Circular Motion

Lots of things in circular motion:

clocks, tires, planets, electrons, stars, galaxies

I. Centripetal Force: force acting inward causing circular motion

A. Newton's First Law: A body at rest will stay at rest and a body in motion will stay in motion, in a straight line, at a constant velocity until acted upon by an outside force!

B. Circular motion must be caused by a force

centripetal force

inward force of string allows circular motion

without it the rock goes straight

C.Newton's Third Law:for every force there is an equal but opposite force, reaction force

centripetal force is inward

centrifugal force is outward, the reaction force

F_{cg} = F_{c}

II.Centripetal Acceleration

A. acceleration = Dv

but velocity is a vector

vectors have 2 parts, magnitude and direction

if you keep the speed the same but change direction

you have changed the velocity!

a_{c} = how fast you are turning

>a_{c} means you are turning
quickly, a small circle

flies around your head

electrons around nucleus

<a_{c} means you are turning
slowly, a big circle

Moon around the Earth

Earth around the Sun

B. The formula

a_{c} = v^{2} /r

and if F = ma then : F = m(v^{2} /r) = F_{c}
=mv^{2} /r

III.A problem:

A. Find the centripetal acceleration of a car making a trun of radius 150 ft at a velocity of 35 mph.

v = 35 x 1.47 = 51.45 ft/sec

ac = v ^{2}/r = (51.45)^{2}/ 150 =
2647.1/150 = 17.65 ft/sec

If the car weighs 2500 lb, find the force need.

m = w/g = 2500/32 = 78.125 slugs

Fc = mv^{2} /r = a_{c}m = 78.125(17.65) =
1378.9 lb

B.The moon is 380,000,000 m away and circles the Earth in 27.3 days. Find the centripetal accleeration.

have r = 3.8 x 10^{8
} m

need v

moon goes AROUND in 27.3 days

distance around a circle is circumfrence

s = 2p r

t = 27.3 x 24 x 60 x 60 =235,872 sec

v = s/t = 2 pr/t= 2 (3.8 x 10 )/235872= 1012.2

a_{c} = v^{2} /r = (1012.2)^{2}/3.8 x 10^{8} =
0.00269 m/s^{2}

If the Moon's mass is 7.36 x 10^{20} kg

then the force required is

Fc = ma_{c}
= 7.36 x 10^{20} (.00269) = 1.9845 x 10^{17} N

IV. Banking of a curve of wing

A. Curves on roads are banked to allow faster travel

Race tracks, interstates

Formula for banking a curve is:

. tan q
= v^{2} /gr

What is important?

Radius of curve

Velocity of car

Gravity we can’t change

What doesn’t matter?

Weight, frictional force

Centrifugal force holds car to road

Works for aircraft wings also

B. Problem:

Taladaga has a track with a large bank to allow for fast racing. If you intend to be able to take the curve at 190 mph, find the bank that is necessary if the turn is of radius 150 yds.

.tan q = v

^{2}/gr

Now find the acceleration you as a driver would feel.

.ac = v

^{2}/rFind the angle of the jet’s wings if it is to make a turn of radius 1 mile at 700 mph. Find the acceleration the pilot will feel.

V. Universal Gravitation

A. Most common example of centripetal force

B.Purposed by Newton in 17

^{th}centuryLaw of Gravity- every object in the universe attracts every other object in the universe with a force directly proportional to each of their masses and inversely proportional to the square of the distance between them.

C. The formula:

F

_{g}= G(m_{1}m_{2)}/r^{2}M

_{1}and M_{2}= masses in kgR = distance between them

G = universal gravitational constant =

6.67 x 10

^{-11}metric= 3.44 x 10

^{-8}britishD.Find the attracting force between a 100 kg male and a 50 kg female sitting 0.25 m away form each other on a couch.

Find the attracting force between a 220 lb male and a 100 lb female sitting 1 ft apart on a couch.

E.Find the mass of the Earth. (to big to put on a scale)

1.The radius of the Earth is 6400000 m.

2.Set an apple on the ground.

3.It is held there by its weight. W = mg

4.This is also the force of gravity between the Earth and the apple.

Fg = G Mm/r

^{2}5.Set the 2 equations equal.

W = Fg

VI.Motion in a Vertical Plane

Reason most objects on Earth travel in a curved path

Curving down due to the pull of Earth’s gravity

A.Look at velocity by components:

V_{x} = horizontial velocity = V cos
q

V_{y} = vertical velocity = V sin
q

B.Look at 2 balls on a table top

1.drop one straight down and roll one off the edge

2.Ball A has V_{x} = 0

V_{y} = 0 but speeds up as it falls,

due gravity

3.Ball B has V_{x} = v
(horizontial vel)

V_{y} = 0 but sspeeds up as it falls

Same as A

4.Both balls stay in the air the same time

hit as same time

from s = v_{o}t + ½ at^{2}

time to hit T = (2h/g)^{1/2}

5.During this time T, ball B will fly vertical at v

a distance of s = vt = vT

C.Problem:

If you drive off a 250 ft cliff at 35 mph, when do you hit and how far have you gone?

VII.Projectile Flight

Ignoring the curvature of the Earth

Ignore the acceleration phase in rockets

A.Breaking things down into components: V_{x}
and V_{y}

1.Time to its peak: like throwing up a stone

Vo = V_{y}
= V sin q

V_{f} = 0

.a = -g

find time using V_{f} = V_{o}
+ at

0 = V sin q + gt

t = Vsin q/g = time to peak

Total time = T = 2Vsin q/g (double it)

2.Range = distance of flight

object keeps going horizontially for this time

V_{x} = V cos q = V

Use s = vt

T = 2Vsin q/g

So

S = (Vcos q) 2Vsin q /g

Turns into:

R = V sin 2q/g

3.maximum range will be using 45 because 2(45) =90^{o}

sin 90 = 1

4.Find the muzzle velocity necessary to fire an object 1 mile.