1. Class Notes for Accelerated Physics
HORIZONTAL CIRCLES
Class Notes for Accelerated Physics
Mr. Lyzinski
CRHS-South

2. Types of Circular Motion Problems
Basic
Horizontal
Vertical
Velocity, Period, Frequency, Centripetal acceleration questions.
Objects (attached to a rope or string) traveling in horizontal circles.
Cars traveling around flat, circular turns.
Cars traveling around banked, circular turns (no friction).
Cars traveling around banked, circular turns (with friction).
Swing Ride Problems
Objects (attached to a rope or string) traveling in vertical circles.
Bike or car doing a loop-de-loop.
Airplane doing a loop de loop (positive g’s)
Airplane doing a loop de loop (negative g’s)
Vehicles going over woop-de-doos

3. The Equations of UNIFORM Circular Motion
period (sec): frequency (Hz): tangential velocity (m/s): centripetal acceleration (m/s2):
(because the velocity is constant.)
If you see RADIANS  in the Holt Physics book, convert to REVS by using the following conversion:

4. Centripetal vs. Centrifugal
Imagine swinging a rock on a rope. A moving object tends to “remain in motion” and move in a straight line (Newton’s 3rd Law!!!!). In order to make the rock move out of a straight line into a circle, you have to keep pulling on the rope. The inward pulling force is the "centripetal" force. But your hand feels the rope pulling it outward. This outward pulling force is called the "centrifugal" force.
HOWEVER, the centrifugal force doesn’t really exist. It is simply the result of the object wanting to remain in motion on its original path (due to inertia). NEVER, EVER use the term or the notion of centrifugal force.
Centripetal force is always directed toward the center of the curve. If centrifugal force did exist (which it doesn’t), it would always be directed away from the center.
Note: Centrifugal means " to flee the center" and Centripetal means "to seek the center"

5. HORIZONTAL CIRCLES
In all examples, the positive direction will be chosen as “INTO THE CENTER” of the circle.

6. A ball on a string swinging in a horizontal circle (on a frictionless table)m
T supplies the centripetal force that keeps the mass in its circular orbit

7. FBD for …. A ball on a string swinging in a horizontal circle on a frictionless table
Fg = W = mg T ac FN R y x

8. A car traveling around a circular track (or curve)Ff m
Friction supplies the centripetal force that keeps the car from slipping off the track (curve)
Continued on next page 

9. FBD for …. A car traveling around a circular track (or curve)mg R Ff FN
The car “wants” to go this way (because of its inertia, it wants to continue in the “straight line” path)
The frictional force supplies the centripetal force necessary to keep the car in the circular path. ac y x

10. mg R Ff FN ac
This is the maximum velocity that a car can travel around the circular path without slipping.
You can also use the following formula to find the minimum coefficient of friction for no slip at a certain velocity.

11. A car traveling around a circular track with a banked turn
Continued on next page 

12. FBD for …. A car traveling around a circular track with a banked turnmg R FN Ff
FN,x
FN,y
Ff,x
Ff,y y x

13. Cars traveling around frictionless banked turnsmg R FN
The normal force (or, rather, a component of the normal force) supplies the centripetal force necessary to keep the car in the circular path. FN ac mg
FN,x
FN,y
Faster slides up
Slower slides down

14. HONORS ONLY!!! Cars traveling around friction(ful ) banked turnsFN ac mg
FN,x
FN,y Ff
Ff,x
Ff,y
Cars traveling around friction(ful ) banked turns y x
HONORS ONLY!!!
Slower slides down

15. HONORS ONLY!!! Cars traveling around friction(ful ) banked turnsFN ac mg
FN,x
FN,y Ff
Ff,x
Ff,y y x
Cars traveling around friction(ful ) banked turns
HONORS ONLY!!!
Faster slides up

16. A planet traveling around another planet
Fg? Is this just “mg”, the force due to gravity? Well …. g = 9.8 m/s2 only holds on the earth’s surface. Soon (next chapter), we will find out how to calculate Fg in different places and for different planets . R Fg m M
The force of gravitational attraction supplies the centripetal force needed to maintain the circular motion.

17. Swing Rides !!!!!

18. FBD for …. a “Simplified” swing ride (a conical pendulum)y r
Fg = mg T ac y x

19. “Simplified” Swing Rides
Fg = mg T ac y x
“Simplified” Swing Rides l y r