
v = (2 a h) 1/2
v = a / t
t = v / a
mg - T = a m
a = R ? I ? = t =T R
T = I ? / R = (1/2) MR2 a / R2
T = Ma/2
mg - Ma/2 = a m
a = mg / (m+M/2) = 2g/3
a = 2g/3 = 2 x 9.8 m/s2 / 3 = 6.5 m/s2
v = (2 a h)1/2 =(2 x 6.5m/s2 10m)1/2 =11m/s
t = v / a = 11 m/s / 6.5 m/s2 = 1.7 s
T = Ma/2 = 10kg 6.5m/s2 / 2 = 32 N
Physics 121,Lecture 18,Pg 20
Connection with CM motion
? Recall what we found out about the kinetic energy of a system of
particles:
??
K T O T ? 12 m i u i2? ? 12 MV CM2
= KREL = KCM
? For a solid object rotating about it’s center of mass,we
now see that the first term becomes:
??
K R E L ? 12 m i u i 2? Substituting ii ru ??
??
K R E L ? 12 ? 2 m i r i? 2but CM
2ii rm I??
22 2121 CMCM MVIK ?? ?
Physics 121,Lecture 18,Pg 21
Connection with CM motion...
? So for a solid object which rotates about its center of mass and
whose CM is moving:
K MVT O T CM CM? ?12 122 2I ?
?
VCM
Active Figure
Physics 121,Lecture 18,Pg 22
Rolling Motion
? Now consider a cylinder rolling at a constant speed,
K MVT O T CM CM? ?12 122 2I ?
VCM CM
The cylinder is rotating about CM and its CM is moving at constant
speed (vCM),Thus its total kinetic energy is given by,
Physics 121,Lecture 18,Pg 23
Rolling Motion
? Consider again a cylinder rolling at a constant speed,
VCM
P
Q
CM
?
At any instant the cylinder is rotating about point P,Its kinetic
energy is given by its rotational energy about that point,
KTOT = 1/2 IP ?2
Physics 121,Lecture 18,Pg 24
Rolling Motion
? We can find IP using the parallel axis theorem
K MVT O T CM CM? ?12 122 2I ?
VCM
P
Q
CM
?
IP = ICM + MR2
KTOT = 1/2 (ICM + MR2 ) ?2
KTOT = 1/2 ICM ?2 + 1/2 M (R2?2 ) = 1/2 ICM ?2 + 1/2 M vCM2 !
Physics 121,Lecture 18,Pg 25
Rolling Motion
? Cylinders of different I rolling down an inclined plane:
h
v = 0
??= 0
K = 0
R
?K = - ?U = Mgh
v = ?R
M K MV
T O T CM CM? ?12 122 2I ?
Physics 121,Lecture 18,Pg 26
Rolling...
? If there is no slipping (due to friction):
v 2v
In the lab reference frame
v
In the CM reference frame
v
Where v = ?R
?
Physics 121,Lecture 18,Pg 27
Rolling...
??
K T O T ? 12 MR 2 ? 2 ? 12 Mv 2 ? 12 ? 1? ? Mv 2
Use v= ?R and I = cMR2,
So,( ) M ghMv1
2
1 2 =+ 11gh2v +=
The rolling speed is always lower than in the case of simple
Sliding since the kinetic energy is shared between CM motion
and rotation.
hoop,c=1
disk,c=1/2
sphere,c=2/5
etc...
c c
c c
K MVT O T CM CM? ?12 122 2I ?
Physics 121,Lecture 18,Pg 28
Example, Rolling Motion
? A cylinder is about to roll down an inclined plane,What is its speed at the
bottom of the plane?
M
?
h
M v?
Ball has radius R
Physics 121,Lecture 18,Pg 29
Example, Rolling Motion
? Use conservation of energy.
Ei = Ui + 0 = Mgh
Ef = 0 + Kf = 1/2 Mv2 + 1/2 I ?2
= 1/2 Mv2 + 1/2 (1/2MR2)(v/R)2
Mgh = 1/2 Mv2 + 1/4 Mv2
v2 = 4/3 g h
v = ( 4/3 g h ) 1/2
Physics 121,Lecture 18,Pg 30
Lecture 18,ACT 3
Rolling Motion
? A race !!
Two cylinders are rolled down a ramp,They have the same
radius but different masses,M1 > M2,Which wins the race to
the bottom?
A) Cylinder 1
B) Cylinder 2
C) It will be a tie
M1
?
h
M?
M2
Active Figure
Physics 121,Lecture 18,Pg 31
Remember our roller coaster.
Perhaps now we can get the ball to go
around the circle without anyone dying.
Note:
Radius of loop = R
Radius of ball = r
Physics 121,Lecture 18,Pg 32
How high do we have to start the ball?
Use conservation of energy.
Also,we must remember that the minimum
speed at the top is vtop = (gR)1/2
E1 = mgh + 0 + 0
E2 = mg2R + 1/2 mv2 + 1/2 I?2
= 2mgR + 1/2 m(gR) + 1/2 (2/5 mr2)(v/r)2
= 2mgR + 1/2 mgR + (2/10)m (gR)
= 2.7 mgR
1
2
Physics 121,Lecture 18,Pg 33
How high do we have to start the ball?
E1 = mgh + 0 + 0
E2 = 2.7 mgR
mgh = 2.7 mgR
h = 2.7 R
h = 1.35 D
(The rolling motion added an extra 2/10 R to the
height)
1
2
Physics 121,Lecture 18,Pg 34
Recap of today’s lecture
? Homework 7,due Friday Nov,4 @ 6:00 PM.
?Chap,7,# 3,11,20,21,25,27,30,40,
46,47,52,and 68.
? Chapter 8:
?Rotation
?Moment of inertia
?Rolling motion