GRAVITATIONAL MOTION

Newton’s Law of Universal Gravitation

states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Gravity is a consequence of mass.The force of gravity between two objects is directly proportional to the product of the objects’ masses.The force of gravity between two objects is inversely proportional to the square of the distance between the two objects’ centers of mass.

Newton’s Law of Universal Gravitation

Universal Gravitational Constant:

Mass of Earth:

Radius of Earth:

Universal Constants

ACCELERATION DUE TO GRAVITY

g

2

2

2

R

GMR

GMmmg

R

GMmF

Planet of Radius

object of mass

Planet the of Mass

Constant nalGravitatio Universal

nGravitatio of Force

R

m

M

G

F

So . . . g = ?Now work it out. Now work it out. Work that problem out. Get that problem right.

The mass of planet Jupiter is 1.9 x 1027 kg and that of the sun is 1.99 x 1030 kg. The mean distance of Jupiter from the sun is 7.8 x 1011m. Calculate the gravitational force which the sun exerts on Jupiter.

Universal Gravitation #2

mr

kgNmG

kgm

kgm

11

2211

302

271

108.7

/1067.6

1099.1

109.1

N

N

m

kgkgkgNm

r

mmGF

23

24

22302711

211

30272211

2

21

101454

104145

104145

1087

10991109110676

.

.

.

).(

).)(.() /.(

Where will the ball go?

What will happen to the Moon?

http://www.mrwaynesclass.com/teacher/circular/CarAndCurve.htm

TORQUE

Torque

Torque

Is the measure of how effectively a force causes rotation.

angleθ

distancer

forceF

torqueτ

sinθFr τ

Torques in equilibrium

When the torques associated with two masses balance each other.

rFτ

:where

0τ-τ

g

21

Torques in equilibrium

Two people with the same mass. Which picture shows torques in equilibrium?

A B

Torques in equilibrium

A – the distances are not the same in B so the torques will not balance.

A B

PROBLEM 1

Alfred weighs 400 N. He sits on one end of a seesaw 1.5 m from the fulcrum. Ann weighs 200 N. How far from the fulcrum must she sit to balance the seesaw?

?

PROBLEM 1

F1 = 400 Nr1 = 1.5 mF2 = 200 Nr2 = ?

1 = 2 F1r1 = F2r2

400(1.5) = 200(r2)r2 = 3 m

1.5 m3 m

CIRCULAR MOTION

Linear Motion

d – distance (in meters)v – velocity (in meters/second)a – acceleration (in meters/second2)

Distance = 2r

Linear/Tangential Velocity

Objects moving in a circle still have a linear velocity = distance/time.This is often called tangential velocity, since the direction of the linear velocity is tangent to the circle.

v

Angular Motion

– angular displacement (in radians) – angular velocity (in radians/second) – angular acceleration (in radians/second2)r – radius of circle (in meters)

Circular Motion Terms

The point or line that is the center of the circle is the axis of rotation.If the axis of rotation is inside the object, the object is rotating (spinning).If the axis of rotation is outside the object, the object is revolving.

Angular Velocity

Objects moving in a circle also have a rotational or angular velocity, which is the rate angular position changes.Rotational velocity is measured in degrees/second, rotations/minute (rpm), etc.Common symbol, w (Greek letter omega)

Dq

𝜔=∆𝜃∆ 𝑡

Rotational & Linear VelocityIf an object is rotating:

All points on the object have the same rotational (angular) velocity.All points on the object do not have the same linear (tangential) velocity.

Linear velocity of a point depends on:The rotational velocity of the point.

More rotational velocity means more linear velocity.The distance from the point to the axis of rotation.

More distance from the axis means more linear velocityIn symbols:

v = r w

vr w

Linear to Angular

d = rvT = r

aT = r

r

Angular & Linear Velocity

If an object is rotating: All points on the object have the same Angular velocity.All points on the object do not have the same linear (tangential) velocity.

Angular & Linear Velocity

Linear velocity of a point depends on:The Angular velocity of the point.

More rotational velocity means more linear velocity.

The distance from the point to the axis of rotation.

More distance from the axis means more linear velocity.

Acceleration

As an object moves around a circle, its direction of motion is constantly changing.Therefore its velocity is changing.Therefore an object moving in a circle is constantly accelerating.

Centripetal Acceleration

The acceleration of an object moving in a circle points toward the center of the circle.This is called a centripetal (center pointing) acceleration. ac

Centripetal Acceleration

2

2

ra

r

va

c

Tc

locityangular ve

pathcircular of radiusr

onaccelerati lcentripetaa

Speed Tangential

c

Tv

The centripetal acceleration depends on:The velocity of the object.The radius of the circle.

A spinning ride at a carnival has an angular acceleration of 0.50 rad/s2. How far from the center is a rider who has a tangential acceleration of 3.3 m/s2?

mr

r

rαaT

6.6

)50(.3.3

Circular Motion #8

What is the tire’s angular acceleration if the tangential acceleration at a radius of 0.15m is 9.4 x 10-2 m/s2?

2/63.

15.094.

srad

raT

Circular Motion #9

A test car moves at a constant speed of 10 m/s around a circular road of radius 50 m. Find the car’s A) centripetal acceleration and B) angular speed.

sradr

vB /2.0

50

10)

222

/250

)10() sm

r

vaA c

Circular Motion #10

A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05 m/s2, what is its tangential speed?

?

/05.8

2.48

:

2

T

c

v

sma

mr

Givens

smv

rav

T

cT

/7.192.4805.8

2.4805.82

Circular Motion #11

A race car moves along a circular track at an angular speed of 0.512 rad/s. If the car’s centripetal acceleration is 15.4 m/s2, what is the distance between the car and the center of the track?

?

/4.15

/512.

:

2

r

sma

srad

Givens

c

ma

r

rr

r

r

va

c

c

7.58)512(.

4.15

)(

22

222

Circular Motion #12

A piece of clay sits 0.20 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 20.5 rad/s, what is the magnitude of the centripetal acceleration of the piece of clay on the wheel?

?

20.

/5.20

:

ca

mr

srad

Givens

2

2

2

/05.84

)5.20)(20(.

sm

rac

Circular Motion #13

tangential and centripetal acceleration are perpendicular to one another.

T

c

cTtotal

a

a

aaa

1

22

tanTa ca

totalaθ

Tangential vs. Centripetal

2

2

1

22

2

if

i

if

tt

t

time(s)t

)on(rad/sacceleratiangular

nt(rad)displacemeangular

/s)locity(radangular ve final

/s)locity(radangular ve initial

2

f

i

Equations for Circular Motion

Angular Acceleration

– angular velocity (radians/second) – angular acceleration (radians/second2)t – time (seconds)

=

A ventilator fan is turning at 600 rev/min when the power is cut off, and it turns 1000 rev while coasting to a stop. Calculate the angular acceleration and the time required to stop.

??

1000

min/0

min/600

t

rev

rev

rev

f

i

Circular Motion #14

2

2

22

min/180

3600002000

)1000(2)600(0

2

rev

if

min33.3

)180(6000

t

t

tif

A bicycle wheel rotates with a constant angular acceleration of 3.5 rad/s2. If the initial speed of the wheel is 2 rad/s at t = 0 s. a) Through what angle does the wheel rotate in 2 s? b) what is the angular speed at t = 2 s?

Circular Motion #15

rad

tti

1174

)2)(5.3(2

1)2(2

2

1

2

2

srad

tif

/972

2)5.3(2

A potter’s wheel moves from rest to an angular speed of 0.20 rev/s in 30s. Find the angular acceleration in rad/s2.

2/?

30

/0

/256.11

220.0

srad

st

srad

sradrev

rad

s

rev

i

f

Circular Motion #16

2/041.

)30(0256.1

srad

tif

A dentist’s drill starts from rest. After 3.20 seconds of constant angular acceleration it turns at a rate of 2.51 x 104 rev/min. a) find the drill’s angular acceleration. b) Determine the angle (radians) through which the drill rotates during this period.

??

2.3

/0

/13.262760

min1

1

2

min25100

st

srad

sradsrev

radrev

i

f

Circular Motion #16

2/821

)2.3(013.2627

srad

tif

rad

tti

52.4203

)2.3)(821(2

10

2

1

2

2

A floppy disk in a computer rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s.A) What is the angular acceleration of the disk, assuming angular acceleration is uniform? B) How many revolutions does the disk make while coming up to speed? C) Find v if r = 4.45cm D) Find at if r = 4.45cm.

??

892.

/4.31

/0

st

srad

srad

f

i

Circular Motion #18

2/2.35

)892(.04.31

)

srad

tA if

revrad

revrad

radttB i

23.22

114

14)892)(.2.35(2

10

2

1) 22

smsradmrvC /4.1)/4.31)(0445(.)

22 /57.1)/2.35)(0445(.) smsradraD t

Centripetal Force

Newton’s Second Law says that if an object is accelerating, there must be a net force on it.For an object moving in a circle, this is called the centripetal force.The centripetal force points toward the center of the circle.

Centripetal Force

In order to make an object revolve about an axis, the net force on the object must pull it toward the center of the circle.This force is called a centripetal (center seeking) force. Fc

Centripetal Force

Centripetal force on an object depends on:The object’s mass - more mass means more force.The object’s velocity - more speed means more force.And…

Centripetal Force

The centripetal force on an object also depends on:

The object’s distance from the axis (radius).

If linear velocity is held constant, more distance requires less force.If rotational velocity is held constant, more distance requires more force.

22

ωmrr

vmmaF t

cc

tvcF

s)speed(rad/ angular

path(m) circular of radiusr

speed(m/s) tangentialv

(kg) massm

(N) force lcentripeta

t

ω

Fc

Centripetal Force

A test car moves at a constant speed of 10 m/s around a circular road of radius 50 m. Find the car’s A) centripetal acceleration and B) angular speed.

sradr

v

smr

va

mr

smv

tc

t

/2.50

10

/2

50

/10

22

Circular Motion #19

A 0.015 kg rubber stopper is attached to 0.93 m length string. The stopper is swung in a horizontal circle, making one revolution in 1.18 seconds. A) find the speed of the stopper. B) find the force the string exerts on it.

Circular Motion #20

sradrev

rad

s

rev

mr

kgm

/32.51

2

18.1

1

93.

015.0

smsradmrvA /95.4/32.593.) NmrFB c 4.)32.5(93.015.) 22

A 50 kg runner moving at a speed of 8.8 m/s rounds a bend with a radius of 25 m. A) Find the centripetal acceleration of the runner B) what supplied the force needed to give this acceleration to the runner?

Circular Motion #21

smv

mr

kgm

/8.8

25

50

2

22

/09.325

)8.8() sm

r

vaA tc

Nr

vmFB t

c 88.15425

8.850)

22

“Centrifugal Force”

“Centrifugal force” is a fictitious force - it is not an interaction between 2 objects, and therefore not a real force.Nothing pulls an object away from the center of the circle.What is erroneously attributed to “centrifugal force” is actually the action of the object’s inertia - whatever velocity it has (speed + direction) it wants to keep.

As a car makes a turn, the force of friction acting upon the turned wheels of the car provide the centripetal force required for circular motion.

As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion.

As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion.

Without a centripetal force, an object in motion continues along a straight-line path.

With a centripetal force, an object in motion will be accelerated and change its direction

The tendency of a moving object to continue in a straight line in the absence of an unbalanced force and to turn in a circle in the presence of a inward-directed force (i.e., centripetal force) has been experienced by any passenger in an automobile. When the car makes a sudden turn, the passengers tend to continue in their straight line path. This straight line motion continues until the presence of a side door or another passenger pushes upon the passenger in order to accelerate him/her towards the center of the turn. The force experienced by the passenger is an inward force; without it, the passenger would slide out of the car.

Feelings of weightlessness and heaviness are associated with the normal force; they have little to do with the force of gravity. A person who feels weightless has not lost weight; the force of gravity acting upon the person is the same magnitude as it always is. Witness in the animation above that the force of gravity is everywhere the same. The normal force however has a small magnitude at the top of the loop (where the rider often feels weightless) and a large magnitude at the bottom of the loop (where the rider often feels heavy). The normal force is large at the bottom of the loop because in order for the net force to be directed inward, the normal force must be greater than the outward gravity force. At the top of the loop, the gravity force is directed inward and thus, there is no need for a large normal force in order to sustain the circular motion. The fact that a rider experiences a large force exerted by the seat upon her body when at the bottom of the loop is the explanation of why she feels heavy. In actuality, she is not heavier; she is only experiencing the large magnitude of force which is normally exerted by seats upon heavy people while at rest.

Banking Curve

When an automobile is driven around a sharp turn on a perfectly level road, friction between the tires and the road provides the centripetal force. If this centripetal force is not adequate, the car may slide off the road. The maximum speed with which a car can negotiate a turn of a given radius.

rgv

r

vg

r

mvmg

FF

s

s

s

cs

2

2

massm

Radius r

)(9.8m/sgravity g

Friction Static oft Coefficien

Force lCentripeta

Friction Static of Force

2

s

c

s

F

F

Banking of a Cyclist

A cyclist provides himself the necessary centripetal force by leaning inward on a horizontal track, while going around a curve. The cyclist should bend through an angle to have the necessary centripetal force while going around a curved path.

)/8.9(

tan

2

21

smgravityg

radiusr

velocityv

rg

v

What is the maximum speed at which an automobile can negotiate a curve of radius 100m with out sliding, if the coefficient of static friction is 0.7?

Circular Motion #22

sm

Rgv

r

vg

r

mvmg

FF

s

s

s

cs

/1.26

)8.9)(100)(7.0(

2

2

A car travels at a constant speed of 13.4 m/s on a level circular turn of radius 50m. What is the minimum coefficient of static friction between the tires and roadway in order that the car makes the circular turn without sliding?

Circular Motion #23

37.8.950

)4.13(

50

)4.13()8.9(

2

2

2

2

s

s

s

s

cs

r

vg

r

mvmg

FF

Find the required banking angle for a curve of radius 300m if the curve is to be negotiated at a speed of 80 km/h without the need of a frictional force.

02

12

1 5.9)8.9(300

2.22tantan

/2.223600

1

1

100080

rg

v

sms

h

km

m

h

kmv

Circular Motion #24

Minimum Velocity Required

Motion in a Vertical CircleThe minimum velocity that the object should possess at the top so that the string doesn’t slack is

The minimum velocity that the object should possess at the bottom so that the string doesn’t slack is

grvtop

grvbottom 5

A person swings a pail of water in a vertical circle with a radius of 1 m. What is the minimum speed the person must maintain at the top and bottom to keep the water from falling out?

smgRv top /.. 13389

smgRv bottom /).( 78955

Circular Motion #25