Uniform Circular MotionUniform Circular MotionMotion in a circular path at

constant speed

Speed constant, velocity changing continually

Velocity changing direction, so there is acceleration

Called centripetal acceleration, since it is toward the center of the circle, along the radius

Value can be calculated by many formulas, first is

ac = v2/r

ExampleExample

A bicycle racer rides with constant speed around a circular track 25 m in diameter. What is the acceleration of the bicycle toward the center of the track if its speed is 6.0 m/s?

ac = v2 =__(6.0 m/s)2 = 36 (m/s)2 = 2.9 m/s2 r 12.5 m 12.5 m

Rotation and RevolutionRotation and RevolutionRotation-Around an Internal Axis-Earth

rotates 24 hours for a complete turn

Linear (tangential) versus rotational speed

Linear is greater on outside of disk or merry-go-round, more distance per rotation

Linear is smaller in middle of disk, less distance per rotation.

Rotational speed is equal for both

Rotations per minute (RPM)

Linear speed is proportional to both rotational speed and distance from the center

Rotation and RevolutionRotation and RevolutionRevolution-Around an

External Axis-Earth revolves 365.25 days per trip around sun

Same relationship between linear and revolutional speeds as with rotational

Planets do not revolve at the same revolutional speeds around the sun

PeriodPeriod

Another important measure in UCM is period, the time for 1 rotation or revolution

Since x=v0t , this implies that vT = 2r and thus T= 2r/ v

Rearranging differently, v= 2r/ T and then inserting it into the acceleration equation

ac = v2/r = 42r/T2

ExampleExample

Determine the centripetal acceleration of the moon as it circles the earth, and compare that acceleration with the acceleration of bodies falling on the earth. The period of the moon's orbit is 27.3 days.

According to Newton's first law, the moon would move with constant velocity in a straight line unless it were acted on by a force. We can infer the presence of a force from the fact that the moon moves with approximately uniform circular motion about the earth. The mean center-to-center earth-moon distance is 3.84 x 108 m.

ExampleExampleac= 42r = 42(3.84 x 108) T= 27.3 da (24 hr/da)(3600 s/hr) = 2.36 x 106 s

T2 (2.36 x 106)2

ac = 2.72 x 10-3 m/s2

The ratio of the moon's acceleration to that of an object falling near the

earth is ac = 2.72 x l0-3 m/s2 = 1 g 9.8 m/s2 3600

FrequencyFrequency

The number of revolutions per time unit

Value is the inverse of the period, 1/T

Units are sec-1 or Hertz (Hz)

Inserting frequency into the ac equation

ac=42f 2r

ExampleExample

An industrial grinding wheel with a 25.4-cm diameter spins at a rate of 1910 rotations per minute. What is the linear speed of a point on the rim?

The speed of a point on the rim is the distance traveled, 2r, divided by T, the time for one revolution. However, the period is the reciprocal of the frequency, so the speed of a point on the rim, a distance r from the axis of rotation, is

v = 2rf

v= (2)(25.4cm/2)(1910/1 min)(1min/60s)

v = 2540cm/s = 25.4 m/s.

Angular VelocityAngular Velocity

Velocity can be defined in terms of multiples of the radius, called radians

There are 2 radians in a circle, and so the angular velocity = v/r

In terms of period = 2/TIn terms of frequency =2f

ExampleExample

At the Six Flags amusement park near Atlanta. The Wheelie carries passengers in a circular path with a radius of 7.7 m. The ride makes a complete rotation every 4.0 s. (a) What is a passenger's angular velocity due to the circular motion? (b) What acceleration does a passenger experience?

a) The ride has a period T = 4.0 s. We can use it to compute the angular velocity as

2= 2 rad = rad/s = 1.6 rad/sT 4.0 s 2.0

(b) Because the riders travel in a circle, they undergo a centripetal acceleration given by

ac= 2r = (/2 rad/s)2(7.7m) = 19m/s2. Notice that this is almost twice the acceleration of a body in free fall.

Angular Velocity and AccelerationAngular Velocity and Acceleration

Any real object that has a definite shape can be made to rotate – solid, unchanging shape

Angular displacement -- -- Radians around circular path

Angular velocity -- --radians per second, angle between fixed axis and point on wheel changes with time

Angular acceleration -- -- increase of , when angular velocity of the rigid body changes, radians per seconds squared

Rotational KinematicsRotational Kinematics

Rotational velocity, displacement, and acceleration all follow the linear forms, just substituting the rotational values into the equations:

ot + 1/2 t2 f

ft ftx/ra/rv/r

Example Example

The wheel on a moving car slows uniformly from 70 rads/s to 42 rads/s in 4.2 s. If its radius is 0.32 m:

a. Find b. Find c. How far does the car go?

a. = (42-70) rads/s = -6.7 rads/s2 t 4.2 s

b. ot + 1/2 t2 = (70)(4.2) + 1/2(-6.7)(4.2)2 = 235 rads

c. = x / r in rads so x = r = (0.32)(235) = 75 m

ExampleExample

A bicycle wheel turning at 0.21 rads/s is brought to rest by the brakes in exactly 2 revolutions. What is its angular acceleration?

= 2 revs = 2(2) radians = 4rads f=0 rads/s o= 0.21 rads/s

Use angular equivalent of vf2 = vo

2 + 2ax which is

f

xrad/s2

Homework: Read Homework: Read pp.898-903 Practice pp.898-903 Practice

Problems 7A, 7BProblems 7A, 7B

Forces in Circular MotionForces in Circular MotionCentripetal Force

Force toward the center from an object, holding it in circular motion

At right angle to the path of motion, not along its distance, therefore does NO work on object

ExamplesGravitation between earth and moon

Electromagnetic force between protons and electrons in an atom

Friction on the tires of a car rounding a curve

Equation is Fc=mac = mv2/r

ExampleExampleApproximately how much force does the earth

exert on the moon? Moon’s period is 27.3 days

Assume the moon's orbit to be circular about a stationary earth. The force can be found from F = ma. The mass of the moon is 7.35 x 1022 kg.

Fc= mac = m 42r

T2

Fc = (7.35 x 1022 kg)42(3.84 x 108m)

((27.3 days)(24 hr/day)(3600 s/hr))2

Fc=2.005 x 1020 N.

Forces in Circular MotionForces in Circular Motion

Centrifugal “Force”

Not a true force, but really the result of inertia

“Centrifugal force effect” makes a rotating object fly off in straight line if centripetal force fails

ExampleExampleImagine a giant donut-shaped space station located so far from all heavenly bodies that the force of gravity may be neglected. To enable the occupants to live a “normal” life, the donut rotates and the inhabitants live on the part of the donut farthest from the center. If the outside diameter of the space station is 1.5km, what must be its period of rotation so that the passengers at the periphery will perceive an artificial gravity equal to the normal ravity at the earth's surface?

The weight of a person of mass m on the earth is a force F = mg.The centripetal force required to carry the person around a circle of

radius r is F =mac = m 42r

T2

We may equate these two force expressions and solve for the period T:

mg =m42r

T2 T=2=2=55s = 0.92 min.r

g

750

9 81 2

m

m s. /

Banked CurvesBanked Curves“Banking” road curves makes turns without skidding possibleFor angle , there is a component of the normal force toward

the center of the curve, thus supplying the centripetal force. The other component balances the weight force.

FN sin = mv2/r FN cos = mg tan = v2/gr thusly = tan-1 (v2/gr) This equation can give the proper angle for banking a curve of

any radius at any linear speed

Banked Curve ExampleBanked Curve Example

A race track designed for average speeds of 240 km/h (66.7 m/s) is to have a turn with a radius of 975 m. To what angle must the track be banked so that cars traveling 240km/h have no tendency to slip sideways?

Determine from

= tan-1 (v2/ g r)

= tan-1 (66.72/9.81(975))= 24.9o

Homework!!Homework!!

Law of Universal GravitationLaw of Universal Gravitation

Newton’s first initiative for the Principia was investigating gravity

From his 3rd law, he proposed that each object would pull on any other object

He likewise noted differences due to distance

His final relationship was that Force was proportional to masses and inversely proportional to distance squared

Using a constant Fg = Gm1m2

r2

Center of GravityCenter of Gravity

Newton found that his law would only work when measuring from the center of both objects

This idea is called the center of gravitySometimes it is at the exact center of the object

Sometimes it may not be in the object at all

All forces must be from the CG of one object to the CG of the other object

Universal Gravitation ConstantUniversal Gravitation Constant

G was elusive to find since gravity is a weak force if masses are small

Cavendish developed a device which made measurement of G possible

The value of G is 6.67 x 10-11 N m2

This puts Fg in Newtons kg2

G can be used then to find values of many astronomical properties

ExampleExampleConsider a mass m falling near the earth's surface. Find its acceleration

in terms of the universal gravitational constant G. The gravitational force on the body is F = GmME

r2

ME = mass of the earth r = the distance of the mass from the center of the earth, essentially the earth's radius.

The gravitational force on a body at the earth's surface is F = mg.

mg= GmME or g =GME r2 r2

Both G and ME are constant, and r does not change significantly for small variations in height near the surface of the earth. The right-hand side of this equation does not change appreciably with position on the earth’s surface, so replace r with the average radius of the earth RE

g = GME

RE2

ExampleExampleShow that Kepler’s third law follows from the law of universal

gravitation. Kepler’s third law states that for all planets the ratio (period)2/ (distance from sun)3 is the same.

Make the approximation that the orbits of the planets are circles and that the orbital speed is constant.

The sun's gravitational force on any planet of mass m is

F= GmM

r2

M =the mass of the sun. Because the mass of the sun is so much larger than the mass of the planet, we can assume, as Kepler did, that the sun lies at the center of the planetary orbit. The circular orbit implies a centripetal force. This net force for circular motion is provided by the gravitational force. Equating these two forces, we get

Fc=GmM = 42mr Rearranging gives T2 = 42

r2 T2 r3 GM

ExampleExampleUse the law of universal gravitation and the measured value of the

acceleration of gravity g to determine the average density of the earth. The density, of an object is defined as its mass per unit volume: = m/V where m is the mass of the object whose volume is V

From a previous example g = GME

RE2

Substitute for M an expression involving , = ME/V.

If we take the earth to be a sphere of radius RE. Then

= ME and ME= 4/3RE3

4/3RE3

The equation for g can then be rewritten in terms of the density as

g = G(4/3RE3 4/3GRE

RE2

Density Example (cont)Density Example (cont)

Upon rearranging, we find the density to be

= 3g 4REG

Inserting the numerical values, we get

= 3(9.81 m/s2) 4(6.38 x 106 m)(6.67 x 10-11 N m2/ kg2

= 5.50 x 103 kg/m3

Moon Period ExampleMoon Period ExampleCalculate the period of the moon’s orbit about the earth,

assuming a constant distance r = 3.84 x 108 m.

The magnitude of the attractive force must equal the centripetal force.

Fc = 42mr T2

In this case the attractive force is the gravitational force between the earth and moon F = G ME m

r2

where ME is the mass of the earth (5.98 x 1024 kg), m is the mass of the moon, and r is the earth-moon distance. We can equate these forces, solve for T and substitute the numerical values.

Period of Moon ExamplePeriod of Moon Example

The centripetal force is provided by the gravitational force, so that

GME m = 42mr

r2 T2

Solving for T gives

T = =

T=2.37x 106 s, or T = 27.4 days.

EGM

r324)1098.5)(/1067.6(

)1084.3(4242211

382

kgxkgNmx

mx

Period of a Satellite ExamplePeriod of a Satellite Example

Estimate the period of an artificial earth-orbiting satellite that passes just above the earth's surface.Set the force required to give a circular orbit--the centripetal force--equal to the gravitational force. The mass of the satellite = m. The mass of the earth ME, the radius of the orbit RE, and the satellite's period T

= T=

Use the result g RE2 = GME, the above

expression for the period becomes

T = =

Notice that the period depends only on the radius of the earth and the acceleration of gravity. Insert the approximate values of 2 =10,

g =10 m/s2, and RE = 6.4 x 106m,

T = = 5100 s = 85 min

2

24

T

Rm E2E

E

R

GmM

E

E

GM

R324

2

324

E

E

gR

Rg

RE24

2

6

/10

)104.6)(10(4

sm

mx

Homework!!Homework!!

Gravitational Field StrengthGravitational Field StrengthGravity works at a distance, and distance limits its strength

At any point in space, the strength of the field would be =F/m0 , where m0 is the test mass

Substituting, we get = GMm0 = GM

r2m0 r2

This picture illustrates that far from a body, the field lines are far apart and thus its strength is reduced.

Gravitation ConsiderationsGravitation Considerations

Orbital Speed--if an object is projected horizontally with enough speed, it remains in orbit around any celestial object

For Earth, this is 8000 m/s

This causes satellites to orbit every 90 min.

Greater radius causes greater period

Stationary orbiting satellite with period 24hrs has radius approx. 23,000 miles

Escape VelocityEscape VelocityEarth spacecraft must get entirely away from the earth

to go on to other planets

This requires giving a spacecraft enough energy to overcome the gravitational potential energy of earth

This gives an equation such that

Where M and R vary

according to the

celestial object involved

vGM

Resc 2

Black HolesBlack Holes

If the escape velocity is equal to the speed of light, gravity will keep even light from escaping--the idea behind the black hole

Conjecture due to observations from space

Theory is a supergiant star collapses in on itself creating super strong gravity at a small point

Black HolesBlack Holes

Gravity is great due to small distance with huge mass

Gravity only great near the object, at distance gravity is no different

Keplers LawsKeplers Laws

First Law: Each planet travels ina an elliptical path around the sun, and the sun is at one of the focal points

Keplers LawsKeplers Laws

Second Law: An imaginary line is drawn from the sun to any planet sweeps out equal areas in equal time intervals.

Keplers LawsKeplers Laws

Third Law: The square of a planet’s orbital period (T2) is proportional to the cube of the average distance (r3) between the planet and the sun or

T2 ∝ r3

Period and speed of an object in Period and speed of an object in circular orbitcircular orbit

Homework!!Homework!!

Major Equations!!Major Equations!!ac=v2/r

ac=42r/T2

ac=42rf2

ac= 2r

=v/r

=2/T

=2f

v=2r/T = 2rf

f=1/T

Fc=mac

Fc=mv2/r

Fg=GMm/r2

T=2

T2 = 42

r3 GM

r

g