file · Web viewThe reason we did this experiment to prove about Newton's first law and Newton's second law. In the experiments we did,

ATWOOD MACHINE

Afiq Agung*), Dirga Ayu, Surya Safitri, Siti Meilani

Fundamental Physics in Laboratory of Physics Department FMIPA

State University of Makassar

Abstract. Restrictions matter of this report is to know about Newton's laws I and II, the

effect of the acceleration of gravity and understanding of the pulley. The purpose of this

experiment is to know Newton's laws, calculate the acceleration of gravity, and know the

pulley system. The methodology used this time no direct experiments using T pole Atwood

machine to conduct experiments. From the results in the can, the authors conclude

experimental Atwood machine is a tool that is often used to observe the laws of mechanics to

uniformly accelerated motion. Simply put Atwood plane is composed of two objects

connected by a wire/rope. Acceleration that occurs is influenced by the mass of the load. The

greater the mass of the load acceleration that occurs would be even greater. T pole should the

correct position - vertical right because if not then there will be an additional force acting on

the Atwood machine experiment.

KEY WORDS: Acceleration, Atwood Machine, Newton’s Law, Pulley.

INTRODUCTION

In practice this time we will learn

two kinds of motion is linear motion and

rotational motion. The cause of this

motion and we will learn the truth about

the laws of motion we will investigate. In

this lab we also menggunakanalat and

materials are quite simple.

Atwood machine is a

experimental tool that is often used to

observe the laws of mechanics in

uniformly accelerated motion. Simply put

Atwood machine is composed of two

objects connected by a wire / rope. When

two equal mass objects, they will be

silent. But if one of the larger (eg m1>

m2). Then both objects will move in the

direction of m1 with accelerated. Style

pullers actual weight of the object 1.

However, because the object 2 also pulled

down (by gravity), then the resultant force

is towing a heavy object 1 minus the

weight of objects 2. Object 1 has weight

m1.g and object 2 has weight m2.g, there

are the resultant force is (m2-m1). G force

is moving both objects. Thus, the

acceleration of the two objects is the

resultant force divided by the mass of the

two objects.

From the above, then we in the

basic physics of this paper will

specifically discuss the plane Atwood by

experimenting directly. The reason we did

this experiment to prove about Newton's

first law and Newton's second law.

In the experiments we did, we

tried to prove whether Newton's law can

be applied to our props, the Atwood

machine. Props consisting of pole-scale R

is at its upper end there is a pulley, rope

hanger whose mass is negligible, two

loads of M1 and M2 cylinder with the

same mass M of each end of the rope tied

to the hanger, two additional burden to the

respective mass m1 and m2 respectively,

and the last one with a spring clasp,

securing the load and the additional load-

bearing cavities. This experiment was

done in order to fulfill our duty physics

lab report after the previous experiments

Atwood machine. Hopefully we've done

experiments that can give us a positive

impact, especially for the practitioner.

THEORY

When two objects of unequal mass

are hung vertically over a frictionless

pulley of negligible mass, as in Figure

1(a), the arrangement is called an Atwood

Machine. The device is sometimes used in

the laboratory to measure the free-fall

acceleration.

Determine the magnitude of the

acceleration of the two objects and the

tension in the lightweight cord.

FIGURE 1. The Atwood machine. (a)

Two objects (m2 > m1) connected by a

massless inextensible cord over a

frictionless pulley. (b) Free-body

diagrams for the two objects.

Conceptualize the situation pictured

in Figure 1(a)—as one object moves

upward, the other object moves

downward. Because the objects are

connected by an inextensible string, their

accelerations must be of equal magnitude.

The objects in the Atwood Machine

are subject to the gravitational force as

well as to the forces exerted by the strings

connected to them—thus, we can

categorize this as a Newton’s second law

problem. To analyze the situation, the

free-body diagrams for the two objects are

shown in Figure 1(b). Two forces act on

each object: the upward force T exerted

by the string and the downward

gravitational force. In problems such as

this in which the pulley is modeled as

massless and frictionless, the tension in

the string on both sides of the pulley is the

same. If the pulley has mass and/or is

subject to friction, the tensions on either

side are not the same and the situation

requires techniques.

We must be very careful with signs

in problems such as this. In Figure 1(a),

notice that if object 1 accelerates upward,

then object 2 accelerates downward. Thus,

for consistency with signs, if we define

the upward direction as positive for object

1, we must define the downward direction

as positive for object 2. With this sign

convention, both objects accelerate in the

same direction as defined by the choice of

sign. Furthermore, according to this sign

convention, the y component of the net

force exerted on object 1 is T – m1.g, and

the y component of the net force exerted

on object 2 is m2.g - T. Notice that we

have chosen the signs of the forces to be

consistent with the choices of signs for up

and down for each object. If we assume

that m2 > m1, then m1 must accelerate

upward, while m2 must accelerate

downward.

When Newton’s second law is

applied to object 1, we obtain

∑ F y=T−m1 g=m1 ay

Similarly, for object 2 we find

∑ F y=m2 g−T=m2 ay

When (2) is added to (1), T cancels and

we have

−m1 g+m2 g=m1 ay+m2 ay

a y=(m2−m1

m1+m2)g

The acceleration given by (3) can

be interpreted as the ratio of the

magnitude of the unbalanced force on the

system (m2 – m1)g, to the total mass of the

system (m1 + m2), as expected from

Newton’s second law.

When (3) is substituted into (1), we

obtain

T=( 2 m1 m2

m1+m2) g

If we have the same mass on both

sides, the system is balanced and it should

not accelerate. Mathematically, we see

that if m1 = m2, Equation (3) gives us ay =

0.

In the case in which one mass is

infinitely larger that the other, we can

ignore the effect of the smaller mass.

Thus, the larger mass should simply fall

as if the smaller mass were not there. We

see that if m1 >> m2, Equation (3) gives us

ay = -g.

Two blocks having masses m1 and

m2 are connected to each other by a light

cord that passes over two identical

frictionless pulleys, each having a

moment of inertia I and radius R, as

shown in Figure 2(a). Find the

acceleration of each block and the

tensions T1, T2, and T3 in the cord.

(Assume no slipping between cord and

pulleys).

FIGURE 2.(a) Another look at Atwood’s

machine. (b) Free-body diagrams for the

blocks. (c) Free-body diagrams for the

pulleys, where mp.g represents the

gravitational force acting on each pulley.

The motion of m1 and m2 is similar

to the motion of the two blocks in that

example. The primary differences are that

in the present example we have two

pulleys and each of the pulleys has mass.

Despite these differences, the apparatus in

the present example is indeed an Atwood

machine.

We shall define the downward

direction as positive for m1 and upward as

the positive direction for m2. This allows

us to represent the acceleration of both

masses by a single variable a and also

enables us to relate a positive a to a

positive (counterclockwise) angular

acceleration a of the pulleys. Let us write

Newton’s second law of motion for each

block, using the free-body diagrams for

the two blocks as shown in Figure 2(b):

m1 g−T 1=m1a

T 3−m2 g=m2 a

Next, we must include the effect of

the pulleys on the motion. Free-body

diagrams for the pulleys are shown in

Figure 2(c). The net torque about the axle

for the pulley on the left is (T1 – T2)R,

while the net torque for the pulley on the

right is (T2 – T3)R. Using the relation Στ =

Iα for each pulley and noting that each

pulley has the same angular acceleration α

we obtain

(T 1−T2 ) R=Iα

(T 2−T3 ) R=Iα

We now have four equations with

five unknowns: α, a, T1, T2, and T3. We

also have a fifth equation that relates the

accelerations, a = Rα. These equations can

be solved simultaneously. Adding

equations (7) and (8) gives

(T 1−T3 ) R=2 Iα

Adding Equation (5) and (6) gives

T 3−T1+m1 g−m2 g=(m1+m2)a

T 1−T 3=(m1−m2) g−(m1+m2) a (10)

Substituting Equation (10) into

Equation (9), we have

[ (m1−m2 )g−(m1+m2 )a ] R=2 Iα

Because α = ɑ/R, this expression can be

simplified to

(m1−m2) g−(m1+m2) a=2 I aR2

a=(m1−m2) g

m1+m2+2 IR2

(11)

Note that if m1 > m2, the acceleration is

positive; this means that the left block

accelerates downward, the right block

accelerates upward, and both pulleys

accelerate counterclockwise. If m1 < m2,

the acceleration is negative and the

motions are reserved. If m1 = m2, no

acceleration occurs at all.

METHODOLOGY OF EXPERIMENT

In this experiment have some

variables, if been identification the

variables so there are time as variable

control, moment inertia, velocity, and

acceleration are variable respond, and

distance as variable manipulation.

Moreover the variables can be definition

like variable operational, there are time is

a magnitude which declare how long the

something happen, moment inertia is a

measure of the inertia of an object to

rotate on its axis, velocity is a vector

quantity that shows how fast things move,

acceleration is the change of velocity in

units of time and distance is a number that

indicates how far an object changes

position through a certain path.

In this experiment required some

specific tools and materials that balance

ohauss 310 g (SSV 0.01 grams) to weigh

the mass of the load, time counter to

calculate how much time it takes to move

objects from C-A and A-B, the ruler (SSV

0.05) for measure the distance C to A and

the distance A to B, and the Atwood

machine consisting of a pole-scale R

which is topped with a pulley at the end of

p, hanger rope whose mass is negligible,

two load M1 and M2 are cylindrical with

the same mass respectively 64, 4 grams is

attached to the ends of the strap hanger, 1

additional burden to the mass 4.15 grams,

with a spring clasp G, B load restraint,

and retaining additional load A

perforated.

Before you make a

measurement, then make sure that the

tools you are using is in good condition

and complete the first step is to weigh the

burden of M1, M2, and m the balance

ohauss 310 grams, then plug grip G, A

retaining additional burden, and retaining

load B, on the pole scale. To investigate

whether the Atwood mechine working

properly, perform the following steps. The

first hanging M1 and M2 at the ends of

the rope and attach the pulley, pairs M1,

the grip G by using springs, investigate

whether mast parallel to the string scale, if

not, set up parallel. Add additional load

on M2 then press G, then M1 will be

released from the grip of G, and moves

upward, while M2 + m will move down.

If the machine is working properly then

the second load will move accelerated,

and when M2 + m through A, m will be

stuck in A, and then the system will move

straight uniform. If this does not happen

does not it lies the additional load restraint

A.

Before performing data collection

there are two activities that must be done

is the first event we have to note the

position of C and A, normalize time

counter, then release M1 and record the

time it takes for the object to move from

C to A. Do it 3 times with the same

distance measurement. Repeat the above

steps to move the position of A in order to

obtain 5 different distance from C to A,

then record the result in the observation

table. Further to the second activity we

determine the position of A and B and

record its position. Position B (under

position A) at a certain distance. Remove

the M1 and record the time it takes for the

object to move from A to B. Do this 3

times repeated measurements with the

same distance from A to B. Repeat the

above steps as many as 5 times the

distance A to B is different.

This experiment has analysis

method for calculate and analysis the data

from experiment result. The analysis

method for activity 1 :

α=2 xt2

=2 xt−2

Δα2=|δαδx

Δx|2+|δα

δtΔt|

2

Δα2=|δ xt−2

δxΔx|

2+|δ xt−2

δtΔt|

2

Δα2=|t−2 Δx|2+|xt−3 Δt|2

Δα2

α 2 = Δx2

x2 + Δt2

t2

Δαα =|√ Δx2

x2 +Δt2

t2 |

Δα=|√ Δx2

x2 + Δt2

t2 |α

Then the analysis method of

activity 2, that are :

v= xt=x .t

Δα2=|δvδx

Δx|2+|δv

δtΔt|

2

Δα2=|δ xt−1

δxΔx|

2

+|δ xt−1

δtΔt|

2

Δα2=|t−1 Δx|2+|xt−2 Δt|2

Δv2

v2 =Δx2

x2 + Δt2

t2

Δvv =|√Δx 2

x2 +Δt2

t 2 |

Δv=|√ Δx2

x2 +Δt2

t 2 |v

Then the experiment has analysis

method for moment inertia. The equation

are used for moment inertia analysis are :

I=|( M 2+m−M1 ) g

α−( M 2+m+M 1 )|R

2

ΔI=|δIδM1

ΔM1|+|δIδM 2

ΔM 2|+|δIδm

Δm|

+|δIδR

ΔR|+|δIδα

Δα|

ΔI=|δ(( M2+m−M 1)g

α −( M 2+m+M 1 ))R2

δM1ΔM1|

+|δ(( M 2+m−M 1 )g

α−(M 2+m+M 1 ))R2

δM2ΔM2|

+|δ(( M 2+m−M 1 )g

α−(M 2+m+M 1 ))R2

δmΔm|

+|δ(( M 2+m−M 1 )g

α−(M 2+m+M 1 ))R2

δRΔR|

+|δ(( M 2+m−M 1 )g

α−(M 2+m+M 1 ))R2

δαΔα|

ΔI=|(−gα−1 R2−R2 )ΔM 1|+|(−gα−1 R2−R2 ) ΔM2|+|(−gα−1 R2−R2 ) Δm|+|(( M 2+m−M1 )2gα−1 R−(M 2+m+ M1 )2 R ) ΔR|

+|(( M 2+m−M1 )gα−2 R2 ) Δα|

ΔI=|(3 gα−1+1 ) R2 ΔM|+|(( M 2+m−M1)2gα−1 R−(M 2+m+ M1)2R ) ΔR|+|(( M 2+m−M1)gα−2 R2 ) Δα|

EXPERIMENT RESULT AND

ANALYSIS DATA

EXPERIMENT RESULT

SSV of Neraca Ohauss 310 g = 0,01 g

Mass M1 = |64 ,40±0 ,01|gram

Mass M2 =|64 ,40±0 ,01|gram

Mass m = |4 ,15±0 ,01|gram

Radius of Pulley =|6 , 00±0 , 05|gram

Mass of Pulley (M) = |64 , 44±0 , 01|gram

SSV of Ruler = 0,1 cm

ACTIVITY 1Tabel 1. Relationship between distance and time for trajectory C to A

N. XCA (cm) TCA (s)

1 |84 ,50±0 ,05|1. |2 ,661±0 ,001|2. |2 , 661±0 ,001|3. |2 ,637±0 ,001|

2 |76 ,50±0 , 05|1. |2 , 474±0 ,001|2. |2 , 436±0 ,001|3. |2 ,442±0 ,001|

3 |72 ,60±0 , 05|1. |2 , 376±0 ,001|2. |2 , 370±0 ,001|3. |2 , 378±0 ,001|

4 |69 , 80±0 , 05|1. |2 , 252±0 , 001|2. |2 ,277±0 ,001|3. |2 , 277±0 ,001|

5 |64 ,20±0 ,05|1. |2 , 215±0 ,001|2. |2 ,147±0 ,001|3. |2 , 117±0 ,001|

ACTIVITY 2

SSV of Ruler = 0,1 cm

XAB = | 32,00±0,05 | cmTabel 2. Relationship between distance and time for trajectory A to B

N. XAB (cm) TAB (s)

1 |68 , 80±0 , 05|1. |3 , 074±0 , 001|2. |3 ,178±0 ,001|3. |3 ,158±0 ,001|

2 |62 , 40±0 , 05|1. |2 , 639±0 ,001|2. |2 ,627±0 ,001|3. |2 , 745±0 , 001|

3 |57 , 00±0 , 05|1. |2 , 337±0 ,001|2. |2 ,371±0 ,001|3. |2 , 315±0 , 001|

4 |52 , 20±0 , 05|1. |2 ,153±0 , 001|2. |2 , 117±0 ,001|3. |2 , 098±0 ,001|

5 |47 , 80±0 , 05|1. |1 , 879±0 ,001|2. |1 ,911±0 ,001|3. |1 , 890±0 ,001|

ANALYSIS DATA

Graph Analysis

ACTIVITY 1 (Rectilinear Changed

Motion)

XCA1 = 84,5 cm

tCA 1=2 ,661+2 ,661+2 ,637

3=2 , 653

s

xCA2 = 76,5 cm

tCA 2=2 ,474+2 , 436+2 ,442

3=2 ,451

s

xCA3 = 72,5 cm

tCA 3=2 , 376+2 , 370+2 ,378

3=2, 375

s

xCA4 = 69,8 cm

tCA 4=2 ,252+2 ,277+2 ,277

3=2 ,263

s

xCA5 = 64,2 cm

t CA 5=2 , 215+2 ,147+2 ,117

3=2 ,160

s

a=2 xt 2

= yx

y1 = 2(84,5) = 169 cm

y2 = 153 cm

y3 = 145,2 cm

y4 = 139,6 cm

y5 = 128,4 cm

x1 = (2,653)2 = 7,0384 s

x2 = 6,0057 s

x3 = 5,6390 s

x4 = 5,1468 s

x5 = 4,6641 s

4 6 8 0

20406080

100120140160180

f(x) = 16.7558671325369 x + 51.5509055304937R² = 0.992882038369204

t² (s²)

2x (c

m)

Graph 1. Rectilinear Changed Motion

y = mx + c

y = 16.756x + 51.551

a=m= yx=2 x

t 2=16 , 756

cm/s2

Degree of Believing

DB=R2×100 %

DB=0 ,9929×100 %

DB=99 ,29 %

Relative Error

RE=100 %−DB

RE=100 %−99 ,29 %

RE=0 ,71% …(4 SF)

Δa=RE×a

Δa=0 , 71100

×16 ,756cm/s2

Δa=0 ,1189 cm/s2

Physics Report

a=|16 , 7560±0 ,1189|cm/s2

ACTIVITY 2 (Rectilinear Motion)

XAB1 = 68,8 cm

t AB1=3 ,074+3 ,178+3 ,158

3=3 ,137

s

xAB2 = 62,4 cm

t AB2=2, 639+2 , 627+2 , 745

3=2 ,670

s

XAB3 = 57 cm

t AB1=2, 337+2 ,371+2 ,315

3=2 ,341

s

xAB4 = 52,2 cm

t AB1=2, 153+2 ,117+2 ,098

3=2 , 123

s

XAB5 = 47,8 cm

t AB1=1 , 879+1 , 911+90

3=1, 893

s

v= xt= y

x

0 2 4 0

10

20

30

40

50

60

70

80

f(x) = 16.929077661211 x + 16.454939865805R² = 0.990917421501634

t (s)

s (cm

)

Graph 2. Rectilinear Motion

y = mx + c

y = 16.929x + 16.455

v=m= yx= s

t=16 ,929

cm/s

Degree of Believing

DB=R2×100 %

DB=0 ,9902×100%

DB=99 , 02 %

Relative Error

RE=100 %−DB

RE=100 %−99 ,02%

RE=0 ,98 % …(4 SF)

Δv=RE×v

Δv=0 , 98100

×16 ,929cm/s

Δv=0 , 1659cm/s

Physics Report

v=|16 ,9290±0 ,1659|cm/s

Calculate Analysis and Indeterminance

ACTIVITY 1

Calculate and compare the moment of

inertia of the two equations

Equation I

I=1159 ,92 g.mc2

Equation II

I=|( M 2+m−M1 ) g

a−( M 2+m+M 1 )|R

2

I=|(64 ,44+4 ,15−64 ,4 )9,816 ,756

−(64 ,44+4 ,15+64 , 4 )|(6 )2

I=4667g.cm2

Calculation of acceleration, moment of

inertia and it’s uncertainty

1. When xCA= 84,5 cm

Δx=12×SSV ruler

Δx=12×0,1

cm

Δx=0 ,05 cm

tCA 1=2 , 661+2 ,661+2 , 637

3=2 , 653

s

δ 1=|t1−t|=|2 ,661−2, 653|=0 , 008s

δ 2=|t 2−t|=|2 ,661−2 ,653|=0 ,008s

δ 3=|t 3− t|=|2 ,637−2 , 653|=0 ,016s

δ max=Δt=0 ,016 s

a= 2xt2 =

2(84 ,5 )(2 ,653 )2=24 , 026

cm/s2

Δa=|√ Δx2

x2 +Δt2

t2 |a

Δa=|√(0 , 05 )2

(84 ,5 )2 +(0 ,016 )2

(2 ,653)2|24 ,026cm/s2

I=12

M pulley R2

I=12(64 , 44 )(6)2

Δa=0 ,159093 cm/s2

Relative Error

RE= Δaa

×100 %

RE=0 , 15909324 ,026

×100 %

RE=0 ,66 % ….(4 SF)

Degree of Believing

DB=100 %−RE

DB=100 %−0 , 66%

DB=99 , 34 %Physics Report

PR=|a±Δa|a=|24 ,02±0 ,15|cm/s2

I=|( M 2+m−M1) g

a−( M 2+m+M 1 )|R

2

I=|(64 , 44+4 , 15−64 ,4 )9,824 ,02

−(64 , 44+4 ,15+64 ,4 )|(6 )2

I = 4275,246 g.cm2

ΔI=|(3 gα−1+1 ) R2 ΔM|+|(( M 2+m−M 1 )2 gα−1 R−( M 2+m+M 1 )2 R ) ΔR|+|(( M 2+m−M 1 )gα−2 R2 ) Δα|

ΔI =|(3 (9,8)(24 ,02)−1+1 )62(0 ,01)|+|((64 ,4+4 , 15−64 ,4 )2(9,8 )(24 ,02 )−1 6− ¿|¿¿

¿

¿

¿

ΔI = 66,5447 g.cm2

Relative Error

RE= ΔII

×100 %

RE=66 ,54474725 ,246

×100 %

RE=1,4 %…..(3 SF)

Degree of BelievingDB=100%−RE

DB=98 ,6 %

Physics ReportPR=|I ±ΔI|g.cm2

I=|4725,2±66 , 5|g.cm2


Δx=0 ,05 cm

tCA=2 ,474+2 , 436+2, 442

3=2 ,451

s

δ1=|t1−t|=|2 ,661−2, 451|=0 , 023s

δ2=|t 2−t|=|2 , 436−2,451|=0 , 015s

δ 3=|t3−t|=|2 ,442−2 ,451|=0 , 009s

δ max=Δt=0 , 023 s

a=2xt2 =

2(76 , 5 )(2 ,451)2=25 ,4685

cm/s2

DB=100 %−1,4%

Δa=|√ Δx2

x2 +Δt2

t2 |a

Δa=|√(0 , 05 )2

(76 , 5 )2 +(0 , 023)2

(2 , 451)2|25 , 4685cm/s2

Δa=0 ,239573 cm/s2

Relative Error

RE= Δaa

×100 %

RE=0 , 23957325 , 4685

×100 %

RE=0 ,94 %….(4 SF)

Degree of Believing

DB=100 %−RE

DB=100 %−0 ,94 %

DB=99 , 06 %

Physics Report

PR=|a±Δa|a=|25 , 46±0 ,23|cm/s2

I=|( M 2+m−M1 ) g

a−( M 2+m+M 1 )|R

2

I=|(64 , 44+4 , 15−64 , 4 )9,825 , 46

−(64 , 44+4 , 15+64 , 4 )|(6 )2

I = 4728,713 g.cm2

ΔI=|(3 gα−1+1) R2 ΔM|+|(( M 2+m−M 1 )2gα−1 R−( M 2+m+M 1 )2R ) ΔR|+|(( M2+m−M 1 )gα−2 R2 ) Δα|

ΔI=|(3 (9,8)(25 , 46 )−1+1)62(0 , 01 )|+|((64 , 4+4 , 15−64 , 4 )2(9,8 )(25 , 46 )−1 6 ¿|¿¿

¿

¿

¿

ΔI = 69,96479 g.cm2

Relative Error

RE= ΔII

×100 %

RE=69,96479 4728,713

×100 %

RE=1 , 47 % …..(3 SF)

Degree of Believing

DB=100%−RE

DB=98 , 53%


I=|472 8,7±69 ,9|g.cm2


Δx=0 ,05 cm

DB=100 %−1 , 47 %

tCA=2 ,376+2 ,370+2 ,378

3=2 ,375

s

δ 1=|t1−t|=|2 ,376−2 ,375|=0 ,001s

δ 2=|t 2−t|=|2 ,370−2 ,375|=0 ,005s

δ3=|t 3− t|=|2 ,378−2 , 375|=0 ,003s

δ max=Δt=0 , 005 s

a=2 xt2 =

2(72,6 )(2 ,375 )2=25 , 741828

cm/s2

Δa=|√ Δx2

x2 +Δt2

t2 |a

Δa=|√(0 ,05 )2

(84 ,5 )2 +(0 ,016 )2

(2 ,653)2|25 ,741828 cm/s2

Δa=0 ,057019 cm/s2

Relative Error

RE= Δaa

×100 %

RE=0 , 05701925 ,74183

×100%

RE=0 ,22 % ….(4 SF)

Degree of Believing

DB=100 %−RE

DB=100 %−0 , 66%

DB=99 , 78 %

Physics Report

PR=|a±Δa|a=|25 ,74±0 ,05|cm/s2

I=|( M 2+m−M1 ) g

a−( M 2+m+M 1 )|R

2

I=|(64 ,44+4 ,15−64 , 4 )9,825 , 74

−(64 , 44+4 , 15+64 , 4 )|(6 )2

I = 4729,323 g.cm2

ΔI=|(3 gα−1+1 ) R2 ΔM|+|(( M 2+m−M 1 )2 gα−1 R−( M 2+m+M 1 )2 R ) ΔR|+|(( M 2+m−M 1 )gα−2 R2 ) Δα|

ΔI=|(3 (9,8)(25 , 74 )−1+1 )62(0 , 01)|+|((64 ,4+4 , 15−64 , 4 )2(9,8 )(25 , 74 )−16 ¿|¿¿

¿

¿

¿

ΔI = 63,66469 g.cm2

Relative Error

RE= ΔII

×100 %

RE=63 ,664694729 ,323

×100%

RE=1,3% …..(3 SF)


DB=98 , 7 %


I=|472 9,3±63 , 6|g.cm2

DB=100 %−1,3 %


Δx=0 , 05cm

t CA=2 ,252+2 , 277+2 ,277

3=2 ,268

s

δ 1=|t1−t|=|2 ,252−2,268|=0 , 016s

δ2=|t 2−t|=|2 ,277−2 , 268|=0 ,009 s

δ 3=|t 3− t|=|2 , 277−2 , 268|=0 , 009s

δ max=Δt=0 ,016 s

a=2 xt 2 =

2(69 ,8 )(2 ,268 )2=27 ,1394

cm/s2

Δa=|√ Δx2

x2 +Δt2

t2 |a

Δa=|√(0 , 05 )2

(69 ,8 )2 +(0 ,016)2

(2 ,268 )2|27 , 1394cm/s2

Δa=0 ,192444 cm/s2

Relative Error

RE= Δaa

×100 %

RE=0 , 19244427 ,1394

×100 %

RE=0,7 %….(4 SF)

Degree of Believing

DB=100 %−RE

DB=100 %−0,7 %

DB=99 , 3 %

Physics Report

PR=|a±Δa|a=|27 ,13±0 ,19|cm/s2

I=|( M 2+m−M1 ) g

a−( M 2+m+M 1 )|R

2

I=|(64 ,44+4 ,15−64 , 4 )9,827 , 13

−(64 ,44+4 ,15+64 , 4 )|(6 )2

I = 4732,252 g.cm2

ΔI=|(3 gα−1+1 ) R2 ΔM|+|(( M 2+m−M 1 )2 gα−1 R−( M 2+m+M 1)2R ) ΔR|+|((M 2+m−M 1 )gα−2 R2) Δα|

ΔI=|(3 (9,8)(27 , 13 )−1+1) 62 (0 , 01)|+|((64 ,4+4 , 15−64 , 4 )2(9,8 )(27 , 13)−1 6 ¿|¿¿

¿

¿

¿

ΔI = 68,86076 g.cm2

Relative Error

RE= ΔII

×100 %

RE=68 , 860764732 ,252

×100 %

RE=1,4 %…..(3 SF)


DB=98 ,6 %

Physics Report

DB=100%−1,4 %

PR=|I ±ΔI|g.cm2

I=|4 73 2,2±68 , 8|g.cm2


Δx=0 , 05cm

tCA=2 ,215+2,147+2 ,117

3=2 , 160

s

δ1=|t1−t|=|2 ,215−2 ,160|=0 , 055s

δ 2=|t 2−t|=|2 ,147−2 ,160|=0 , 013s

δ 3=|t3− t|=|2 ,117−2,160|=0 ,043s

δ max=Δt=0 ,055 s

a= 2xt2 =

2(64 ,2 )(2 ,160 )2=27 , 5205

cm/s2

Δa=|√ Δx2

x2 +Δt2

t2 |a

Δa=|√(0 , 05 )2

(64 ,2 )2 +(0 , 055)2

(2 , 451)2|27 , 5205cm/s2

Δa=0 ,701081 cm/s2

Relative Error

RE= Δaa

×100 %

RE=0 ,70108127 ,5205

×100 %

RE=2 , 51 %….(3 SF)

Degree of Believing

DB=100 %−RE

DB=100 %−2 , 51%

DB=97 , 49%

Physics Report

PR=|a±Δa|a=|27 ,5±0,7|cm/s2

I=|( M 2+m−M1 ) g

a−( M 2+m+M 1)|R

2

I=|(64 ,44+4 ,15−64 , 4 )9,827 , 52

−(64 ,44+4 ,15+64 , 4 )|(6 )2

I = 4732,999 g.cm2

ΔI=|(3 gα−1+1 ) R2 ΔM|+|(( M 2+m−M 1 )2 gα−1 R−( M 2+m+M 1 )2R )ΔR|+|(( M 2+m−M 1 )gα−2 R2 ) Δα|

ΔI=|(3 (9,8)(27 , 52)−1+1 )62(0 ,01 )|+|((64 , 4+4 , 15−64 , 4 )2(9,8 )(27 , 52)−1 6−(64 , 4+4 , 15+64 , 4 )2(6 ))0 , 05|+|((64 , 4+4 , 15−64 , 4 )(9,8 )(27 ,52 )−2(6 )2 )0,7|

ΔI = 85,18401 g.cm2

Relative Error

RE= ΔII

×100 %

RE=85,18401 4732,999

×100 %

RE=1 , 79%…..(3 SF)

Degree of Believing

DB=100 %−RE

DB=98 , 21%

DB=100 %−1 ,79%


I=|4 73 2,9±85 ,1|g.cm2

ACTIVITY 2Accounting Velocity a and indeterminancy 1. with XAB = 68,80 cm

Δx = 12 x SSV of the ruler

=12 x 0,1 = 0,05 cm

t AB1=3 ,074+3 , 178+3 ,158

3=3 ,137

sδ1 = |t1 – t∨¿ = |3,074 – 3,137| = 0,063 sδ2 = |t2 – t∨¿ = |3,178 – 3,137| = 0,041 sδ3 = |t3 – t∨¿ = |3,158– 3,137| = 0,021 sδ max = ∆t = 0,063 s

v = xt

v=68,803,137 = 21,93178 cm/s

Δv =|√ Δx2

x2 +Δt2

t 2 | vΔv = |√ 0,052

68,82 +0,0632

3,1372| 21,93178 cm/sΔv = 0,4407 cm/s

Relative Error

RE = Δvv x 100%

RE = 0,440721,9317 x 100%

RE = 2,009 % ….. ( 3 SF )

Degree of BelievingDB = 100% - KRDB = 100% - 0,4%DB = 97,99 %Physics ReportPR= | v ± Δv| v = | 21,9 ± 0,4∨cm /s

2. with XAB = 62,40 cm


=12 x 0,1 = 0,05 cm

t AB2=2,639+2 ,627+2 ,745

3=2 ,670

sδ1 = |t1 – t∨¿ = |2,639 – 2,670| = 0,031 sδ2 = |t2 – t∨¿ = |2,627 – 2,670| = 0,043 sδ3 = |t3 – t∨¿ = |2,745 – 2,670| = 0,075 sδmax= ∆t = 0,075 s

v = xt

v = 62,42,67 = 23,3707 cm/s

Δv =|√ Δx2

x2 +Δt2

t 2 | vΔv = |√ 0,052

62,42 +0,0752

2,672 | 23,3707

cm/sΔv = 0,65675 cm/s

Relative Error

RE = Δvv x 100%

RE = 0,24829,88 x 100%

RE = 2,81 % …. ( 3 SF )

Degree of BelievingDB = 100% - KRDB = 100% - 0,83%DB = 97,18%

Physics ReportPR = | v ± Δv| v = | 23,3 ± 0,6∨cm /s



=12 x 0,1 = 0,05 cm

t AB1=2, 337+2 ,371+2 ,315

3=2 ,341

s

δ1 = |t 1 – t|= |2,337 – 2,341| = 0,004 sδ2 = |t 2 – t|= |2,371 – 2,341| = 0,030 sδ3 = |t 3 – t|= |2,315 – 2,341| = 0,026 sδ max= ∆t = 0,03 s

v = xt

v = 572,341 = 24,34857 cm/s

Δv =|√ Δx2

x2 +Δt2

t 2 | v

Δv = |√ 0,052

572 +0,032

2,3412| 24,34857 cm/sΔv = 0,312758 cm/s

Relative Error

RE = Δvv x 100%

RE = 0,31275824,34857 x 100%

RE = 1,284 % ….. ( 3 SF )

Degree of BelievingDB = 100% - KRDB = 100% - 1,284%DB = 98,7155 %




=12 x 0,1 = 0,05 cm

t AB1=2,153+2 ,117+2 ,098

3=2 , 123

s

δ1 = |t 1 – t|= |2,153 – 2,123| = 0,03 s

δ2 = |t 2 – t|= |2,117 – 2,123| = 0,006 sδ3 = |t 3 – t|= |2,098 – 2,123| = 0,025 sδ max = ∆t = 0,03 s

v = xt

v = 52.22,123 = 24,58785 cm/s

Δv =|√ Δx2

x2 +Δt2

t 2 | vΔv = |√ 0,052

52,22 +0,032

2,1232| 32,28

cm/sΔv = 0,348247 cm/s

Relative Error

RE = Δvv x 100%

RE = 0,33632,28 x 100%

RE = 1,41 % … ( 3 SF )





=12 x 0,1 = 0,05 cm

t AB1=1 ,879+1 , 911+1 ,890

3=1 ,893

s

δ1 = |t 1 – t|= |1,879 – 1,893| = 0,014 sδ2 = |t 2 – t|= |1,911 – 1,893| = 0,018 sδ3 = |t 3 – t|= |1,890 – 1,893| = 0,003 sδmax = ∆t = 0,018 s

v = xt

v = 47,81,893 = 25,25 cm/s

Δv =|√ Δx2

x2 +Δt2

t 2 | vΔv = |√ 0,052

47,82 +0,032

1,8932| 25,25

cm/sΔv = 0,401 cm/s

Relative Error

RE = Δvv x 100%

RE = 0,45233,43 x 100%

RE = 1,58 % ….. ( 3 SF )



DISCUSSION

In this experiment has concluded

that if the object has been observed

moving uniformly accelerated and in

accordance with the principles of

Newton's second law the body of mass M1

= 64,4 grams moving upward and M2 + m

= (64,4 + 4,15) grams moving down the

accelerated 16,756 cm/s2 and the activity

2 that these objects experiencing uniform

rectilinear motion and in accordance with

the principles of Newton's first law

discovered in activity 2 with a uniformly

accelerated motion principle do with the

speed of movement of 16.,929 m/s at the

time of object with mass m that had

united the body of mass M2 off and stop

moving.

CONCLUSION

After doing this experiment finally

concluded that Newton's laws apply in the

working principle Atwood Machine. That

is Newton's Second Law in activity 1 with

the principle of uniformly accelerated

motion and Newton's law on the principle

of rectilinear motion activity 2.

Besides the moment of inertia

affect the acceleration of objects on the

working principle Atwood Machine.

Because of the greater moment of inertia

of the pulley then the acceleration of a

moving object will be smaller.

REFERENCEE

[1] Halliday & Resnick. 2012.

Fundamental of Physics (Nineth ed.).

John Willey & Sons, Inc.

[2] Tim Dosen Fisika Dasar I. 2013.

Penuntun Praktikum Eksperimen Fisika

Dasar I Unit Laboratorium Fisika Dasar

Jurusan Fisika FMIPA UNM.

file · Web viewThe reason we did this experiment to prove about Newton's first law and Newton's second law. In the experiments we did,

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