BAB 5

Bab 5
Circular Motion

Hello readers!!!!!

Before discussing this topic, do you still remember about circumference???? I hope yo won't forget it. Because we're going to use this circumference material for this topic. So, what is circumference?

Circumference is can be said as perimeter of the circle to determine the length from a point of a circle to this point in the circle. if we want to determine that, don't forget this formula....

s = 2πR

1) Uniform Circular Motion 

    Uniform Circular motion is the motion of an object traveling at a constant speed of a circular motion. 

    priod is the time requuired to travel once around the circle to make one revolution. 

      

       V = 2πR / T    

    If expressed in units of radiian, one complete circle covers an angle of Ө = 2π radians. Therefore, the angular velocity of the object making a uniforom circular motion is 

ω= Ө/T = 2π / T

     * ω = angular velocity 

we get a relationship between speed and angular velocity in a circular motion as : 

V = ωR

Before we start to the next, these are the notes that we also have to know. 

     

1 rpm = 2πrad/60 rad/s

2) Centripetal acceleration 

     Magnitude ac of the centripetal acceleration on the speed v of the object and the radius r of the circular path. 

Ac = V^2 / R

3) Applications 

•      Safe Driving 

             

μs ---- dry : 0.9 

μs ---- icy : 0.1 

V = √μgr

μ = friction 

For the safe riding, the velocity must be lower than √μgr. 

• Satellites in circular orbit

    

G = universal gravitation

    = 6.67 x 10^-11 N m^2/k^2

BAB 4

Bab 4 

Projectile Motion 

Before we continue to this topic, lets see this table first.....

After finish reading this table, let's continue to this topic again!

• Initial speed of the projectile :

            1. Vxo = Vo cos Ө (In horizontal direction)

            2. Vyo = Vo sin Ө (In vertical direction)

•  Speed of the projectile at arbitrary time :

           1. In horizontal direction :

                Vx = Vxo

           2. In vertical direction : 

                Vy = Vyo - gt 

• Position of projectile at arbitrary time :

             1. In horizontal direction : 

                 X = Xo + Vxo t

             2. In vertical direction :

                 y = yo + Vyot - 1/2 gt^2

• Maximum height :

           1.  Time required by the projectile to arrive the peak of its path is :

                                         Tm = Vyo : g 

                     * g = gravity 

            2. The maximum heigth of the path of the projectile is : 

                                          Hm = 1/2 x Vyo^2 : g

• Range and Maximum Range 

            1. Time to reach its path : 

                     T = 2tm

            

            2. Because the horizontal motion is a motion with uniform speed, we find the range of the                       projectile is : 

                    

                    R = VxoT = Vo^2 x sin 2Ө : g

                            

                             Which is equal To....

                  

                    R = Vo^2 x 2 x (cos Ө) x (sin Ө) : g 

Example :

A projectile was shot with initial speed of 100 m/s making an elevation angle of 30。with respect horizontal direction. determine : 

a) Components of initial velocity of the projectile. 

         Vxo = Vo cos Ө = 100 x cos 30 = 100 x √3 /2 = 50√3 m/s

         Vyo = Vo sin Ө = 100 x sin 30 = 100 x 1/2 = 50 m/s

b) Compoents of velocity and position of the projectile one seconds after shooting.

         Vx = Vox = 50√3 m/s

         Vy = Vyo - gt = 50 - 10 x 1 = 40 m/s

    Components of position at t = 1 s

         Xo = 0

         yo = 0

       X = Xo + Vxot = 0 + 50√3 x 1 = 50√3 m

       Y = Yo + Vyot - 1/2 gt^2 = 0 + 50 x 1 - 1/2 x 10 x 1^2 = 45 m 

c) Time required to reach the peak of its path. 

        Tm = Vyo/g = 50/10 = 5 sec

d) The maximu height of the path 

        Hm = 1/2 x Vyo^2/g = 1/2 x 50^2/10 = 125 m

e) The time required by the projectile to reach a height of one fourth its maximum height of the path is 

              h = 1/4 x Hm = 1/4 x 125 = 31.25 m

             

   Use equation 

              h = y - yo = Vyot - 1/2 gt^2

    31.25 = 50 x t - 1/2 x 10 x t^2

    31.25 = 50t - 5t^2 or t^2 - 10t + 6.25 = 0

         T1 = - (-10) - √(-10)^2 - 4 x 1 x 6.25 / 2 x 1 = 10 - 5√3 / 2 s

         T2 = - (-10) - √(-10)^2 - 4 x 1 x 6.25 / 2 = 10 + 5√3 / 2 s

 From this calculation, we found moments in time when the projectile reached one fourth of if maximum height. The smaller time is required when the projectile is moving up and the longer one is the time required when the projectile is moving back down to the ground. 

f) the time required to reach the ground again :

        T = 2Tm = 2 x 5 = 10 sec

g) The range of the projectile : 

        R = VyoT = 50 x 10 = 500 m

BAB 3

Bab 3

Linear Motion

>An object is said to make a linear motion if its path form a straight line. 

>Therefore, linear motion is also called one dimensional.

1. Quantities of linear motion :

   1. Velocity

   2. Accelaeration

   3. Decceleration

   4. Speed

   5. Length

   6. Displacement

   7. Average value of velocity and acceleration

   8. Instantenous of velocity and acceleration.

   9. Distance

   >Example :

      Distance : 6+8 = 14

      Displacement : Displacement = 10 Km

 

   >The difference between speed and velocity

      Speed = there is direction

      Velocity = there's no direction

   >How to measure ?

      Average velocity = displacement/time duration

      Average speed = distance traveled/time duration

      Instantenous = V=DeltaX/DeltaT

2. Linear Motion woith constant velocity

    If an object moves with zero acceleration, the velocity of the object remains constant which is             exactly equal to the instantaneous velocity.

3. Linear motion with constant acceleration

    At this motion, the velocity of the acceleration remains constant, the average acceleration is                 exactly equal to the instantaneous acceleration.

4. Equation

    1. Provides the relation between initial velocity, find velocity, the time taken to reach it , and the acceleration.

        initial velocity = Vo

        Final velocoity = Vi

        Time taken = t

        Acceleration = a

        Change of vellocity = Vi - Vo

       

                   a = (Vi - Vo)/t     or      Vi = Vo + at

   2. Relation between distance traveled, initial velocity, acceleration, and time

       S = average velocity x time

          = ((Vo + Vi)/2) x t

       Vi = Vo + at

       Substitute for Vi, we get S

       S = ((Vo + Vo + at)/2) x t = Vot + at^2/2 

       S = Vot + at^2/2

   

   3. Realtion between initila velocity, find velocity, acceleration and distane. There's no factor in it.

       Vi = Vo + at

      - Squaring, we get 

         Vi^2 = (Vo + at)^2

                  = Vo^2 + 2Voat + a^2t^2

                  = Vo^2 + 2a (Vot + a t^2/2)

                  = Vo^2 + 2as 

         Vi^2 = Vo^2 + 2as

5. Vertical Object

    This topic discusses about free falling. When an object is free falling, this object will also has the       acceleration which is caused by the G (gravitation). The constant value of acceleration which is           caused by gravitation is 9.8m/s^2 or 10m/s^2. 

    Equation : 

                       Vi = Vo + gt

                       S = Vot + 1/2 gt^2

                       Vi2 = Vo^2 + 2gs

6. Position VS Time 

 

1.

2. 

Answer : 

1. 

2. 

    

7. Velocity : + Vs -

  

BAB 2

Bab 2

Vector

What is vectors?

= It is completely defined if both magnitude and direction are mentioned.

On the other hand, quantities which are completely defined only by magnitude are known as scalar.

1. Symbolism of vector

    

d  2. Addition of vector

       a.  Triangular Method

        

   

c     b. Parallelogram Method

      > Determine the of resultants of vectors by analytical method

            Using Consinus formula

      > Before continue to the consinus formula, we have to know how to get the resultant by using this          graph.

     

     > Subtraction of  two vectors is identical to the addition of a vector with the negative of the other           vector. The subtraction is C = A- B

        

        

     > Determine the resultant of vectors  

       

                                                                     

 

       > Vector multiplication 

             1. Dot product

             2. Cross Product

            

             1. Dot Product

                 = Geometrically, the dot product of two vector is the magnitude of one times the projection of                 the second onto the first.

          

       2. Cross Product

          = Geometrically the cross product of two vectors is the area of the parallelogram between them. 

         

       The way to find the direction in cross is using the right rule. These are the steps :

      1.Hold your right hand flat with your thumb perpendicular to your fingers. Do not bend your thumb at anytime.

2. Point your fingers in the direction of the first vector.

3. Orient your palm so that when you fold your fingers they point in the direction of the second vector.

4. Your thumb is now pointing in the direction of the cross product.

         Example question 

         1. 

          2. 

BAB 1

Bab 1

Quantity, Unit, and Measurement

1. Quantities

a. Base Quantities

    Characteristic :

• They are not derivations of other quantities.
• They can produce or derive other quantities.

        Base Quantities and Function

• Length (L) = To express the length, width, etc. (metre (m))
• Mass (M) = To express mass of an object (kilogram (kg))
• Time (T) = To express time duration of an event (second (S))
• Electric current (I) = To express the strength of electric current (ampere)
• Temperature = To express the warmness of things (kelvin)
• Light intensity (J) = To express the brightness of light irradiation (candela)
• Amount of substance (N) = To express the amount of particles composing things (mol)

b. Derived Quantities

    Definition : All physical quantities except the base quantities

    Example : Volume, mass density, velocity, etc.

    volume = combination of length, length, and length

    mass density = combination of  mass, length, length, and length

    velocity = combination of length and time

c. Measurement

    Definition = is the comparison of the magnitude of quantity with the value of measuring devices

2. Units

a. S1 units

Base quantities	Unit	Abbreviation
Length	Meter	M
Mass	Kilogram	Kg
Time	Second	S
Electric Current	Ampere	A
Temperature	Kelvin	K
Light intensity	Candela	Cd
Amount of substance	mole	Mol

b. Derived S1

    Because derived quantities are combination of base quantities, the units of the derived quantities are combinations of units of the corresponding base quantities. For example : unit of velocity=m/s

c. Prefixes of units

Prefixes	Symbols	Multiplication Factors
	More than One
Kilo	K	10^3
Mega	M	10^6
Giga	G	10^9
Tera	T	10^12
Peta	P	10^15
exa	E	10^18

	Less than one
Milli	m	10^-3
Micro	Micro	10^-6
Nano	n	10^-9
Piko	p	10^-12
Femto	f	10^-15
Atto	a	10^-18

	Other frequently used prefixes
Deci	d	10^-1
Centi	c	10^-2
angstrom	A	10^-12

2. Measurement

·         Measuring is the process to know the quantities of the object in standard units

·         The tools are ruler, vernier caliper, micro meter

a.       Ruler

Smallest scale = mm (1 mm)

Uncertainty : ½ smallest scale (05 mm)

Precision : same calculation and uncertainty

b.      Vernier Caliper

Smallest scale : 0.1 mm

Uncertainty : ½ smallest scale (0.05Precision : same calculation with Uncertainty

Contuns base scale and nonius scale

c.       Micro meter

Smallest scale : 0.01 mm

Uncertainty : ½ x 0.01 : 0.005 mm

Precision : ½ x 0.01 : 0.005 mm

·         Factors are different tools or device, human error, environment , factor superstition

·         Report the measurement result :

              X=Xoriginal +/-  deltaX

              Xoriginal : calculation result

              Delta x : Uncertainty

·         Current measurement result

3. Dimension

Base Quantity	Dimension
Length	(L)
Mass	(M)
Time	(T)
Current	(I)
Temperature	(Teta)
Light intensity	(J)
Number of substance	(N)

Example :

·         3 kg + 2 N = 3.2 Kg

      (M) + (M) . (L) (T)^-2

      (M)^2(L)(T)

·         (v) = (s)/(t) = (L)/(t) = (L)(T)^-1

·         (a) = (v)/(t) = (L)(T)^-1/(T) = (L)(T)^-2

4. Significant figure

Rules :

  Ø  All non zero digits are considered significant

Ex : 91 has two significant figure

        12345 has five significant figure

  Ø  Zeros appearing anywhere between two non-zero digits are significant

Ex : 104.2356 has seven significant figures

  Ø  All number in scientific notation are scientific figures 

  Ø  Leading zero are not significant figure

Ex : 0.00000236 has 3 significant figures.

  Ø  Trailing zero in a number containing a decimal point are significant.

Ex : 13.2500 has six significant figures

        0.0000125600 still has six significant figures

        150.00 has five significant figures since it has three trailing zero.

Multiplication and Division

0.628 cm x 2.2 cm = 1.38226 cm^2

                                = 1.4 cm^2 (2 sf)

4.554 x 105 kg : 3.00 x 10^2 kg = 1.5718 x 10^5-2

                                                     = 1.52 10^3 (3 sf)