Uniform Circular Motion

Motion 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

a

c

=

v 2 /r

Example 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?

a

c =

v

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

Rotation and Revolution

Rotation -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 Revolution

Revolution 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

Period

Another important measure in UCM is period, the time for 1 rotation or revolution Since x=v 0 t , this implies that vT = 2



r and thus T= 2



r/ v Rearranging differently, v= 2



r/ T and then inserting it into the acceleration equation

a

c = v

2

/r = 4

 2

r/T

2

Example

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 10 8 m.

Example

a

c = 4  2 r = 4  2 (3.84 x 10 8 ) T 2 (2.36 x 10 6 ) 2

a

c = 2.72 x 10 -3 m/s 2 T= 27.3 da (24 hr/da)(3600 s/hr) = 2.36 x 10 6 s The ratio of the moon's acceleration to that of an object falling near the earth is

a

c = 2.72 x l0 -3 m/s 2

g

9.8 m/s 2 = 1 3600

Frequency

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 a c

a

c =4

 2

f

2 r equation

Example

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, 2  r , 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

= 2  r

f v

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

v =

2540cm/s = 25.4 m/s.

Angular 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

w =

v/r In terms of period

w = 2  /T

In terms of frequency

w =2  f

Example

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/s T 4.0 s 2.0

(b) Because the riders travel in a circle, they undergo a centripetal acceleration given by a c = w 2 r = (  /2 rad/s) 2 (7.7m) = 19m/s 2 . Notice that this is almost twice the acceleration of a body in free fall.

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

Angular displacement - q -- Radians around circular path Angular velocity - w --radians per second, angle between fixed axis and point on wheel changes with time Angular acceleration - a -- increase of w , when angular velocity of the rigid body changes, radians per seconds squared

Rotational Kinematics

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

o t + 1/2

a

t 2

w

f

=w 0 q=

x/r

+a

t

a=

a/r

w f 2 = w o 2 q=(w 0 w=

v/r

+ 2 aq +w

f

)

t

/ 2

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 a a. a = D w D

t

4.2 s b. Find q c. How far does the car go?

= (42-70) rads/s = -6.7 rads/s 2 b

.

q= w

o t + 1/2

a

t 2 =

(70)(4.2) + 1/2(-6.7)(4.2) 2 = 235 rads c. q =

x / r

in rads so

x =

q

r =

(0.32)(235) = 75 m

Example

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?

q = 2 revs = 2(2  ) radians = 4  rads w f =0 rads/s w o = 0.21 rads/s Use angular equivalent of

v

f 2 =

v

o 2 + 2

ax

which is w f 2 = w o 2 + 2 aq (0) 2 = (0.21) 2 + 2 a (4  ) a = -(0.21) 2 = -1.8 x 10 -3 rad/s 2 2(4  )

Homework: Read pp.898-903 Practice Problems 7A, 7B

Forces in Circular Motion

Centripetal 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 Examples Gravitation 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 F c =ma c = mv 2 /r

Example

Approximately 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 =

m

a.

The mass of the moon is 7.35 x 10 22 kg.

F c = m

a

c = m 4  2 r F c = (7.35 x 10 22 T 2 kg)4  2 (3.84 x 10 8 m) ((27.3 days)(24 hr/day)(3600 s/hr)) 2 F c =2.005 x 10 20 N.

Forces 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

Example

Imagine 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 = m

g.

The centripetal force required to carry the person around a circle of radius

r

is

F =

m

a c =

m 4  2 r T

2

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

T:

m

g

= m 4  2 r

r

750

m

T

2

T=2 

g

=2  .

/ 2 = 55s = 0.92 min.

Banked Curves

“Banking” road curves makes turns without skidding possible For angle q , 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.

F N sin q = mv 2 /r F N thusly q = tan -1 (v 2 cos /gr) q = mg tan q = v 2 /gr This equation can give the proper angle for banking a curve of any radius at any linear speed

Banked 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 q =

tan

-1 q from

(

v

2

/ g r

) = tan

-1

(

66.7

2

/9.81(975)) =

24.9

o

Homework!!

Law 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 F g = Gm 1 m 2 r 2

Center 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 gravity

Sometimes 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 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 m 2 This puts F g in Newtons kg 2

G

can be used then to find values of many astronomical properties

Example

Consider 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 =

GmM E r

2

M E = 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= GmM E r 2

or

g =GM r 2 E

Both G and M E 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 R E

g = GM E R E 2

Example

Show 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 r 2

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

F c

gravitational force. Equating these two forces, we get

=GmM =

4  2 mr Rearranging gives

T 2 =

4  2

r 2 T 2 r 3 GM

Example

Use the law of universal gravitation and the measured value of the acceleration of gravity

g

The density

,

r to determine the average density of the earth. of an object is defined as its mass per unit volume: r

= m/V

where

m

is the mass of the object whose volume is

V

From a previous example

g = GM E

R E 2

Substitute for M an expression involving r , r

= M E /V

.

If we take the earth to be a sphere of radius

R E .

Then r =

M E

4/3



R E 3

and M E =

4 / 3



R E 3

r The equation for

g g

=

G

(

4/3



R E 3

can then be rewritten in terms of the density as r ) =

4 / 3 G



R E

r R E 2

Density Example (cont)

Upon rearranging, we find the density to be r =

3g

4 

R E G

Inserting the numerical values, we get r = 3(9.81 m/s 2 ) 4  (6.38 x 10 6 m)(6.67 x 10 -11 N m 2 / kg 2 r = 5.50 x 10 3 kg/m 3

Moon Period Example

Calculate the period of the moon’s orbit about the earth, assuming a constant distance

r

= 3.84 x 10 8 m.

The magnitude of the attractive centripetal force.

F c

force must equal the

= 4



2 mr

T 2

In this case the attractive force is between the earth and moon the gravitational force

F

=

G M E

m r 2 where M E is the mass of the earth mass of the moon, and

r

is the (5.98 x 10 24 kg),

m

is the earth-moon distance. We can equate these forces, solve for

T

and substitute the numerical values .

Period of Moon Example

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

GM E m = 4



2 mr r 2 T 2

Solving for T gives T = 4  2

r

3 =

GM E

4  2 ( 3 .

84

x

10 8

m

) 3 ( 6 .

67

x

10 11

Nm

2 /

kg

2 )( 5 .

98

x

10 24

kg

) T=2.37x 10 6 s, or T = 27.4 days.

Period of a Satellite Example

Estimate the period of an

m

4 

T

2 2

R E R E

2

E

T= 4  2

GM R

3

E E

artificial earth-orbiting satellite that passes just above the earth's surface.Set the force required to give a Use the result g R E 2 = GM E , the above expression for the period becomes 4  2

R E

3 T = =

gR E

4  2

g R E

Notice that the period depends only circular orbit--the on the radius of the earth and the centripetal force--equal to the gravitational force. The acceleration of gravity. Insert the approximate values of  2 =10, mass of the satellite =

m

. g =10 m/s 2 , and R E = 6.4 x 10 6 m, The mass of the earth M E , the radius of the orbit

R E ,

and the satellite's period

T

4 ( 10 )( 6 .

4

x

10 6

m

)

Homework!!

Gravitational Field Strength

Gravity works at a distance, and distance limits its strength At any point in space, the strength of the field would be G =F/m 0 , where m 0 is the test mass Substituting, we get G = GMm 0 = GM r 2 m 0 r 2 This picture illustrates that far from a body, the field lines are far apart and thus its strength is reduced.

Gravitation 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 Velocity

Earth 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 2

GM

Where M and R vary

v esc

= according to the

R

celestial object involved

Black 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 Holes

Gravity is great due to small distance with huge mass Gravity only great near the object, at distance gravity is no different

Keplers Laws

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

Keplers Laws

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

Keplers Laws

Third Law: The square of a planet’s orbital period (T 2 ) is proportional to the cube of the average distance (r 3 ) between the planet and the sun or T 2 ∝ r 3

Period and speed of an object in circular orbit

Homework!!

Major Equations!!

a

c =v 2 /r f=1/T a c =4  2 r/T 2 a c =4  2 rf 2 a c = w 2 r w =v/r F c =ma c F c =mv 2 /r F g =GMm/r 2 w =2  /T w =2 

f

T=2 

r g

v=2  r/T = 2  rf T 2 r 3 = 4  2 GM