Monday, Apr. 13, 2009
Dr. Jae hoon Yu
• Center of Mass
• Fundamentals of Rotational Motion
• Equations of Rotational Kinematics
• Relationship Between Linear and Angular
Quantities
• Rolling Motion
Today’s homework is HW #10, due 9pm, Tuesday, Apr. 21!!
Monday, Apr. 13, 2009 PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
1

• 2 nd term exam
– 1 – 2:20pm, Wednesday, Apr. 22, in SH103
– Non-comprehensive exam
– Covers: Ch. 6.1 – what we complete this Wednesday, Apr. 15
(somewhere in Ch. 9)
– A help session in class Monday, Apr. 20 by Humphrey
– One better of the two term exams will be used for final grading
• Reading assignments
– Ch. 7 – 9 and 7 – 10
Monday, Apr. 13, 2009 PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
2

• Derive the formula for the final velocity of two objects which underwent an elastic collision as a function of known quantities m
1
, m
2
, v
01 and v
02 in page 14 of the
Apr. 8 lecture note in a far greater detail than the note.
– 20 points extra credit
• Show mathematically what happens to the final velocities if m motion.
1
=m
2 and describe in words the resulting
– 5 point extra credit
• Due: Start of the class this Wednesday, Apr. 15
Monday, Apr. 13, 2009 PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
3

We’ve been solving physical problems treating objects as sizeless points with masses, but in realistic situations objects have shapes with masses distributed throughout the body.
Center of mass of a system is the average position of the system’s mass and represents the motion of the system as if all the mass is on that point.
What does above statement tell you concerning the forces being exerted on the system?
The total external force exerted on the system of r
 entire mass of the system is on the center of mass.
m
1 x
1 x
CM
Monday, Apr. 13, 2009 m
2 x
2
Consider a massless rod with two balls attached at either end.
The position of the center of mass of this system is the mass averaged position of the system x
CM
 m x
1 1
 m x
2 2 m
1
 m
2
CM is closer to the heavier object
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
4

Diver performs a simple dive.
The motion of the center of mass follows a parabola since it is a projectile motion.
Monday, Apr. 13, 2009
Diver performs a complicated dive.
The motion of the center of mass still follows the same parabola since it still is a projectile motion.
Jaehoon Yu
The motion of the center of mass
5

Thee people of roughly equivalent mass M on a lightweight (air-filled) banana boat sit along the x axis at positions x
1 x
3
=6.0m. Find the position of CM.
=1.0m, x
2
=5.0m, and
Using the formula x
CM for CM

 i
 i m i m i x i

M

1.0
5.0
M M M
Monday, Apr. 13, 2009
6.0
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu

12.0
M
3 M

6

v
 x cm
 cm

 
1 1 m
2
 x
2 m
1
 m
2
 x cm
 t
 m
1
  
1 m m
1
 m
2
2
 x
2
 t
 m v
1 1
 m v
2 2 m
1
 m
2
In an isolated system, the total linear momentum does not change, therefore the velocity of the center of mass does not change.
Monday, Apr. 13, 2009 PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
7

Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54-kg woman and one is a 88-kg man. The woman moves away with a speed of +2.5 m/s. v
10

0 m s v
20

0 m s v
 cm 0 m v
1 1
 m v
2 2 m
1
 m
2

0 v
1 f
 
2.5
m s v
2 f
 
1.5
m s v cmf


54
 m v
1 1 f m
1
 m
2
2.5

 m v
2 2
88
 f
1.5


3
142

Monday, Apr. 13, 2009
0.02

0
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu m s
8

In the simplest kind of rotation, points on a rigid object move on circular paths around an axis of rotation.
The angle swept out by a line passing through any point on the body and intersecting the axis of rotation perpendicularly is called the angular displacement.
     o
It’s a vector!! So there must be directions…
How do we define directions? +:if counter-clockwise
-:if clockwise
The direction vector points gets determined based on the right-hand rule.
These are just conventions!!
Jaehoon Yu
9

Arc length 
(in radians)
 
Radius r s
Dimension? None
For one full revolution:
 
2
 r
How many degrees is in one radian?
360
1 radian is 1 rad
 
1 rad
2

How radians is one degree?
And one degrees is
1

2

360

1 r



180

180

How many radians are in 10.5 revolutions?
Since the circumference of a circle is 2  r

1

1 rad

10.5
rev


3.14
180 o
180 o
3.14

1 o



1 rad
10.5
rev

2

0.0175
rad rad rev


360

57.3
o
21
  rad

PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
10

Synchronous satellites are put into an orbit whose radius is 4.23
× 10 7 m. If the angular separation of the two satellites is 2.00 degrees, find the arc length that separates them.
What do we need to find out? The Arc length!!!

(in radians)

Arc length
Radius
 r s
Convert degrees to radians s

2.00 deg

 360 deg



0.0349 rad r
 

 7
  

1.48 10 m (920 miles)
Monday, Apr. 13, 2009 PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
11

The diameter of the sun is about 400 times greater than that of the moon. By coincidence, the sun is also about 400 times farther from the earth than is the moon. For an observer on the earth, compare the angle subtended by the moon to the angle subtended by the sun and explain why this result leads to a total solar eclipse.

(in radians)

Arc length

Radius r s
I can even cover the entire sun with my thumb!! Why?
Monday, Apr. 13, 2009
Because the distance (r) from my eyes to my
12
Jaehoon Yu

Angular displacement is defined as
    f
  i
How about the average angular velocity, the rate of change of angular displacement?
 
Unit? rad/s Dimension?
[T -1 ]
By the same token, the average angular acceleration, rate of change of the angular velocity, is defined as…
 
Unit? rad/s 2 Dimension?
[T -2 ] t f
 f t f
 f
 i
 f
  i
 t i
  i
 t i


 
 t
 
 t
When rotating about a fixed axis, every particle on a rigid object rotates through the same angle and has the same angular speed and angular acceleration.
Monday, Apr. 13, 2009 PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
13

A gymnast on a high bar swings through two revolutions in a time of 1.90 s. Find the average angular velocity of the gymnast.
What is the angular displacement?
 
2.00 rev
 
12.6 rad

1 rev
Why negative? Because he is rotating clockwise!!
 

12.6 rad
 
6.63rad s
1.90 s
Monday, Apr. 13, 2009 PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
14

As seen from the front of the engine, the fan blades are rotating with an angular speed of -110 rad/s. As the plane takes off, the angular velocity of the blades reaches -330 rad/s in a time of 14 s. Find the angular acceleration, assuming it to be constant.




 f
  i t
 t f i

330 rad s

 
14 s
 
 t
 
110 rad s

 
16 rad s
2
Monday, Apr. 13, 2009 PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
15

The first type of motion we have learned in linear kinematics was under a constant acceleration. We will learn about the rotational motion under constant angular acceleration, because these are the simplest motions in both cases.
Just like the case in linear motion, one can obtain
Angular velocity under constant angular acceleration:
Linear kinematics v
 v o
 at
Angular displacement under constant angular acceleration:
Linear kinematics x
 f x
0
 v t o

1
2 at 2

 f f
  
0
 t
  
0

0 t

1
2
 t
2
One can also obtain
Linear kinematics v
2
 f v
2 o

2
 f
 x i


2 f
 
2
0

2
 f
 
0

Monday, Apr. 13, 2009 PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
16

• Visualize the problem by drawing a picture
• Decide which directions are to be called positive (+) and negative (-).
• Write down the values that are given for any of the five kinematic variables and convert them to SI units.
• Verify that the information contains values for at least three of the five kinematic variables. Select the appropriate equation.
• When the motion is divided into segments, remember that the final angular velocity of one segment is the initial velocity for the next.
• Keep in mind that there may be two possible answers to a kinematics problem.
Monday, Apr. 13, 2009 PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
17

The blades are whirling with an angular velocity of
+375 rad/s when the “puree” button is pushed in.
When the “blend” button is pushed, the blades accelerate and reach a greater angular velocity after the blades have rotated through an angular displacement of +44.0 rad. The angular acceleration has a constant value of +1740 rad/s 2 . Find the final angular velocity of the blades.
θ
+44.0rad
α
+1740rad/s 2
ω
?
ω o
+375rad/s t
Which kinematics eq?

2  
2 o

2

   
2 o

2

  
375 rad s

2 

2
 
44.0rad
  
542 rad s
Which sign?
  
542rad s
Monday, Apr. 13, 2009
Why? Because it is accelerating in counter-clockwise!
PHYS 1441-002, Spring 2009 Dr.
Jaehoon Yu
18