Chapter 5Dynamics of Uniform Circular

Motion

Circular Motion

If the acceleration is in the same direction as the velocity i.e., parallel to the velocity, the magnitude of the velocity increases i.e., the object moves faster.

If the acceleration is in the opposite direction as the velocity i.e., antiparallel to the velocity, the magnitude of the velocity decreases i.e., the object moves slower.

If the acceleration is non-parallel to the velocity, The direction (and possibly the magnitude) of the velocity changes.

The effect of an object’s acceleration on its velocity depends on their relative directions.

In the special case where the acceleration is perpendicular ( ) to the velocity the object will follow a uniform circular path. The direction of the velocity changes while the magnitude of the velocity remains constant.

The acceleration is called the centripetal acceleration which means “center-seeking” and always points toward the center of the circular path.

The magnitude of the centripetal acceleration, ac, depends on the magnitude of the object’s velocity and the radius of the circular path.

ac =v2r

If an object travels in a uniform circle there must be a net force acting on it to produce the centripetal acceleration. This net force is called the centripetal force.

From Newton’s Second Law:Fnet =ma

Fc =mac

Fc =mv2r

NOTE :The centripetal force is not a new type of force but rather the net force that results in uniform circular motion.

The source of the centripetal force varies from case to case.

Examples of Sources of Centripetal Force:

1. Car travelling around a curve:

2. Planet travelling around the sun:

3. Electron travelling around the nucleus of an atom:

4. Ball on a string being swung in a circle:

5. Plane circling an airport:

Friction between tires and road

Gravitational force

Electrostatic force

Tensional force

Component of lift on the wing

NOTE: Planetary orbits are not truly circular but elliptical.

rv

Consider a car following a circular curve in the broad.

Static friction between the tires and the road must supply the necessary centripetal force.

Fc =μsmg=mv2r

μsg = v2

rThe maximum speed at which the car can safely travel around the curve is given by:

vmax = μsgr

θ

The dependence on friction can be reduced by “banking” the curve.

mg

FN

θ

FN cosθ

FN sinθ

θ

mg

FN

θ

FN cosθ

fs

FNsinθ can supply some of the centripetal force.

θfs cosθ

fscosθ can also supply some of the centripetal force.

Fc =FN sinθ + fscosθFN cosθ =mg

FN = mgcosθ

⎧ ⎨ ⎪

⎩ ⎪

FN sinθ = mgcosθ sinθ

FN sinθ =mgtanθ

⎧ ⎨ ⎪

⎩ ⎪

fs =μsFN

fs =μsmgcosθ

fscosθ =μsmgcosθ cosθ

fscosθ =μsmg

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

Fc =mgtanθ +μsmg

mv2

r =mgtanθ + μs( )

v2

r =g tanθ +μs( )

vmax = tanθ +μs( )gr

θmg

FNcosθ

w/o “banking

vmax = μsgr

vmax,banking

vmax,w /o banking=

tanθ + μs( )grμsgr

vmax,banking

vmax,w /o banking=

tanθ +μs( )μs

= tanθμs

+1 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

The track at the Daytona National Speedway is banked at 31°. The typical coefficient of friction between rubber and asphalt is 0.5. How much faster can cars travel around the track due to its banking? vmax,banking

vmax,w /o banking= tan31°

0.5 +1 =1.48

About 50% faster

Satellites

For a satellite in a circular orbit the gravitational force provides the required centripetal force.

FG =Fc

Gmsmo

rs,o2 =

msv2

rs,o

Where:ms = mass of satellite

mo= mass of object it is orbiting

rs,o = distance from satellite to the center

of the orbited body

The velocity of a satellite in a stable circular is given by:

v =Gmors,o

What is the velocity of a satellite orbiting above the earth, me=5.98x1024kg and re=6.38x106m, at an altitude of (a) 5mi. (8.05x103m), (b) 1000mi. (1.61x106m)?

(a) rs,o =re+h=6.38×106m+8.05×103m

rs,o =6.39 ×106m

v =Gmors,o

= 6.67×10−11 Nm2

kg2⋅5.98 ×1024kg

6.39×106m

v =7.9×103 ms ≈17,700mihr

(b) rs,o =re+h=6.38×106m+1.61×106m

rs,o =7.99 ×106m

v =Gmors,o

=6.67×10−11 Nm2

kg2⋅5.98 ×1024kg

7.99 ×106m

v =7.06×103 ms ≈15,800mihr

Frequently the time required for a satellite to complete one orbit is of interest. This time is called the period, T, of the satellite’s orbit.

Since the satellite is travelling in a circle one orbit equals a distance of: 2r.

Since in a circular orbit the velocity of the satellite is constant:

Substituting v = Gmors,o

gives:

T =2 r3 2

Gmo

T =2 rv

What is the orbital period of the moon whose average distance from the earth is 3.84x108m.

T =2 r3 2

Gmo= 2 π(3.84 × 108m)3 2

6.67 × 10−11 Nm2

kg2 ⋅5.98 × 1024 kg

T =2.367 ×106s=27.3 days

This period of time is called the Sidereal Month. It is not the period of time on which our calendar months are based.

Upon what period of time are our calendar months based?

A geosynchronous satellite remains directly above a given point on the earth’s surface. Therefore, its period must be the same as the earth’s period of rotation (24 hours = 8.64x104s)

At what altitude must a geosynchronous satellite be placed?

Historical Models of the Solar System

Aristotelean-Geocentric Model:

The earth is stationary and at the center of the Solar System. All celestial bodies revolve around the earth in perfectly circular orbits.

Copernican-Heliocentric Model:

The sun is at the center of the Solar System. The earth is a planet that revolves around the sun in a circular orbit. The earth spins on its axis.

Planetary Orbits

Terrestrial Planets Orbits

Earth-Mars Orbits

Kepler’s Laws of Planetary Motion

In the late 15th and early17th centuries, Johannes Kepler a believer in the Copernican sun-centered universe derived three laws of planetary motion.

Kepler used the extremely precise measurements of the position of Mars made by the Danish astronomer Tycho Brahe.

Kepler’s First Law

I. The orbits of the planets are ellipses, with the Sun at one focus of the ellipse.

II. The line joining the planet to the Sun sweeps out equal areas in equal times as the planet travels around the ellipse.

Kepler’s Second Law

A planet moves faster when it is nearer the Sun. Thus, a planet executes elliptical motion with constantly changing angular speed as it moves about its orbit. The point of nearest approach of the planet to the Sun is termed perihelion; the point of greatest separation is termed aphelion. Hence, by Kepler's second law, the planet moves fastest when it is near perihelion and slowest when it is near aphelion.

Kepler’s Third Law

II. The ratio of the squares of the revolutionary periods for two planets is equal to the ratio of the cubes of their semimajor axes:

T12

R13 =

T22

R23

Kepler's Third Law implies that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit. Thus, we find that Mercury, the innermost planet, takes only 88 days to orbit the Sun but the outermost planet (Pluto) requires 248 years to do the same.

T =2 r3 2

Gmo

T2

r3= 2

Gmo=constant