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
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
Fcg = Fc
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!
ac = how fast you are turning
>ac means you are turning quickly, a small circle
flies around your head
electrons around nucleus
<ac means you are turning slowly, a big circle
Moon around the Earth
Earth around the Sun
B. The formula
ac = v2 /r
and if F = ma then : F = m(v2 /r) = Fc =mv2 /r
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 = mv2 /r = acm = 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 108 m
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
ac = v2 /r = (1012.2)2/3.8 x 108 = 0.00269 m/s2
If the Moon's mass is 7.36 x 1020 kg
then the force required is
Fc = mac = 7.36 x 1020 (.00269) = 1.9845 x 1017 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 = v2 /gr
What is important?
Radius of curve
Velocity of car
Gravity we canít change
What doesnít matter?
Weight, frictional force
Centrifugal force holds car to road
Works for aircraft wings also
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 = v2 /gr
Now find the acceleration you as a driver would feel.
.ac = v2 /r
Find the angle of the jetís 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 17th century
Law 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:
Fg = G(m1m2)/r2
M1 and M2 = masses in kg
R = distance between them
G = universal gravitational constant =
6.67 x 10-11 metric
= 3.44 x 10-8 british
D.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/r2
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 Earthís gravity
A.Look at velocity by components:
Vx = horizontial velocity = V cos q
Vy = 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 Vx = 0
Vy = 0 but speeds up as it falls,
3.Ball B has Vx = v (horizontial vel)
Vy = 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 = vot + Ĺ at2
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
If you drive off a 250 ft cliff at 35 mph, when do you hit and how far have you gone?
Ignoring the curvature of the EarthIgnore the acceleration phase in rockets
A.Breaking things down into components: Vx and Vy
1.Time to its peak: like throwing up a stone
Vo = Vy = V sin q
Vf = 0
.a = -g
find time using Vf = Vo + 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
Vx = V cos q = V
Use s = vt
T = 2Vsin q/g
S = (Vcos q) 2Vsin q /g
R = V sin 2q/g
3.maximum range will be using 45 because 2(45) =90o
sin 90 = 1
4.Find the muzzle velocity necessary to fire an object 1 mile.