ucm and gravitation by nabih

Uniform Circular MotionAnd Gravitation

Table of contents.1- UCM• Definition• Angular abscissa • Frequency in rotation• Angular velocity• Acceleration• Force2. Gravitation

04/13/23 IB Physics (IC NL) 1

1- Uniform Circular Motion

What does it mean?

How do we describe it?

What can we learn about it?


What is UCM?

Motion in a circle with: Constant Radius R Constant Speed |vv| Velocity is NOT constant

(direction is changing) There is acceleration!


How can we describe UCM?

In general, one coordinate system is as good as any other: Cartesian:

(x,y) [position] (vx ,vy) [velocity]

Polar: (R,) [position] (vR ,) [velocity]

In UCM: R is constant (hence vR = 0) (angular velocity) is constant. Polar coordinates are a natural way to describe Polar coordinates are a natural way to describe

UCM!UCM!04/13/23 IB Physics (IC NL) 4

Polar Coordinates:

The arc length s (distance along the circumference) is related to the angle in a simple way: s = R, where is the angular displacement. units of are called radians.

For one complete revolution:2R = Rc

c = 2

has period 2.

1 revolution = 21 revolution = 2radiansradians

RR

vv

x

y


Polar Coordinates...

x = R cos y = R sin

RR

x

y

(x,y)

2 3/2 2

-1

1

0

sincos


Polar Coordinates...

In Cartesian coordinates, we say velocity dx/dt = v. x = vt

In polar coordinates, angular velocity d/dt = . = t has units of radians/second. Remember it is a derivative of abscissa.

Displacement s = vt. but s = R = Rt, so:

RR

vv

x

y

st

v = R04/13/23 IB Physics (IC NL) 7

Period and Frequency Recall that 1 revolution = 2 radians

frequency (f) = revolutions / second (a) angular velocity () = radians / second (b)

By combining (a) and (b) = 2 f

Realize that: Period (T) = seconds/revolution So T = 1 / f = 2/ Or f = /T

RR

vv

s


Acceleration in UCM: Even though the speedspeed is constant, velocityvelocity is notnot constant since the

direction is changing: there must be acceleration! Consider average acceleration in time t

aav = vv / t

* Bold letter means vector

tvv1

vv2 vv

vv1vv2

RR

vv (hence vv/t )

points at the origin!

We see that aa points

in the - R R direction.

aa = dvv / dt


Acceleration in UCM:

This is called This is called Centripetal Acceleration. Now let’s calculate the magnitude:

vv2

vv1

vv1vv2

vv

RRRR

v

v

R

RSimilar triangles:

But R = vt for small t

So: v

v

v t

R

v

t

v 2

R

a v 2

R


Centripetal Acceleration

UCM requires an acceleration: Magnitude: a = v2 / R Direction: (toward center of circle)

raa

r̂


Uniform Circular Motion

A force, Fr , is directed toward the center of the circle.

This force is associated with an acceleration, a.

Applying Newton’s Second Law along the radial direction gives


Uniform Circular Motion, cont

A force causing a centripetal acceleration acts toward the center of the circle

It causes a change in the direction of the velocity vector

If the force vanishes, the object would move in a straight-line path tangent to the circle


Motion in a Horizontal Circle

The speed at which the object moves depends on the mass of the object and the tension in the cord

The centripetal force is supplied by the tension

T mv2

r


Centripetal Force

The force causing the centripetal acceleration is sometimes called the centripetal force

This is not a new force, it is a force that causes circular motion.

The centripetal force is a resultant force and not an acting external force.


Question

If the rotation is flipped from clockwise to anti-clockwise, what happens to the direction of the centripetal force? Remains the same Flips direction


Conical Pendulum

The object is in equilibrium in the vertical direction and undergoes uniform circular motion in the horizontal direction

v is independent of m

T cos mg

F T sin mac mv2

r

Dividing: tan v2

rg v rg tan


Horizontal (Flat) Curve The force of static

friction supplies the centripetal force

The maximum speed at which the car can negotiate the curve is

Note, this does not depend on the mass of the car

fs mv2

r


Banked Curve (not IB)

These are designed with friction equaling zero

There is a component of the normal force that supplies the centripetal force

ncos mg

nsin mv2

r


Loop-the-Loop

This is an example of a vertical circle

At the bottom of the loop (b), the upward force experienced by the object is greater than its weight


Loop-the-Loop, Part 2

At the top of the circle (c), the force exerted on the object is less than its weight


Non-Uniform Circular Motion

The acceleration and force have tangential components

Fr produces the centripetal acceleration

Ft produces the tangential acceleration

F = Fr + Ft


Vertical Circle with Non-Uniform Speed

The gravitational force exerts a tangential force on the object Look at the

components of Fg

The tension at any point can be found


Top and Bottom of Circle

The tension at the bottom is a maximum

The tension at the top is a minimum

If Ttop = 0, then


Motion in Accelerated Frames

A fictitious force results from an accelerated frame of reference A fictitious force appears to act on an

object in the same way as a real force, but you cannot identify a second object for the fictitious force


“Centrifugal” Force

From the frame of the passenger (b), a force appears to push her toward the door

From the frame of the Earth, the car applies a leftward force on the passenger

The outward force is often called a centrifugal force

It is a fictitious force due to the acceleration associated with the car’s change in direction


Fictitious Forces, examples

Although fictitious forces are not real forces, they can have real effects

Examples: Objects in the car do slide You feel pushed to the outside of a

rotating platform Centrifuge Drying machine


Fictitious Forces in Linear Systems

The inertial observer (a) sees

The non-inertial observer (b) sees


Fictitious Forces in a Rotating System

According to the inertial observer (a), the tension is the centripetal force

The non-inertial observer (b) sees


Question:

Why do airplanes make banked turn?

Answer: To generate the centripetal force required for the circular motion.


2- GravitationNewton’s universal law

Every mass attracts every other mass through the force called gravity.

The force of attraction is directly proportional to the product of their masses.

The force of attraction decreases with the square of the distance between the mass centers. This is called an inverse square law.


The Gravitational Force Fg


Newton’s Law of Universal Gravitation

2

12

21

r

mmFg 2

12

21

r

mmGFg

1110673.6 G22 kgmN

People have been very curious about the stars in the sky, making observations for a long time. But the data people collected have not been explained until Newton has discovered the law of gravitationEvery 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.

How would you write this principle mathematically?

With GWith G

G is the universal gravitational constant, and its value is

Unit?

This constant is not given by the theory but must be measured by experiment.

This form of forces is known as an inverse-square law, because the magnitude of the force is inversely proportional to the square of the distances between the objects.


More on Law of Universal Gravitation

It means that the force exerted on the particle 2 by particle 1 is attractive force, pulling #2 toward #1.

Consider two particles exerting gravitational forces to each other.

Gravitational force is a field force: Forces act on object without physical contact between the objects at all times, independent of medium between them.

12221

12 r̂r

mmGF

The gravitational force exerted by a finite size, spherically symmetric mass distribution on a particle outside the distribution is the same as if the entire mass of the distributions was concentrated at the center.

m1

m2

r

F21

F12

12r̂ Two objects exert gravitational force on each other following Newton’s 3rd law.

Taking as the unit vector, we can write the force m2 experiences as

12r̂

What do you think the negative sign mean?

gF

How do you think the gravitational force on the surface of the earth look?

2E

E

R

mMG


Example for GravitationUsing the fact that g=9.80m.s-2 at the Earth’s surface, find the average density of the Earth.

g

Since the gravitational acceleration is

So the mass of the Earth is

G

gRM E

E

2

Therefore the density of the Earth is

2E

E

R

MG 2

111067.6E

E

R

M

E

E

V

M

3

2

4E

E

R

GgR

EGR

g

4

3

33

6111050.5

1037.61067.64

80.93

kgm


Free Fall Acceleration & Gravitational Force

Weight of an object with mass m is mg. Using the force exerting on a particle of mass m on the surface of the Earth, one can get

•The gravitational acceleration is independent of the mass of the object•The gravitational acceleration decreases as the altitude increases•If the distance from the surface of the Earth gets infinitely large, the weight of the object approaches 0.

What would the gravitational acceleration be if the object is at an altitude h above the surface of the Earth?

mg

What do these tell us about the gravitational acceleration?

2E

E

R

mMG

g2E

E

R

MG


Example for Gravitational ForceThe international space station is designed to operate at an altitude of 350km. When completed, it will have a weight (measured on the surface of the Earth) of 4.22x106N. What is its weight when in its orbit?

The total weight of the station on the surface of the Earth is

Therefore the weight in the orbit is

GEF

OF

Since the orbit is at 350km above the surface of the Earth, the gravitational force at that height is

MEE

OF

mg 2E

E

R

mMG N61022.4

'mg 2hR

mMG

E

E

GE

E

E FhR

R2

2

GE

E

E FhR

R2

2

N66256

26

1080.31022.41050.31037.6

1037.6


Variation of g with height

The object is at a height z above earth What is the value of gz relative to go the value of g at

the surface of the Earth?


MEE

ucm and gravitation by nabih

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