ln_revision of circular motion and gravitational field

7/27/2019 LN_Revision of Circular Motion and Gravitational Field

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Revision of Circular Motion

and Gravitational Field


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Introduction1. Angular velocity, , of an object is defined as the rate of change of

angular displacement.

1

2 A (at t 1)

B (at t 2)

2 1

2 1

22

d f

dt t t T

Linear Speeds r

v r t t

2: Furthermore

Where s is the arc length


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Exercise


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Centripetal Force and Acceleration1. Newtons first law states that an object continues in its state of motion in

a straight line unless acted upon by a force.

2. For circular motion to occur, a force must be acting on the object as itsdirection is kept on changing with time.

3. This object is therefore undergoing acceleration. Hence, there is a forceacting on the object that is performing circular motion.

4. For uniform circular motion, the magnitude of the velocity is constantwhich means there is no acceleration along direction of motion, theacceleration is perpendicular to the direction of motion. Thus, accelerationis directed towards the center of the circle which implies that there mustbe a resultant force directed towards the center of the circular motion.

F

F1

Question: Based on the diagram on the right, explainwhy the centripetal force must be perpendicular to thevelocity in a uniform circular motion?


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Centripetal Force and Acceleration

22va r

r r

v

5. It can be shown, for an object move in a circular motionwith radius r and a speed of v , the centripetal acceleration

is as shown below. Direction of the acceleration is alwaystowards the center of the circle

6. Hence, the RESULTANT FORCE needed to keep an object moving in acircle (also known as centripetal force) is given by

22v F ma m mr

r

Note: It is important to identify whether the object is moving in ahorizontal circle or vertical circle.

It is also important to draw a free-body diagram to indicate all forcesacting on the object. But remember, centripetal force is the resultant

of other forces


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Exercise


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Exercise


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4 A pendulum bob of mass 2.0kg is attached to one end of a string of length1.2m. The bob moves in a horizontal circle in such a way that the string is

inclined at 300

to the vertical. Calculate(a) the tension in the string(b) the period of the motion. 30 o

Exercise


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5. A mass of 0.50kg is rotated in a horizontal circle by a string 1.0mlong. The maximum tension in the string before it breaks is 50N.Determine the greatest number of revolution per second of theobject.

Exercise


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w

w w

wT

T T

T

Centripetal Force and Acceleration2. The following diagram shows the vehicle move INSIDE a circulartrack.(a) Force acting on the water in a bucket when the bucketwhirled around in a vertical circle.(b) Force acting on the passenger in a rollercoaster.

A

B

C

D

At point A and C:2

2mvT mr

r

At point B2

2mvT W mr r

At point D

22mvT W mr

r


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Centripetal Force and Acceleration2. The following diagram shows the vehicle move OUTSIDE a circulartrack. N3

N2

N1 N4

W

W

W W

Circular MotionNot Possible

A

C

B


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Non-Uniform Circular Motion1. Many of the times, object does not travel with a constant speedthroughout the circular path. A common example is the loop-the-loop inamusement parks. The car is moving at a very high speed at the bottom of the loop and its speed decreases gradually as it climbs up the loop.

2. As speed keeps on changing, it would be most convenient to use energyconservation to help to solve problem.

Total Energy at A = Total Energy at B

A

B


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6. Consider a loop-the-loop ride. If the loop is of radius of 10 m and thepassenger has a weight of 50kg.(a) Calculate the minimum speed the passenger must have at the top of theloop.(b) Calculate the minimum speed of the passenger entering the loop (IgnoreFriction)

Exercise


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7. A pilot banks the wing of his aircraft so as to travel at a speed of 360km/h in a horizontal circular path of radius 5.0km. Calculate thebanking angle of the aircraft.

Exercise


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8. A circular curve of highway is designed for traffic moving at 60km/h.If the radius of the curve is 150m, calculate the angle of banking of the

road such that vehicles will be able to negotiate the bend withoutskidding when the road is slippery.

Exercise


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9. The figure below shows a toy runway. A small model car is released at X, rundowns the slope, loop -theloop and travels on towards Z. The radius of the loop is

0.25m.

h

Z

PX (a) What is the minimum speed

with which the car must pass pointP at the top of the loop if it is toremain in contact with the runway?

(b) Calculate the value of h toachieve this minimum speed.

Exercise


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Gravitational Field

m1 m2

r

F F

1 22

m m F G

r Where G = 6.67 x 10 -11 Nm 2kg -2

Fr

F

g m

2

Mm F G

r 2GM

g r

As Therefore

(1)

(2)


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Gravitational FieldKeplers Law states that for a planet orbiting round the sun, the square of its period of revolution is proportional to the cubes of their mean distances from the sun.

22

2

2

3

2

3

2 3 2

22 3

2

2

4

4

Mm vG m mr

r r GM

r

T

GM T r

r GMT

T r GM


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Geostationary Orbit1. A geostationary orbit refers to a circular orbit around the Earth where an orbiting

satellite would appear stationary to an observer on the Earths surface. This implies (a) The period T for a geostationary satellite must be 24 hours(b) The satellite must rotate from west to east, as the earth moves from west to

east.(c) The satellite must be placed vertically above the equator. If the orbit does not

coincide with the plane of the equator, it will sometimes be in the northern or southern hemisphere, hence cannot be geostationary.

2. Since period T is only dependent on the radius of the orbit, all geostationary orbit for earth will have the same radius, implying only 1 geostationary orbit; hence limiting

parking space for orbiting geostationary satellites.


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Gravitational Potential 1. The gravitational potential , at a point is defined as the work done in bring a unit

mass from infinity to that point.

GM

r

2. The negative sign indicates that negative work is done. In other words, the potentialat infinity is greater than the potential at a point nearer to the source mass.Conventionally, the potential at infinity is taken to be zero.

3. The gravitational potential energy, U, of a mass, at a point in a gravitational field isdefined as the work done in bringing the mass from infinity to that point.

r GM

U m

GMm

U r


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Total Energy of A SatelliteA satellite in orbit has both potential and kineticenergies. The total energy is the sum of the potentialand kinetic energies:

2

2

2

2

.

12

And

1Therefore .

2 2

,2 2

E U K E

GMmmv

r GMm mv

r r GMm

K E mvr

GMm GMm GMm Hence Total Energyr r r

Energy

K.E

G.P.E

Total Energy

r


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Relationship Between Terms


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Escape Speed1. Escape Speed is the minimum speed at which an object can be

projected so that it reaches infinity.

2. On Earth Surface:

3. At an infinite distance from the Planet:Minimum kinetic energy = 0

Minimum gravitational potential energy = 0

2

2 2

1 0 021 2

,2

2

GMmmv R

GMm GM mv v

R R

GM v

R

21Kinetic Energy2

mv

Gravitational Potential EnergyGMm

r

Where v is the launching speed

Hence:


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Binary (Double) Star System

1. A binary star is a star system consisting of two stars orbiting around their common centre of mass. The brighter star is called the primary and the other is its

companion star.2. The gravitational force of attraction between the two stars provide the centripetalforce that enable them to orbit around the common centre.

3. The period of revolution of the two stars are the same.

4. Due to different radius of the revolution, their linear velocities are different.

22

2

22

2

For small star ( )

For big star ( )

s

b

v MmG mr mr R r

V MmG MR M

r R R

M

Star H

Star L

Common CentreR

r


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ln_revision of circular motion and gravitational field

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