1. PHYS 172: Modern Mechanics
Fall 2012
Lecture 16 – Multiparticle Systems, Moment of Inertia
Chapter 9.3 – 9.5
Exam 2 tomorrow night 8 pm. Next door in Rm 112
EVENING EXAM II :00-9:30 PM TUES. OCT Room 112
Covers through Chapter 8 and through Lecture 14, inclusive
You may bring one sheet of paper, two sides, with equations, notes, etc. to use during the exam
DEMOS FOR THIS LECTURE:
Two hovercrafts mounted by two students (small weight), they both hold a long stick from two sides and pull together or push away from each other. They move, but center of mass stays. Note – hovercrafts are noisy! They could also throw a heavy ball to each other.
Rolling wheel. This is heavy wheel with an axle and base made of two boards that has parabolic shape on top. The wheel can be placed between the two boards so that it can roll on its axle, it is very slow oscillation since wheel has to spin a lot. We use it in a modified way to demonstrate Clicker questions with lolling and sliding wheel (after them). We will use an additional inclined board and the original wheel base turned upside-down and placed on it (so that boards are straight). We can now roll the wheel down on its outer rim on flat surface of inclined board – fast, but not as fast if it would be sliding since some energy goes into spinning. Then we let it roll on its axle – very small translational energy since most of mgh goes into spin
A slope and cart and a wheel, to show that cart gets down first.

2. Rotational Kinetic Energy
Consider a rigid system rotating on an axis
All atoms are rotating at the same “angular speed”

3. Moment of Inertia
Note that the presumed AXIS of rotation passes through the point where all the “r┴” lines emerge, and AT RIGHT ANGLES TO THE PLANE of the figure. r┴ are NOT 3-Dimensional distances, they are “closest distances to the CHOSEN rotation axis” and depend both on the direction and the placement of the rotation axis.

4. Some Moments of Inertia About the specified axes, and about the CM
A disc is just a very short cylinder h
Might want to do a derivation or two
NOTE THAT ABOUT AN AXIS ALONG THE CENTER OF THE CYLINDER, OBJECT (1) HAS A MOMENT OF INERTIA OF ½MR2
About the blue axis, the disc has I = 1/12 Mh2 + ¼ MR2

5. Rigid Rotation about a Point Not the Center of Mass
In General
In this case
What is Ktrans?

6. What is Ktrans?
Consider limits: r_cm=0, and that of a point particle (I_cm = 0)

7. Moment of Inertia for off-center axis
For any object, if you choose a different axis, the moment of inertia changes. Even if the new axis is parallel to the old one, as the axis moves a distance D = rCM from the CM, the effective moment of inertia of the object becomes
I = MD2 + ICM
Consider limits: r_cm=0, and that of a point particle (I_cm = 0) 7

8. Some Moments of Inertia About axes not through the CM
1/12 + 1/4 = 1/3
+1/4 ML2
+MR2
+MR2
Might want to do a derivation or two
About the red axes, just add the red terms to get I 8

9. Point particle system – translational motion
For both, real and point system:
Point particle system: .
See derivation in the book
Note however that a point system cannot have a moment of inertia, and idealizing the top figure to a point is completely invalid for its rotational motion 9

10. Application: Jumping up
Point particle system:
Difference: point particle does not change shape . .
Real system – work done on the system, change in potential energy etc.
Point particle system – work calculated using cm gives only translational kinetic energy
Real system: FN is pushing your feet which do not move! No work is done by floor, BUT…
+lost Chem E
Chemical energy is used by muscles, doing + work on CM. Gravity does negative work on CM. The net (gain from Chem energy –Mgh) appears as KE of the CM. 10

11. Application: Stretching a spring
Real system: .
Point particle system:
Real system – work done on the system, change in potential energy etc.
Point particle system – work calculated using cm gives only translational kinetic energy
In this real system: each force does work, involves displacement of the point to which the force is applied 11

12. Example: hockey pucks
Note carefully that the translational K comes from the CM motion under the influence of FT
This is not necessarily the same amount of motion as the motion of one of the hands (which hand?) d1 d1
Force FN acts on hands, but hands do not move: electric repulsion and reciprocity
Notice motion of a center of mass.
Notice: no sideways motion, momentum principle!
Work: energy must be different! d2 . 12

13. Example: a box containing a spring
System:
Ball with mass mball
Box with mass mbox<<mball
Spring’s mass << mball
M  mball b s
a) How fast is the ball moving immediately after it sticks to the box? cm
sticky
b-s = a
b) What is the increase in thermal energy of the ball?
assume Q= (no heat flow)
= Fs – ½ kss2

14. Example: a box containing a spring
System:
Ball with mass mball
Box with mass mbox<<mball
Spring’s mass << mball
M  mball b s
a) How fast is the ball moving immediately after it sticks to the box? cm
sticky
b-s = a
Let’s look at extreme cases of this example:
ks=0 Then ball won’t move until it hits box! So at that instant, v=0 How long does it take for the box to hit the ball? What’s the impulse?
In the zero mass limit for box and spring, the box will accelerate instantly and (unrealistically) move the distance b immediately. So the impulse on the SYSTEM is zero since Δt is zero. This is consistent with v=0 due to no force acting on the ball (which carries all of the system’s mass)

15. Example: a box containing a spring
System:
Ball with mass mball
Box with mass mbox<<mball
Spring’s mass << mball
M  mball b s
a) How fast is the ball moving immediately after it sticks to the box? cm
sticky
b-s = a
Let’s look at second extreme case of this example:
2. ks is large in relation to F, M, and size of box – such that ball NEVER touches box!! The ball will accelerate on average at a rate F/M
The ball will oscillate with angular frequency sqrt(ks/M) This oscillation energy takes the place of heating the putty. There will also be an average stretch of the spring, a new equilibrium position given by s=F/ks about which the oscillation takes place.

16. Clicker question 1 1 2 Wheel 1 of mass M rolls down from a slope.
Wheel 2 of the same mass M slides down from the same slope (ignore friction)
Which wheel will acquire larger total kinetic energy? 1
Wheel 1 (rolling)
Wheel 2 (sliding)
The same 2 16

17. Clicker question 2 1 2 Wheel 1 of mass M rolls down from a slope.
Wheel 2 of the same mass M slides down from the same slope (ignore friction)
Which wheel will acquire larger translational kinetic energy? 1
Wheel 1 (rolling)
Wheel 2 (sliding)
The same 2 17

18. Clicker question 3 1 2 Wheel 1 of mass M rolls down a slope.
Wheel 2 of the same mass M slides down the same slope (ignore friction)
Which of the wheels will get down first? 1
Show DEMO with rolling wheel on tracks (a lot of spin) and on its outer rim
Wheel 1 (rolling)
Wheel 2 (sliding)
Both will get down in the same time 2 18

19. Center of mass demo: hovercrafts
for v << c
Two kids on howercraft holding a stick on two sides can pull closer or push away – show that center of mass stays at the same point in the absence of external force.
The experiments does not work very well (slope and uneven blow), but it is close and students like it…
Two equal-mass people on frictionless hovercrafts pull on a rope between them. No external forces, in the absence, for example, of unwanted floor friction on one of the hovercraft !! means that the center of mass will not move, and the two people will approach each other meeting halfway. 19

20. Flywheel Energy Storage
Let’s consider a 1 (metric) Ton flywheel of radius R=1 m in the form of a disc. I = ½ 1x103 kg m2
Consider 6000 rpm = 100 Hz revolution rate (60 seconds in a minute)
K = ½ I ω where ω = 200π rad/s = 628 rad/s and vrim = 628 m/s
K = 9.87 x 107 J = 98.7 MJ (~10 sticks of dynamite)
How fast would a 2 Ton car be going with that K = ½ m v2 ?
197MJ/1000kg = v2 v= 444 m/s = 1,600 km/hr
What’s the g force on the rim of this flywheel?
ac = v2/R = 394 kN =40,240 g’s (so the apparent weight of the iron at the rim is 40,240 times it’s actual static weight at the Earth’s surface.)
Let’s hope the material of the flywheel is strong enough. Maybe we want carbon fiber, or nanotube fabric when it becomes available, with a more favorable strength to mass ratio than, say, steel.
Note that the stored energy and the g forces scale proportionally to each other and to v2
Two kids on howercraft holding a stick on two sides can pull closer or push away – show that center of mass stays at the same point in the absence of external force.
The experiments does not work very well (slope and uneven blow), but it is close and students like it… 20