Assume stopper is at constant 2 m/s.

• Is it accelerating.

• Does it have Fnet.

• What causes Fnet?

• Direction of Fnet?

• Direction of acceleration?

• Velocity direction at a point?

Newton’s Newton’s Laws of MotionLaws of Motion::

A net force is necessary A net force is necessary

for acceleration.for acceleration.

Centripetal forceCentripetal force

causes causes CENTRIPETAL ACCELERATIONCENTRIPETAL ACCELERATION. .

ThisThis

acceleration results from a change in acceleration results from a change in

direction,direction,

And not (necessarily) a change in And not (necessarily) a change in

speed,speed,

Uniform Circular Motion

• Object moves in curved/circular path at constant speed & constant acceleration.

Fc needed to keep the object moving in a curved or circular path. If no other forces

Fc = Fnet

• Any force that causes a curved path is called a centripetal force.

• The term centripetal force simply means a Fnet that causes curved motion (hence acceleration).

What forces provide centripetal acceleration for the following?

• - planets orbiting the sun- planets orbiting the sun

• - car rounding a curve- car rounding a curve

• - swinging a mass on a string- swinging a mass on a string

Vector Directions

Velocity Direction what happens when force is turned off. Cut

string.

• http://www.mrwaynesclass.com/teacher/circular/TargetPractice/home.html

Equations of Circular Motion

Objects move with circular velocity m/s or cm/s.

The path is circular or curved with radius r.

Since uniform motion has constant speed.

• v = d/t.

Since the path is circular &:

vc = dt

vc = 2r.

T

R = radius m or cm.

T = period of revolution (s)

Circular vc = Tangential speed or velocity

The formula for The formula for centripetal accelerationcentripetal acceleration is: is:

aacc == v v22//rr

ac – m/s2 centripetal accelerationv – m/s velocityr –m radius

Check ref table.

1. A car moves around a circular track at a constant speed. If the car is 48.2 m from the center and has a centripetal acceleration of 8.05 m/s2, what is its speed?

• 19.7 m/s

Ex 2: The distance from the moon to earth is 3.8 x 108 m.

Moon’s ac = 2.8 x 10-3 m/s2.

a. What is the moon’s velocity?

b. What is moon’s period in seconds and days?

The net force that causes centripetal acceleration is called centripetal force.

•Fnet = ma.

• Fc = mac.

•Fc = m v2/r.

FFcc == mmaacc == mvv22//rr

m = mass in kgm = mass in kgv = linear velocity in m/sv = linear velocity in m/s

FFcc = centripetal force in N = centripetal force in N

r = radius of curvature in mr = radius of curvature in maacc = centripetal acceleration in m/s = centripetal acceleration in m/s22

3. A pilot is flying at 30 m/s in a circle of radius 100 m. If a force of 635 N is needed to maintain the pilot’s motion, what is his mass?

• 70.6 kg.

Calculation of Fc tells you how much force is needed to travel around a

curve at a particular speed & radius.

• Which is more significant: radius or speed when determining force?

• Since F = mv2/r & v is doubled.

• We need v2 or (2)2 = 4x more force.

• 4 x 100 N = 400 N is the new force.

• 4. If 100-N of force are needed for a car traveling 5 m/s to complete a turn of radius 4-m, how much force is required for the same turn when the car increases its speed to 10 m/s.

• F = mv2/r & r is doubled.

• We need 1/(r) or ½ x force to make the turn.

• ½ x 400-N = 200-N of force needed.

• 5. If 400-N of force are needed for a car traveling 10 m/s to complete a turn of radius 4-m, how much force is required for the same speed when the radius increases to 8 m?

You feel acceleration!!That is because of your

inertia.

The car is turning due to the inward force, you feel as though you are being forced leftward or outward. The car is beginning its turning motion (to the right) while you continue in a straight line path.

Mud sticks to tire.

What can you say for sure about the mud on this tire?

• It is accelerating for its entire trip.

• Fnet = 0 when it’s flung from the tire.

• It could complete the whole circle if the radius of the tire were halved.

• There Fnet on the mud is too small for to complete the circle.

Sometimes a measured in g’s.Multiples of Earth’s a of gravity.

1g = 9.81 m/s2.2g = 19.6 m/s2.3g = 29.4 m/s2.

.

.

.etc.

6. An 80-kg astronaut experiences a force of 2890-N when orbiting Earth. How many g’s does he feel?

• ac = F/m

• 2890 N / 80-kg = 36 m/s2.

• 36 m/s2 / 9.81 m/s2

• = 3.7 g.

• Real world centripetal force 8 min.

• http://www.youtube.com/watch?v=PBpe_LLlQJw

• How to create artificial gravity 4 min.

• https://www.youtube.com/watch?v=jkgrVL69mmA

• 2001 space station 2 min.

• https://www.youtube.com/watch?v=1wJQ5UrAsIY

Film Clips

Fc set equal to providing force

• Since the Fnet, Fc must be provided by one of the forces, it is sometimes helpful to set them equal.

• If a car turns a corner, Fc is provided by Ff, so:

• Fc = Ff.

• mv2/r = Fn.

• mv2/r = Fg On horizontal surface.

• mv2/r = mg If given mass in kg.

• Notice mass drops out so,• v2/r = g. is not dependent on mass!

Ex 1: A car turns a corner on a horizontal road with radius 5-m at a speed of 5 m/s. What is the minimum coefficient of friction, , needed between the tires and the road to allow the car to make the turn?

• Fc = Ff

• mv2/r = Fn

• mv2/r = mg

• v2/r = g

• 0.5

Ex 2: A truck with rubber tires wishes to make a turn on dry concrete while traveling 17 m/s. What is the minimum radius of the curve that will allow the truck to make the turn safely?

• Fc = Ff

• mv2/r = mg

• v2/r = g

• v2/ g = r

• (17 m/s)2/ (0.9)(9.81 m/s2)

• r ~ 33 m

Hwk Wksht “Forces Providing Fc”.

• Real world centripetal force 8 min.

• http://www.youtube.com/watch?v=PBpe_LLlQJw

• How to create artificial gravity 4 min.

• https://www.youtube.com/watch?v=jkgrVL69mmA

• 2001 space station 2 min.

• https://www.youtube.com/watch?v=1wJQ5UrAsIY

Film Clips

Velocity and Satellites

Derive an equation to express the minimum

velocity a satellite must have to stay in orbit.

Fg keeps the satellite in orbit. Set Fc equal to gravitational force.

Fg = Fc Gmem mv2. r2 r

Gme = v2. r

v = (Gme)1/2. (r) 1/2.

What is the velocity needed for a satellite to stay in orbit 200 km above the Earth’s surface?

First find the total distance between their centers.

6.37 x 106 m.

2.0 x 105 m.

6.57 x 106m total r.