1. Wednesday(09/23): Introduction to Chap. 6, Vectors, and Projectiles.

HW: Read pages 147-154 and then solve prob. 51, 53, 56, and 61 on

pages 165-6.

2. Thursday(09/24): Solving Projectile Motion Problems. HW: Read

pages 154-159 and solve prob. 62, 63, 68, 70, and 78 on page 166-7.

3. Friday(09/25): The Inclined Plane. Uniform Circular Motion.

HW: Complete Review Handout.

4. Monday(09/28): No School - Fall Holiday. HW: Continue to work on

all assignments.

Homecoming Week. Dress Up!

5. Tuesday(09/29): (Dress Sailor) Lab on Projectile Motion.

HW: Process Lab Data.

6. Wednesday(09/30): (Dress Beachware) Post-Lab Discussion and

Problem-solving. HW: Complete Lab Report and write Abstract (due on

Friday).

7. Thursday(10/01): (Jammin' in Jamaica - Class Colors) Finish Post-

Lab Discussion. HW: Complete Lab Report and write Abstract

(due on Friday).

8. Friday(10/02): (Bermuda Triangle Blackout) Review I for Ch.6.

HW: Complete Review Handout.

9. Monday(10/05): Review II for Ch.6. HW: Complete Review Handout.

10. Tuesday(10/06): TEST on Ch.6 - Motion in Two Dimensions.

HW: Go to web-site for notes on Ch.7 - Gravitation.

In two-dimensional motion, the horizontal and vertical components of the

motion must be regarded independently. For these two directions we use

x and y, respectively. For example, if an object is projected from the

ground with a velocity v_i at an angle of elevation θ_i , then we can use

SohCahToa to find out how fast it is moving in the x and y directions.

1. An object launched from the ground at some angle θ_i is called a

projectile. The path it travels is an inverted parabola called its trajectory.

A classic example would be the motion of golf ball when struck with a

golf club. Can you think of a few more?

2. The initial velocity in the x direction is v_ix = v_i·cos(θ_i). The

acceleration in the x direction is a_x = 0. The velocity of the object in the

y direction is v_iy = v_i·sin(θ_i). The acceleration is that of gravity which

acts only in the y direction. So, we can say that a_y= g = -9.8 m/s².

3. We still have the five motion formulas from the study of kinematics

developed by Galileo (1564-1642). We know them as:  Δd = v_avg·Δt ,

v_avg = (v_i+v_f)/2 ,     v_f = v_i + a·Δt ,     v_f² = v_i² + 2a·Δd ,

Δd = v_i ·Δt + ½a·Δt² . The task now is to adjust these for the separate

x and y directions.

4. In the absence of air resistance a projectile has a constant horizontal

velocity and a constant downward free-fall acceleration which effects the

vertical velocity, subtracting 9.8m/s from it on the way up, and adding

9.8m/s to it on the way down.

5. Newton's second law applied to a particle moving in uniform circular

motion states that the net force must be toward the center. This is the

Centripetal Force, F_c.

6. Uniform circular motion occurs when an acceleration of constant

magnitude is perpendicular to the tangential velocity and the object

maintains a constant speed but is accelerated toward the center of

the circle.

7. This introduces the concept of centripetal acceleration, a_C = v²/r ,

and, by Newton's second law, centripetal force, F_C = mv²/r .

8. The central force acting on an object that provides the centripetal

acceleration could be have its origin in the following: (i) the force of

gravity (as in satellite motion), (ii) the force of friction (as in a car

rounding a curve), or (iii) a force exerted by a string (motion in a

horizontal circle).

9. In the case of motion in a vertical circle, the force of gravity

provides the tangential acceleration and part or all of the

centripetal acceleration. At the top of the circle, the net Force on the

object is zero so that F_c = F_G which implies mv²/r = mg .  At the bottom

F_NET = mv²/r + mg .

10. In the case of a car rounding an unbanked curve, the force of static

friction is the central force.

11. When the curved roadway is banked at an angle, then the horizontal

component of the normal force is centripetal.

12. A frame of reference is a coordinate system for specifying the precise

location of objects in space. Maybe you have heard the expression, "It

depends on your frame of reference."

13. To two observers moving relative to each other there would not be

agreement on the displacements and velocities of an object in motion

when each is using his/her own frame of reference.

14. For example, a person standing in a moving subway car, and facing

towards the back of the car, drops a book. According to the frame of

reference of the person in the car, the book fell in a straight line to the

floor. An observer standing outside on the subway platform as the car

goes by, sees the book traveling in a parabolic path toward the floor.

15. Therefore, the motion of an object depends on your frame of reference.

This is also occurs when boats travel in moving streams and when planes

encounter moving air masses. Also, sometimes you hear about certain

records in track and field that are not allowed if it is determined that

athletes were "wind aided."

16. And still, we need these steps to solve any problem in Physics:

(i) read the problem and identify the given variables

(ii) determine what you are asked to solve for

(iii) find the correct motion formula to use

(iv) use algebra to isolate the unknown

(v) substitute-in the given information and simplify.

ARCHIVES:  CH. 1     CH. 2&3     CH. 4&5