pala bana

December 16, 2018 | Author: AliMubarak | Category: Pendulum, Experiment, Physical Quantities, Classical Mechanics, Mechanical Engineering
Share Embed Donate


Short Description

a good read...

Description

BACHELOR OF ENGINEERING (HONS) MECHANICAL APPLIED MECHANICS LAB (DYNAMIC) MEC424

EXPERIMENT:

Physical Pendulum –  Pendulum – Wooden Wooden Pendulum LECTURER:

MISS ELI NADIA GROUP:

EMD4M3A

GROUP MEMBERS: NAME Paan senoi Baah Baah 2

MATRIX NO.

SIGNATURE

RESULT

Angle of oscillations is 10° (Point 1) Point



Time (s)

Average ( ),(s) 3

1

14.03

14.25

14.22

14.17

2

14.19

14.25

14.18

14.21 ∑ 1 =14.19

t 1

10

(s)=

14.19

10

=1.419s Angle of oscillations is 10° (Point 2) Point



Time (s)

Average ( ),(s) 3

1

14.31

14.16

14.22

14.23

2

14.19

14.18

14.31

14.22 ∑ 2 =14.23

t 2

10

(s)=

14.23

=1.423s

10

EXPERIMENTAL CALCULATION

Sample of calculations to find L 1 and L2





 L

2 

 g 

2

  T     L1   1   g   2   

2

  T     L2   2   g   2  

2

 1.419   L1     9.81    2      L1  0.5004m

2

 1.423   L2     9.81   2     L2  0.5032m

Sample of calculations to find  R G

 RG



 RG



 RG





 x  L2  L1

  x

  L2 



2 x



0.73 0.5032  0.73



 

0.5004  0.5032  2 0.73 0.363m

Sample of calculations to find I o1 and Io2

 I 0

T   2  

T   2  

mgRG 2

  T     I 02     mgr G  2   

2

2

 1.419  0.363  I 01     0.69.81   2     4  I 01  0.1089m

 1.423  0.363  I 02     0.69.81   2     4  I 02  0.1096m

Sample of calculations to find I G1 and IG2

 L1





 

2

m r G

mr G

 I G1

   0.5004 0.6 0.363  0.60.363

 I G1



 I G1

 L2



 L1 mr G



m r G

2

2



0.0299m

 I G 2



4



m  x  r G



2

 I G 2

    L m x  r    m x  r    0.5032 0.6 0.7  0.363  0.6 0.7  0.363 

 I G 2



 I G 2

m  x  r G

2

2

G

G

2

0.0336m

4

mgRG 2

  T     I 01     mgr G  2   

 I G1

 I 0

Theoretical calculation    

m v 0.6

   

   0.042  0.01    4  

(0.8  0.08  0.01)  (0.45  0.01  0.01)       1030.16 kg 

m3

m1    v1 m1  1030.16(0.8  0.08  0.01) m1  0.659kg  m2  1030.16(0.45  0.01  0.01) m2  0.046kg 

   0.042  0.01    4  

m3  1030.16 m3  0.013kg 

 I 1



 I 1



 I 2



 I 2



 I 3



 I 3



1 12

(0.659)(0.08 2



0.8 2 )



0.12 )

0.0355kgm 2 1 12

(0.046)(0.45 2

0.00081kgm kg m2 1 12

(0.013)(0.02)

4.33  10 7 kgm kg m2 

2

 I 2   I 3 

 I G



 I 1

 I G



(0.0355)  (0.00081  (4.33 x10 7 )

 I G



0.0347kgm 2





 No

V(m3)

x(m)

y(m)

xV(m4)

yV(m4)

1

6.4 x 10-4

0.04

0.4

2.56x10-5 

2.56x10-4

2

- 4.5 x 10-5

0.04

0.2775

-1.8x10-5 

-1.24875x10-4

3

-1.26x10-5

0.04

0.7555

-5.04x10-7 

-9.5193x10-5

∑xV = 7.096 x 10-6

∑yV = 2.26318x10-4

∑ V = 5.824x10-4

X =

 x V  V 

X= 0.012m

Y  yV 

=

V 

Y = 0.3886m

d = 0.05 + 0.698 –  0.698 –  0.3886  0.3886 = 0.3594 m

md 2

 I 0



 I G

 I 0



0.0347  (0.6)(0.3594) 2

 I 0



0.1122kgm 2



Percentage of error  I 0

=

 I 01

 I 01

(theoritical   exp exp erimental ) exp exp erimental  0.1122  0.1089

=

0.1089



 x 100

 I 02

100

 I 02

=3.03%

 I G

(theoritical   exp exp erimental ) =

 I G1

 I G1

exp exp erimental  0.0347  0.0299

=

0.0299

 100

=16.0%

0.0347  0.0336

 I G 2

=

 I G 2

=3.27%

0.0336

 100

 x 100

0.1122  0.1096

 =

0.1096

 = 2.37 %

 100

DISCUSSION

After finished this this experiment we get to determine determine the value of L1, L2, RG, I01, I02, IG1, and IG2. Based on the experiment that we have done we get the value of IG1 IG1 is 0.0358 m4 and IG2 is 0.0406 m4 . The theoretical IG for wooden pendulum is 0.0347 kgm2. The experimental value of IG1 and IG2 that we get is in between the theoretical value. Then we compare the value of I0 for experimental and theoretical. The value of I0 that we get in this experiment ex periment is I01 is 0.1048 m4 and I02 is 0.1107 m4 for. for. The theoretical value, for I0 I0 is 0.1122 kgm2. The values for theoretical and experimental are also slightly different. The values of theoretical values versus the experiment values for IG and Io has h as showed a bit different. This occurs because som e errors during this experiment have been done. don e. In our experiment our eyes e yes maybe not sharply parallel with the angle that we have ha ve state. Besides, it also occurs when we me asured the length of the wooden pendulum using a meter rule. We need to measure the length for every side to know the volume of wooden pendulum. The circle at wooden pendulum is not circle at all. But, in this experiment, we just consider it as a circle. As a result, it will affect our result. The wooden  pendulum is very sensitive. We need to release it steadily. But, in this experiment, maybe we release it with a little push. So, it will takes more time for ten oscillations of wooden pendulum. It makes our results results quite different. Besides that, we need to release the wooden pendulum and simultaneously start start the stopwatch. In that’s case maybe the stopwatch is started a little bit early or after the wooden pendulum pendu lum was released. So, from this error, it will influence the readings taken. In order to avoid the errors during this experiment, make sure we avoid doing parallax errors. Our eyes must be are at same level with the reading points. We also have to make sure that we must release a wooden pendulum and at the same time, start the stopwatch. We have taken three readings for time of ten oscillations of wooden pendulum to get an accurate value.

CONCLUSION

As conclusions the experiment that has been carried out was achieved the objectives and conducted very well. The different between the theoretical value and experiment has showed a  bit different which we already know the factor influence the result. This result will will be more accurate if the control method as what has been discussed earlier. From this experiment we h ave  been exposed how to determine the mass moment of inertia (at center of gravity, IG and at suspension point Io) by oscillation even the experimental and theoretical values of that quite different. The difference between experimental and theoretical values because some errors during conducted this experiment. Overall, this experiment shows that the values does not vary var y so much and only little differences appear while c omparing the values obtained both from experimental and theoretical values. Thus, I can say that this kind of experiment is a success even there are some errors occurred but bu t the effects is only in a small percentage. It is hard to get 100% accuracy in any an y experiment performed. Lastly, in my opinion, if a mo re suitable devices or apparatus are used in this experiment most probably p robably better or more accurate results will be obtained. So the objective is achieved.

THEORY:

Physical pendulum In this case, a rigid body –  body –  instead  instead of point mass - is pivoted to oscillate as shown in the figure. There is no requirement of string. As a result, there is no tension involved in this case. Besides these physical ramifications, the working of compound pendu lum is essentially same as that of simple pendulum except in two important aspects:



Gravity acts through center of mass of the rigid bod y. Hence, length of pendulum pendu lum used in equation is equal to linear distance between pivot and center of mass (“h”).



The moment of inertia of the rigid body bod y about point suspension is not equal to “mL2”

*

as in the case of simple pendulum. The time period of compound pendulum, pe ndulum, therefore, is given by :



In case we know MI of the rigid body, we can evaluate above expression of time  period for the physical pendulum. For illustration, let us consider a uniform rigid rigid rod,  pivoted from a frame as shown in the figure. Clearly, center of mass is at a distance “L/2” from the point of suspension :

h= L/2  Now, MI of the rigid rod about its center is:

We are, however, required to evaluate MI of the rod about the point of suspension, i.e. “O”. Applying parallel axes theorem,

Putting in the equation of time period, we have:

The important thing to note about this relation is that time period is still independent of mass o f the rigid body. However, time period is not independent of mass distribution of the rigid bod y. A change in shape or size or change in mass distribution will change MI of th e rigid body about  point of suspension. This, in turn, will change time period.

Further, we should note that physical pendulum is an effective device to measure “g”. As a matter of fact, this device is used extensively in gravity surveys around the world. We only ne ed to determine time period or frequency to dete d etermine rmine the value of “g”. Squaring Squa ring and rearranging,

Point of oscillation

We can think of physical pendulum pen dulum as if it were a simple pendulum. For this, we can consider the mass of the rigid body to be concentrated at a single point as in the case of simple pendulum such that time periods of two pendulums are same. Let this point be at a linear line ar distance “Lo “from the point of suspension. Here,

The point defined by the vertical distance, "Lo “, from the point of suspension is called point of oscillation of the physical pendulum. Clearly, point of oscillation will change if point of suspension is changed.

View more...

Comments

Copyright ©2017 KUPDF Inc.
SUPPORT KUPDF