1. PHYS 1443 – Section 003 Lecture #11
Monday, Oct. 21, 2002
Dr. Jaehoon Yu
Collisions in Two Dimension
Center of Mass
Definition
CM of a Rigid Object
Center of Mass and Center of Gravity
Motion of a Group of Particles
Rocket Propulsion
Fundamentals on Rotation
Today’s homework is homework #12, due 12:00pm, next Monday!!
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

2. Announcements 2nd Term exam
Wednesday, Oct. 30, in the class
Covers chapters 6 – 10
Mixture of Multiple choice and Essay problems
Review on Monday, Oct. 28
Magda Cortez, please come and talk to me after the class
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

3. Two dimensional Collisions
In two dimension, one can use components of momentum to apply momentum conservation to solve physical problems. m1
v1i
x-comp. m2
y-comp. m1
v1f q
Consider a system of two particle collisions and scatters in two dimension as shown in the picture. (This is the case at fixed target accelerator experiments.) The momentum conservation tells us: m2
v2f f
What do you think we can learn from these relationships?
And for the elastic conservation, the kinetic energy is conserved:
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

4. Example 9.9
A 1500kg car traveling east with a speed of 25.0 m/s collides at an interaction with a 2500kg van traveling north at a speed of 20.0 m/s. After the collision the two cars stuck to each other, and the wreckage is moving together. Determine the velocity of the wreckage after the collision, assuming the vehicles underwent a perfectly inelastic collision. m1
v1i
The initial momentum of the two car system before the collision is m2
v2i m1 vf m2
The final momentum of the two car system after the perfectly inelastic collision is
Using momentum conservation
X-comp.
Y-comp.
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

5. Example 9.10
Proton #1 with a speed 3.50x105 m/s collides elastically with proton #2 initially at rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and proton #2 deflects at an angle f to the same axis. Find the final speeds of the two protons and the scattering angle of proton #2, f. m1
v1i
Since both the particles are protons m1=m2=mp.
Using momentum conservation, one obtains m2 m1
v1f q
x-comp.
y-comp. m2
v2f f
Canceling mp and put in all known quantities, one obtains
From kinetic energy conservation:
Solving Eqs. 1-3 equations, one gets
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

6. Center of Mass
We’ve been solving physical problems treating objects as sizeless points with masses, but in realistic situation 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 the point.
The total external force exerted on the system of total mass M causes the center of mass to move at an acceleration given by as if all the mass of the system is concentrated on the center of mass.
What does above statement tell you concerning forces being exerted on the system? m1 m2 x1 x2
Consider a massless rod with two balls attached at either end.
xCM
The position of the center of mass of this system is the mass averaged position of the system
CM is closer to the heavier object
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

7. Center of Mass of a Rigid Object
The formula for CM can be expanded to Rigid Object or a system of many particles
The position vector of the center of mass of a many particle system is
A rigid body – an object with shape and size with mass spread throughout the body, ordinary objects – can be considered as a group of particles with mass mi densely spread throughout the given shape of the object
Dmi ri
rCM
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

8. Center of Mass and Center of GravityCM
Axis of symmetry
The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry, if object’s mass is evenly distributed throughout the body.
One can use gravity to locate CM.
Hang the object by one point and draw a vertical line following a plum-bob.
Hang the object by another point and do the same.
The point where the two lines meet is the CM.
How do you think you can determine the CM of objects that are not symmetric?
Since a rigid object can be considered as collection of small masses, one can see the total gravitational force exerted on the object as
Center of Gravity
Dmi
Dmig
The net effect of these small gravitational forces is equivalent to a single force acting on a point (Center of Gravity) with mass M.
What does this equation tell you?
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

9. Example 9.12
A system consists of three particles as shown in the figure. Find the position of the center of mass of this system.
Using the formula for CM for each position vector component m1
y=2
(0,2)
(0.75,4)
rCM m2
x=1
(1,0) m3
x=2
(2,0)
One obtains If
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

10. Example 9.13
Show that the center of mass of a rod of mass M and length L lies in midway between its ends, assuming the rod has a uniform mass per unit length.
The formula for CM of a continuous object is L x dx
dm=ldx
Since the density of the rod (l) is constant;
The mass of a small segment
Therefore
Find the CM when the density of the rod non-uniform but varies linearly as a function of x, l=a x
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

11. Motion of a Group of Particles
We’ve learned that the CM of a system can represent the motion of a system. Therefore, for an isolated system of many particles in which the total mass M is preserved, the velocity, total momentum, acceleration of the system are
Velocity of the system
Total Momentum of the system
Acceleration of the system
External force exerting on the system
What about the internal forces?
System’s momentum is conserved.
If net external force is 0
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

12. Rocket Propulsion
What is the biggest difference between ordinary vehicles and a rocket?
The force that gives propulsion for normal vehicles is the friction between the surface of the road and the tire. The system in this case consists of the tire and the surface of the road. Newton’s 3rd law and the momentum conservation of an isolated system.
Since there is no road to push against, rockets obtain propulsion from momentum conservation in the system consists of the rocket and gas from burnt fuel.
M+Dm v
Initial momentum before burning fuel M
v+ Dv Dm
v-vg
Final momentum after burning fuel and ejecting the gas
From momentum conservation
Since dm is the same as –dM, one can obtain
Thrust is the force exerted on the rocket by the ejected gas
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

13. Example 9.18
A rocket moving in free space has a speed of 3.0x103 m/s relative to the Earth. Its engines are turned on, and fuel is ejected in a direction opposite the rocket’s motion at a speed of 5.0x103 m/s relative to rocket. A) What is the speed of the rocket relative to the Earth once its mass is reduced to one-half the mass before ignition?
M+Dm v
Precisely the case we’ve discussed in the previous slide. M
v+ Dv Dm
v -vg
Find the thrust on the rocket if it burns fuel at the rate of 50kg/s?
Since the thrust is given proportional to the rate of mass change or the rate the fuel burns as given in the formula
One can obtain
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

14. Fundamentals on Rotation
Linear motions can be described as the motion of the center of mass with all the mass of the object concentrated on it.
Is this still true for rotational motions?
No, because different parts of the object have different linear velocities and accelerations.
Consider a motion of a rigid body – an object that does not change its shape – rotating about the axis protruding out of the slide. q r P s O
The arc length, or sergita, is
Therefore the angle, q, is And the unit of the angle is in radian.
One radian is the angle swept by an arc length equal to the radius of the arc.
Since the circumference of a circle is 2pr,
The relationship between radian and degrees is
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

15. Angular Displacement, Velocity, and Acceleration
Using what we have learned in the previous slide, how would you define the angular displacement?
How about the average angular speed? qi qf
And the instantaneous angular speed?
By the same token, the average angular acceleration
And the instantaneous angular acceleration?
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, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

16. Rotational Kinematics
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 acceleration, because these are the simplest motions in both cases.
Just like the case in linear motion, one can obtain
Angular Speed under constant angular acceleration:
Angular displacement under constant angular acceleration:
One can also obtain
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

17. Example 10.1
A wheel rotates with a constant angular acceleration pf 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at ti=0, a) through what angle does the wheel rotate in 2.00s?
Using the angular displacement formula in the previous slide, one gets
What is the angular speed at t=2.00s?
Using the angular speed and acceleration relationship
Find the angle through which the wheel rotates between t=2.00 s and t=3.00 s.
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

18. Relationship Between Angular and Linear Quantities
What do we know about a rigid object that rotates about a fixed axis of rotation?
Every particle (or masslet) in the object moves in a circle centered at the axis of rotation. ri P q O x y
When a point rotates, it has both the linear and angular motion components in its motion.
What is the linear component of the motion you see? vt
The direction of w follows a right-hand rule.
Linear velocity along the tangential direction.
How do we related this linear component of the motion with angular component?
The arc-length is
So the tangential speed vt is
What does this relationship tell you about the tangential speed of the points in the object and their angular speed?:
Although every particles in the object has the same angular speed, its tangential speed differs proportional to its distance from the axis of rotation.
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu
The farther away the particle is from the center of rotation, the higher the tangential speed.

19. How about the Accelerations?
How many different linear accelerations do you see in a circular motion and what are they? r P q O x y at
Two a ar
Tangential, at, and the radial acceleration, ar.
Since the tangential speed vt is
The magnitude of tangential acceleration at is
Although every particle in the object has the same angular acceleration, its tangential acceleration differs proportional to its distance from the axis of rotation.
What does this relationship tell you?
The radial or centripetal acceleration ar is
What does this tell you?
The father away the particle from the rotation axis the more radial acceleration it receives. In other words, it receives more centripetal force.
Total linear acceleration is
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu

20. Example 10.2
Audio information on compact discs are transmitted digitally through the readout system consisting of laser and lenses. The digital information on the disc are stored by the pits and flat areas on the track. Since the speed of readout system is constant, it reads out the same number of pits and flats at the same time interval. In other words, the linear speed is the same no matter which track is played. a) Assuming the linear speed is 1.3 m/s, find the anfular speed of the disc in revolutions per minute when the inner most (r=23mm) and outer most track (r=58mm) are read.
Using the relationship between angular and tangential speed
b) The maximum playing time of a standard music CD is 74 minutes and 33 seconds. How many revolutions does the disk make during that time?
c) What is the total length of the track past through the readout mechanism?
d) What is the angular acceleration of the CD over the 4473s time interval, assuming constant a?
Monday, Oct. 21, 2002
PHYS , Fall 2002
Dr. Jaehoon Yu