dinamik.docx

7/27/2019 dinamik.docx

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OBJECTIVE

1.To calculate angular acceleration.

2. To determine the relations between angular acceleration and load.

LEARNING OUTCOME

At the end of this laboratory session, students should be able to :

1.Apply mathematical formulae of rotational motion of a rigid body.

2.Identitify factors affecting the angular acceleration of a body.

THEORY

Rotational motion is more complicated than linear motion, and only the motion of rigid

bodies will be considered here. A rigid body is an object with a mass that holds a rigid shape,

such as a phonograph turntable, in contrast to the sun, which is a ball of gas. Many of the

equations for the mechanics of rotating objects are similar to the motion equations for linear

motion The angular displacement of a rotating wheel is the angle between the radius at the

beginning and the end of a given time interval. The SI units are radians. The average angular

velocity(, Greek letter omega), measured in radians per second, is

The angular acceleration(, Greek letter alpha) has the same form as the linear quantity

and is measured in radians/second/second or rad/s2.

The kinematics equations for rotational motion at constant angular acceleration are

f= 0tAngular velocity as a function of time

Angular displacement as a function of velocity and time

Angular displacement as a function of time


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f = 0 + 2Angular velocity as a function of displacement

Consider a wheel rolling without slipping in a straight line. The forward displacement of the

wheel is equal to the linear displacement of a point fixed on the rim. As can be shown in

Figure1, d= S= r

Figure 1A wheel rolling without slipping.

In this case, the average forward speed of the wheel is v = d/ t= ( r)/t= r, whereris the

distance from the center of rotation to the point of the calculated velocity. The direction of

the velocity is tangent to the path of the point of rotation.

The average forward acceleration of the wheel is aT = r(f o)/ t= r. This component of

the acceleration is tangential to the point of rotation and represents the changing speed of the

object. The direction is the same as the velocity vector.

The radial component of the linear acceleration is ar= v2/ r= 2r.

When a body spin, it experiences angular acceleration based on the value of force that

accelerates it. According to Newtons Second Law,forceFcan be related to accelaeration as

F=ma

m=mass of body

a =linear acceleration

As the body experiences angular acceleration the value of linear acceleration needs to be

converted to angular acceleration based on the position of force. By identifying that force is

applied at a certain radius r from the center of rotation ,its angular acceleration can then be

calculated using the following equation

= a / r
http://www.cliffsnotes.com/study_guide/Rotational-Motion-of-a-Rigid-Body.topicArticleId-10453,articleId-10419.html#huetinck3831c01-fig-0028http://www.cliffsnotes.com/study_guide/Rotational-Motion-of-a-Rigid-Body.topicArticleId-10453,articleId-10419.html#huetinck3831c01-fig-0028http://www.cliffsnotes.com/study_guide/Rotational-Motion-of-a-Rigid-Body.topicArticleId-10453,articleId-10419.html#huetinck3831c01-fig-0028http://www.cliffsnotes.com/study_guide/Rotational-Motion-of-a-Rigid-Body.topicArticleId-10453,articleId-10419.html#huetinck3831c01-fig-0028


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When the rotation is given in unit of rpm n,it must be converted to angular velocity with

unit (rad/s) which can be calculated by using the equation

=2n/60

Angular acceleration for a predetermined period can also be calculated by tacking the

difference of angular velocity within the time divided by the time period.

=2-1/t


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APPARATUS

1. Rotational Apparatus2. Mass hanger (5 g)3. Set of weights (15 gram to 95 gram)4. The software EM028


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PROCEDURE

1. Base plate of the rotational apparatus adjusted so that it stands horizontal.2. EM208 programme switched on.3. Rope around the driving unit winded up and it was let pass through the pulley.

Height of the pulley was adjusted.

4. A weight hanger with a 15 gram weight was hold to the end of the rope.5. Key and weight was released at same instant.6. When the rope was unrolled completely pressed again.7. Readings of

a. Initial angular velocity(SPPED0)b. Final angular velocity (SPEED1)c. Initial time (TIME0)d. Final time (TIME1)

Recorded.

8. Step 3 to 7 was repeated twice and average was calculated.9. By winding up the same driving unit, step 3 to 8 repeated by replacing 15 g weight

with 35 g, 55 g, 75 g and finally 95 g.


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EXPERIMENTAL DATA

EXPERIMENTAL RESULT

Load, (g) reading initial angular Final angular initial time Final time

(include mass velocity velocity TIME0,t (s) TIME1,t(s)

hanger) SPEED0, (rpm) SPEED1, (rpm)

1 3.4 7.5 194.4 231.8

20 2 2.5 7.5 320.8 361.4

3 4 7.9 418.5 460.3

1 9.3 18.2 505.3 520.7

40 2 4 17.3 555.9 575.6

3 4.5 18.2 610.8 630.6

1 7.1 24.2 667.9 682.2

60 2 7.1 24.2 709.7 724

3 7.1 23.7 752.5 766.8

1 9.3 29.2 860.2 871.2

80 2 7.5 28.7 897.5 909.6

3 8.3 28 939.3 950.3

1 7.1 33.2 986.5 997.5

100 2 8.3 32.2 1023.9 1033.8

3 10.1 33.2 1064.5 1074.4

Table 2 Calculated data for angular velocity,time dif ferent and angular acceleration

Load,(g) Reading initial angular Final angular Time di fferent Angular Acceleratio

(include mass velocity (rad/s) velocity (rad/s) t (s) (rad/s)

hanger) average

1 0.356 0.785 37.4 0.011

20 2 0.262 0.785 40.6 0.013 0.011

3 0.419 0.827 41.8 0.0098

1 0.974 1.906 15.4 0.061

40 2 0.419 1.812 19.7 0.071 0.68

3 0.471 1.906 19.8 0.072

1 0.744 2.534 14.3 0.125

60 2 0.744 2.534 14.3 0.125 0.124

3 0.744 2.482 14.3 0.122

1 0.974 3.058 11 0.18980 2 0.785 3.005 12.1 0.183 0.187

3 0.869 2.932 11 0.188

1 0.744 3.477 11 0.248

100 2 0.869 3.372 9.9 0.278 0.254

3 1.058 3.477 9.9 0.244


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DISCUSSION

1. Calculate the angular velocity of each speed-reading and show all yourcalculations.

Speed to angular velocity

= =2n/60

Load,g Reading Initial angular

Velocity,(rad/s)

Final angular

Velocity,(rad/s)

20 1 2()(3.4)/60

=0.356

2()(7.5)/60

=0.785

2 2()(2.5)/60

=0.262

2()(7.5)/60

=0.785

3 2()(4.0)/60

=0.419

2()(7.9)/60

=0.827

40 1 2()(9.3)/60

=0.974

2()(18.2)/60

=1.906

2 2()(4.0)/60

=0.419

2()(17.3)/60

=1.812

3 2()(4.5)/60

=0.471

2()(18.2)/60

=1.906

60 1 2()(7.1)/60

=0.744

2()(24.2)/60

=2.534

2 2()(7.1)/60

0.744

2()(24.2)/60

=2.534

3 2()(7.1)/60

=0.744

2()(23.7)/60

=2.482

80 1 2()(9.3)/60

=0.974

2()(29.2)/60

=3.058

2 2()(7.5)/60

=0.785

2()(28.7)/60

=3.005


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2. Calculate the angular acceleration of each reading and the average agularacceleration of each load and show all your calculation.

=2-1/t

reading Time different,

s

Angular acceleration

(rad/s2)

average

1 231.8-194.4

=37.4

0.785-0.356/37.4

=0.011

2 361.4-320.8

=40.6

0.785-0.262/40.6

=0.013

0.011+0.013+0.0098/3

=0.011

3 460.3-418.5

=41.8

0.827-0.419/41.8

=0.0098

1 520.7-505.3

=15.4

1.906-0.974/15.4

=0.061

2 575.6-555.9

=19.7

1.812-0.419/19.7

=0.071

0.061+0.071+0.072/3

=0.068

3 630.6-610.8

=19.8

1.906-0.471/19.8

=0.072

1 682.2-667.9

=14.3

2.534-0.744/14.3

=0.125

2 724-709.7

=14.3

2.534-0.744/14.3

=0.125

0.125+0.125+0.122/3

=0.124

3 766.8-725.5

=14.3

2.482-0.744/14.3

=0.122

1 871.2-860.2 3.058-0.974/11

3 2()(8.3)/60

=0.869

2()(28.0)/60

=2.932

100 1 2()(7.1)/60

=0.744

2()(33.2)/60

=3.4772 2()(8.3)/60

=0.869

2()(32.2)/60

=3.372

3 2()(10.1)/60

=1.058

2()(33.2)/60

=3.477


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=11 =0.189

2 909.6-897.5

=12.1

3.005-0.785/12.1

=0.183

0.189+0.183+0.188/3

=0.187

3 950.3-939.3=11

2.932-0.869/11=0.188

1 997.5-986.5

=11

3.477-0.744/11

=0.248

2 1033.8-1023.9

=9.9

3.372-0.869/9.9

=0.278

0.248+0.278+0.244

=0.257

3 1074.4-1064.5

=9.9

3.477-1.058/9.9

=0.244

3. Plot graph average angular acceleration(rad/s2) versus load (g) and discuss theresult.

0

0.05

0.1

0.15

0.2

0.25

0.3

0 20 40 60 80 100 120

Graph load versus average angular

acceleration

Graph load versus

average angular

acceleration


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4.Explain why there are difference in readings for the same load.

When taking the reading parallax error occurs.Key not pressed exactly the

same time when the rope unrolled completely. Hence, the time taken varies some

fraction of a seconds. Other than that, there are also factors such as air resistance

that resist motion of rope in the air and friction force of the pulley in which the

rope unrolled. These factors causes the reading to be different even though the

load is same. Taking the average reading will help to justify these errors.

5.Explain what are the factors affecting the angular acceleration of body.

According to Newtons second law, the greater a bodys mass, the greater its resistance to

linear motion. Therefore, mass is a bodys inertial characteristic for considerations relative to

linear motion. Resistance to angular acceleration is also a function of a bodys mass. The

greater the mass, the greater the resistance to angular acceleration.However, the relative ease

or difficulty of initiating or halting angular motion depends on an additional factor - the

distribution of mass with respect to the axis of rotation. The more closely mass is distributed

to the axis of rotation, the easier it is to initiate or stop angular motion.

COCLUSION

From this experiment students are exposed to angular acceleration. Angular

acceleration can be calculated by taking the difference between angular velocity

within time divided by the time period. The relation between angular acceleration and

load is also determined. The larger the load the bigger the angular acceleration.

Objectives are accepted.

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