∆v ∆θ∆θ v 1 v 2 ∆θ∆θ ∆r r 1 r 2 P O Q v 2 ∆v v 1 v 2 ∆v = – v 1 Velocities at two different positions P and Q have the equal magnitude (v 1 = v 2.

Data booklet reference:

•

•

•

∆v∆θ

v1

v2

∆θ

∆r r1

r2

P

O

Q v

2

∆v v1 v2∆v = –

v1 Velocities at two different positions P and Q have the equal magnitude (v1 = v2 = v),

but different directions.

clim Δv

a =Δt 0 Δt

An object is moving in circular path at constant speed.

Goal is to obtain an expression for the magnitude and direction of centripetal acceleration.

v2=v1+∆ v

v1

ca∆v

a

centripetal or radial acceleration

∆v∆θ

v1

v2

∆θ

∆r r1

r2

P

Q v

2

∆v v1 v2∆v = –

v1 Velocities at two different positions P and Q have the equal magnitude (v1 = v2 = v),

but different directions.

clim Δv

a =Δt 0 Δt

1 2v , v , Δv

~ ∆ 1 2r , r , Δr

∆

Δv Δr Δr = Δv = v

v r r

c

Δrlim va =

Δt 0 r Δt

magnitude of ca Δrlim

v = Δt 0 Δt

2

c

va =

r 2-1

2c

m/sa = = m/s

m

∆θ

∆r∆v ∆θ

r1 r2

v2

P

O

Q v1

v2

v1_

you must be able to sketch this.

A “slightly” misleading sketch

2

c

va =

r

v2=v1+∆ v should be added to to change it to but if we add it at point P would not point toward center !!!!!.

velocity – tangent to the circle centripetal acceleration – points toward the center

velocity and centripetal acceleration vectors are always perpendicular to each other

If centripetal acceleration were suddenly to disappear, there would be nothing to change the direction of the velocity. The object would continue to move in a straight line with constant speed. In that case that object moves “inertially” (because of inertia of the body).

Centripetal force and acceleration

A particle is said to be in uniform circular motion if it travels in a circle (or arc) with constant speed v.

2

c c

mv F = ma =

r

Object undergoing uniform circular motion is accelerating with centripetal acceleration ac , so it has a force acting upon it:

it is not separate force – it is simply one of our familiar forces acting in the role of causing circular motion

Many forces can force an object to move in circular path, therefore becoming centripetal force:

Moon around the Earth ……… gravity

object sitting on a rotating table (strawberries sitting on a seat of a turning car)(a car moving in a circular path) ………… friction

a ball whirling in a circle at the end of a string ……… tension in the string

a person pressed against the inner wall of a rapidly rotating circular room in an amusement park …………. normal force from the wall

Ferris–wheel rider passes through the lowest point of the ride ……. normal force from the seat and the force of gravity

useful relations: - speed(linear) and angular speed

2πRv =

T

2πω =

T

constant speed

period T: time required for one complete revolution (s)

speed v = distance/time

angular speed ω = angle swept/time

Variable Tran. Variable Rot. Relationship

distance sangle swept (rad)

s = r tangential velocity

angular velocity (rad/s)

v = r ω

tangentialacceleration a

angular acceleration (rad/s2) a = r

Speed depends on position (r) but angular speed does not

v = s / t = ( r ) / t = r

= 2 / T = 2f

A CAR ON A LEVEL (UNBANKED) SURFACE

Think of what would happen if there were no friction. Have you ever found yourself driving on ice? If you turn steering wheel nothing happens because there is little or no friction between the ice-covered road and the car's tires. You need frictional force to turn.

A 1200 kg car rounds the curve on a flat road of radius 45 m. If the coefficient of static friction between tires and the road is s = 0.82, what is the greatest speed the car can have in the curve without skidding?

So if you turn the steering wheel you are pushing the ground in the black direction and the ground is pushing back on you in the orange direction. That is the friction force that the road exerts on the car's tires pointing toward the center of the turn, responsible for turning the car.

Fn = mg Ffr = s Fn = s mg

2

fr cv

F = ma = ma = mr

2

sv

μ g = r

v = 19 m/s

A CAR ON A BANKED SURFACE…

can a car negotiate a curve without friction (on ice)? YES, but the curve has to be banked!

If a roadway is banked at the proper angle, a car can round the corner without any assistance from friction between the tires and the road. The horizontal component of the normal (reaction) force is the net force. This force is centripetal force to turn the car.

Find the appropriate banking angle for a 900 kg car traveling at 20.5 m/s in a turn of radius 85.0 m so that NO friction is required.

𝐹 𝑛 sin𝜃=m𝑣2

𝑟⇒ tan𝜃= 𝑣2

𝑟𝑔

for given speed v and radius r of the turn we get angle of the bank independent of mass m

20v

θ = arctan = 26.7rg

𝐹 𝑛cos𝜃=m𝑔⇒𝐹𝑛=𝑚𝑔

cos𝜃

for   given   θ   and    r :

if   v  <√rg tanθ the   car  would   slide   down  a   frictionless   banked   curveif   v  >√rg tan θ the   car   would  slide   off   the   top

HORIZONTAL CIRCULAR MOTION………

centripetal force = tension

cord cannot be horizontal, only for

T will act nearly horizontally and provide the force necessary to give the ball its centripetal acceleration

mg << T, ∑ F ≈ T

22max max

max maxmv T rv

T = m T = v =r r m

CONICAL PENDULUM…

2

T cos θ = mg

vTsinθ = m

r rg

vtan

2

L

r

𝑟=𝐿sin 𝜃

𝑣=√𝑔𝑟 tan𝜃=√𝑔𝐿 sin 𝜃 tan 𝜃

P2πr L cosθ

period T = = 2πv g

independent of m

ROLLER COASTER

Top of a Hill 2 2

n nmv mv

mg - F = F = mg - R R

T v = Rg

minimum speed required to stay alive is mathematically given with Fn = 0 (the rider just barely looses contact with the seat

Upside-down at the Top of a loop

Fn = 0 danger when feeling weightless

min speed required to stay alive and not to fall down:

T v = RgBottom of a Valley 2 2

n nmv mv

F - mg = F = mg + R R

The rider's hands and arms are hard to move. The rider's blood is even hard to move. Airplane pilots are in this situation as they pull out of a dive. Apparent weight of 6-7 times one's real weight -- can mean that not enough blood will be circulated to the brain and a pilot -- or other passenger -- may pass out.

𝐹 𝑛+𝑚𝑔=𝑚𝑣2

𝑅𝐹𝑛=

𝑚𝑣2

𝑅−𝑚𝑔

AIRPLANE MAKING A TURN - BANKING

Lift force (L) generated by the airplane wings is not equal to mg anymore, but greater.

𝐿 cos𝜃=𝑚𝑔 𝐿sin 𝜃=𝑚 𝑣2

𝑅

𝑅= 𝑣2

𝑔 tan𝜃

As the plane banks (rolls), the lift vector begins to have a horizontal component.

ARTIFICIAL SATELLITE 2E E

2

mm mmvG = v= G

r rr

A geostationary satellite (communications satellite) is placed at an altitude of approximately 35,800 kilometers (22,300 miles) directly over the equator, that revolves in the same direction the earth rotates (west to east)

given r (we want satellite there) v precise and independent of m

Emv= G

r

23 E

2

Gm T2πrv = r =

T 4π

r = 4.23x107 m = 42,3000 km from the Earth center

rE = 6,380 km h = 36,000 km

- we launch satellites above the atmosphere with rockets, then tilt them over and give them enough horizontal speed that it can orbit the Earth. - for slightly higher speed but the same distance the orbit will be elliptical - if the speed is too high, the spacecraft will not be confined by the Earth’s gravity, and will escape never to return (v ≈ 11.1 km/s ) - if the speed is too low, it will fall back to the Earth

EXAMPLE: Explain how an object can remain in orbit yet always be falling.SOLUTION:Throw the ball at progressively larger speeds.In all instances the force of gravity will draw the ball toward the center of the earth.When the ball is finally thrown at a great enough speed, the curvature of the ball’s path will match the curvature of the earth’s surface.The ball is effectively falling around the earth!

PRACTICE: Find the angular speed of the minute hand of a clock, and the rotation of the earth in one day.

SOLUTION:The minute hand takes 1 hour to go around one time. Thus = 2 / T = 2 / 3600 s = 0.00175 rad s-1.

The earth takes 24 h for each revolution so that = 2 / T = ( 2 / 24 h )( 1 h / 3600 s ) = 0.0000727 rad s-1.

This small angular speed is why we can’t really feel the earth as it spins.

EXAMPLE: The Foucault pendulum is a heavy pendulum on a very long cable that is set in oscillation over a round reference table. Explain how it can be used to tell time.

SOLUTION:

The blue arcs represent the motion of the pendulum bob relative to the universe at large.

The the green lines represent the plane of motion of the pendulum relative to the building.

Since the building is rotating with the earth at = 0.0000727 rad s-1, each hour the green line rotates by = t = 0.0000727(3600)

= 0.262 rad (360/ 2 rad) = 15.0.

FYI This solution only works when the pendulum is at one of the poles. See the Wiki for a general solution.

EXAMPLE: Find the apparent weight of someone standing on an equatorial scale if his weight is 882 N at the north pole.

SOLUTION: Recall that = 0.0000727 rad s-1 anywhere on the earth.

The blue arcs represent the lines of latitude.

The white line R represents the earth’s radius.

The yellow line r represents the radius of the circle a point at a latitude of follows.

Note that r = R cos , and that at the equator, = 0˚ and at the pole, = 90˚.

Rr

0˚

90˚

Thus, at the equator, r = R, and at the pole, r = 0. Furthermore, R = 6400000 m.

Then, at the equator,

ac = r 2 = 6400000 0.00007272 = 0.0338 ms-2.

Then, at the pole,

ac = r 2 = 0 0.00007272 = 0.000 ms-2.

Make a free-body diagram at the equator…

Scales read the normal force R:

F = ma

R – W = - mac

R = W – mac

Then, R = 882 – ( 882 / 9.8 ) 0.0338 = 879 N.

The man has apparently “lost” about 3 N!

W R

ac

kx = FC = mv 2/ r implies that as v increases, so does the centripetal force FC needed to move it in a circle. Thus, x increases.

F = kx (k = CONST).

kx = F k = F / x = 18 / 0.010 = 1800 Nm-1.FC = kx = 1800( 0.265 – 0.250 ) = 27 N.FC = v 2/ r v 2 = r FC = 0.265(27) = 7.155v = 2.7 ms-1.

Use v = r ( = CONST).

Use a = r 2 ( = CONST).

At Pr = R

v = R a = R 2

At Qr = 2Rv = 2R = 2va = 2R 2 = 2a

Objects moving in uniform circular motion feel a centripetal (center-seeking) force.