Circular Motion When an object travels about a given point at a set distance it is said to be in circular motion.

Circular Motion

When an object travels about a given point at a set distance it is

said to be in circular motion

Cause of Circular Motion1st Law…an object in motion stays in motion in a straight line at a constant speed unless acted on by an outside force.

2nd Law…an outside (net) force causes an object to accelerate in the direction of the applied force.

THEREFORE, circular motion is caused bya force acting on an object pulling it out of itsinertial path in the direction of the force.

Circular Motion Analysis

r

v1

r

v2

q

Circular Motion Analysisv1

v2

rr

v1

v2

v = v2 - v1

or v = v2 + (-v1)

(-v1) = the opposite of v1

v1

(-v1)

v1

v2

rr

v2

(-v1)

v = v2 - v1

or v = v2 + (-v1)

(-v1) = the opposite of v1

v1

(-v1)

v1

v2

vNote how v is directed toward the center of thecircle

q

v1

v2

rr

v1

v2

v2

(-v1)

v

l

Because the two triangles aresimilar, the angles are equal andthe ratio of the sides areproportional

v1

v2

rr

v1

v2

v2

(-v1)

v

l

Therefore,

v/v ~ l/r and v = vl/r

now, if a = v/t and v = vl/r

then, a = vl/rt since v = l/t

THEN, a = v2/r

Centripetal Acceleration

ac = v2/r

now, v = d/t and, d = c = 2r

then, v = 2r/t and, ac = (2r/t)2/r

or, ac = 42 r2/t2/r or, ac = 42r/T2

The 2nd Law and Centripetal Acceleration

Fc

ac

vt

F = ma

ac = v2/r = 42r/T2

therefore,

Fc = mv2/r or,

Fc = m42r/T2

Motion in a Vertical Circle

A

B

Fw TA

Fw

TB

Vertical circle

Fw TA

Fw

TB

A

B

Top of Circle

at vmin TA = 0 and Fw = Fc

therefore, TA + mg = mv2/r

because TA = 0, mg = mv2/r

and v2 = rg

Vertical Circle

Fw TA

Fw

TB

A

B

Bottom of Circle

vmax at bottom

therefore,

TB + mg = mv2/r

Fc = TB + Fw

or

or, TB = mv2/r - mg

Cornering on the HorizontalWhen an object is caused to travel in a circular path because of the force of friction,then,...

Fc = FF

car

Fw

FN

FF

Cornering on the Horizontal

Fc = FF

car

Fw

FN

FF

Therefore, mv2/r = FN

on horiz., FN = Fw = mg,

mv2/r = mg … or,

= v2 /rg

```````

Cornering on a Banked Curve

car

Fw FN

Fc

Fc

Cornering on a Banked Curve

`car

Fw FN

Fc

FN

Fc

Note how FN isthe Resultant

Fw

Fw FN

Fc

If we want to know the anglethe curve has to be at to allowthe car to circle without friction,then we have to analyze theforces acting on the car.

Sin = Fc/FN Fc = SinFN

Fc = mv2/r SinFN = mv2/r

therefore, mv2/r= Sin FN

Fw FN

Fc

Sin = Fc/FN Fc = SinFN

Fc = mv2/r SinFN = mv2/rtherefore, mv2/r = Sin FN

Cos = Fw/FN FN = Fw/Cos

Fw = mg FN = mg/Cos

mv2/r = SinFN

FN = mg/Cos

or, mv2/r Sin = mg/Cos

tan = v2/rg

FC

FWFN

Note! FN is resultant

FC

FW

FN

Cos = FW/FN and Sin = FC/FN

FN = FW/Cos mg/Cos and FN = FC/Sin mv2/r Sin

mg/Cos = mv2/r Sin

Sin /Cos = mv2/rmg

Tan = v2/rg

FN supplies FC for circular motion, no FF needed

OR

Universal Gravitation

E

M

Ah, the same force that pullsthe apple to the ground pulls moon out of its inertial pathinto circular motion aroundthe earth!

Therefore, the forces must beproportional to each other!

E

M Now, if the earth pulls the appleat a rate of 9.8 m/s2, then, the same earth must pull to moon at a proportional rate to that.

60re

If the moon is 60 x further from the apple, and all forms of energy obeythe Inverse Square Law, then, the acceleration of the moon should be1/602 of that of the apple, or9.8 m/s2 x 1/602 = 0.0027 m/s2

And, in one second it should fall d = 1/2 (0.0072m/s2)(1sec)2

or, 0.0014 m

Universal Gravitation

FE

Force ofEarth onmoon

FM Force ofMoon onEarth

FE = FM

3rd Law

Universal Gravitation

FE

Force ofEarth onmoon

FM Force ofMoon onEarth

FE = FM

3rd Law

Because of the3rd Law and theInverse SquareLaw :

F = Gm1m2/r2

Universal Gravitation

F = Gm1m2/r2

If “F” is the weight of an object, “Fw”, then,Fw = m2gand, m2g = Gm1m2/r2

or, g = Gm1/r2

or, m1 = gr2/G

Universal GravitationIf gravity is the force that causes anobject to travel in circular motion, then,

F = Fc or, Gm1m2/r2 = m2v2/r

or, m1 = v2r/G or, r = Gm1/v2

or, v2 = Gm1/r

or, m1 = v2r/G

Universal GravitationIf gravity is the force that causes anobject to travel in circular motion, then,

F = Fc or, Gm1m2/r2 = m2v2/r

or, Gm1m2/r2 = m242r/T2

transpose extremes T2/r2 = m242r/Gm1m2

divide by “r” and cancel m2

T2/r3 = 42/Gm1

Universal Gravitation

T2/r3 = 42/Gm1

Note that for any objectcircling a superior objectthat 42/Gm1 remainsconstant!!!!

Therefore, T2/r3 is also constantfor all objects circlingthat superior object

Universal Gravitation

T2/r3 = “k” for all circling objects

Therefore, for two objects circling the same superior object...

T12/r1

3 = T22/r2

3 or (T1/T2)2/(r1/r2)3

Kepler’s Laws1st Law…all planets circle the Sun in

ellipital paths with the Sun at onefocus

Sun

planet

F2F2

Kepler’s Laws1st Law…all planets circle the Sun in

ellipital paths with the Sun at onefocus

2nd Law…Each planet moves around the sun in equal area sweep inequal periods of time

Kepler’s Laws

2nd Law…Each planet moves around the sun in equal area sweep inequal periods of time

1

2 3

4

ba

Area 12a = Area 43b

Kepler’s Laws1st Law…all planets circle the Sun in

ellipital paths with the Sun at onefocus

2nd Law…Each planet moves around the sun in equal area sweep inequal periods of time

3rd Law…the ratio of the squares of theperiods to the cube of their orbitalradii is a constant

Kepler’s Laws

3rd Law…the ratio of the squares of theperiods to the cube of their orbitalradii is a constant

T2/r3 = “k” for all circling objects

Therefore, for two objects circling the same superior object...

T12/r1

3 = T22/r2

3 or (T1/T2)2/(r1/r2)3

Sample Problems

What is the gravitational attraction between the Sunand Mars?

F = ?ms = 1.99 x 1030 kgmm = 6.42 x 1023 kgrm = 2.28 x 1011 m F = Gmsmm/rm

2

F = 6.67 x 10-11 N m2/kg2(1.99 x 1030kg)(6.42 x 1023kg)(2.28 x 1011 m)2

F = 1.64 x 1021 N

Sample Problems

What velocity does Mars circle the Sun at?

v = ?ms = 1.99 x 1030 kgmm = 6.42 x 1023 kgrm = 2.28 x 1011 m

F = Fc

Gmsmm/rm2 = mmv2/rm

v2 = Gms/r

v2 = 6.67 x 10-11Nm2/kg2(1.99 x 1030 kg)/2.28 x 1011m

v = 2.4 x 104 m/s

Sample Problems

What is the period of Mars as it circles the Sun?

T = ?ms = 1.99 x 1030 kgmm = 6.42 x 1023 kgrm = 2.28 x 1011 m

F = Fc

Gmsmm/rm2 = mm42r/T2

T2 = 42r3/Gms

T2 = 42(2.28 x 1011 m)3 /6.67 x 10-11)1.99 x 1030 kg

T = 5.9 x 107 s

or, T = 685 days

Sample Problems

What is the period of Mars? This time use Kepler’s 3rd Law to find it!

Tm = ?Te = 365.25 dare = 1.5 x 1011 mrm = 2.28 x 1011 m

F = Fc

Gmsmm/rm2 = mm42r/T2

T2/r3 = 42/Gms

Tm2/rm

3 = Te2/re

3

Tm2/(2.28 x 1011 m)3 = (365.25 da)2/(1.5 x 1011m)3

Tm = 684 days