lab pendulum 2

U N I V E R S I T I T E K N O L O G I M A R A A p p l i e d M e c h a n i c s L a b (MEC4 2 4 ) F a c u l t y O f M e c h a n i c a l E n g i n e e r i n g

TITLE:

Physical Pendulum - Wooden Pendulum

OBJECTIVE:

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

1) To determine the relation between the period (T) of oscillation of a simple pendulum

with verify 3 types of different specimens.

2) To determine the center of gravity of a connecting rod, as well as the radius of

gyration about the center of gravity by using compound pendulum.

3) Determine the mass moment of inertia, (Ig & Io ) by oscillation and manual calculation.

APPARATUS: Universal Vibration System Apparatus

1) Wooden pendulum

2) Vee support, cylindrical support

3) Ruler

4) Stopwatch

The apparatus used for the experiment.

1 | P a g e


PROCEDURE:

1. The dimension for the wooden pendulum is taken ( thickness, width and length )

2. Vee support is inserted into the hole of the wooden pendulum.

3. The wooden pendulum is hanging at the testing apparatus to swing it.

4. The wooden pendulum was swing from right at 15o , and time for 10 complete cycle

was taken.

5. Step 4 was repeated for swinging the pendulum from left.

6. Step 2 to 5 was repeated for cylindrical support for another hole of wooden

pendulum.

THEORY:

Physical pendulum

In this case, a rigid body – 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 pendulum is essentially same

as that of simple pendulum except in two important aspects:

Gravity acts through center of mass of the rigid body. Hence, length of pendulum

used in equation is equal to linear distance between pivot and center of mass (“h”).

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

* as in the case of simple pendulum. The time period of compound pendulum,

therefore, is given by :

2 | P a g e


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

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:

3 | P a g e


The important thing to note about this relation is that time period is still independent of mass

of the rigid body. However, time period is not independent of mass distribution of the rigid

body. A change in shape or size or change in mass distribution will change MI of the 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

need to determine time period or frequency to determine the value of “g”. Squaring and

rearranging,

Point of oscillation

We can think of physical pendulum 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

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.

4 | P a g e


SAMPLE OF CALCULATIONS:

Experimental Result

For

Point T1 T2

1 14.65 14.1

2 14.5 14.0

3 14.5 14.1

Average 14.53 14.07

1 period of oscillation , T1 = 14.53 / 10

= 1.453 s

T2 = 14.07 / 10

= 1.407 s

Point 1

(1.453)2 = (2 Л)2 ( L1 / 9.81 )

L1= 0.5246 m

Point 2

(1.459)2 = (2 Л)2 ( L2 / 9.81 )

L2 = 0.529 m

5 | P a g e

Point

1

Point

2


Component 1 Component 2 Component 3

So, Total volume, VT = V1 – ( V2 + V3 )

= 640 – ( 45 + 4.9 )

= 590.1 cm3

Given mass, m = 0.6 kg

Density, ϼ = m / vtotal

= 0.6 kg / 590.1 × 10-6

= 1016.78 kg/m3

To obtain the mass for each component,

m 1 = ϼ × v m 3 = ϼ × v

= 1016.78 × (640×10-6) = 1016.78 × (45×10-6)

= 0.6507 kg = 0.0458 kg

m 2 = ϼ × v

= 1016.78 × (4.9×10-6)

= 4.982 × 10-3 kg

6 | P a g e

Component 1 Component 3

Length 80cm 1cm

Height 1cm 45cm

Width 8cm 1cm

Volume 640cm3 45cm3

Component 2

Diameter 2.5cm

Height 1cm

Volume 4.9cm3


X= 73 cm

= 15

rG = x (L2 – x) / (L1 + L2 - 2x)

= 0.73 ( 0.529 – 0.73 ) / ( 0.5203 + 0.529 - 1.46 )

= -0.1467 / -0.4107

= 0.3572 m

to obtain I0 :

At point 1

(T1)2 = (2 Л)2 ( I01 / mgrG )

(1.453) 2 = (2 Л)2 ( I01 / 0.6 × 9.81 × 0.3572)

I01 = 0.1124 kg.m 2

At point 2

(T2)2 = (2 Л)2 ( I02 / mgrG )

(1.407) 2 = (2 Л)2 ( I02 / 0.6 × 9.81 × 0.3572 )

I02 = 0.1054 kg.m 2

7 | P a g e


To obtain IG :

Thus, IG = mrG (L – rG)

At point 1

IG1 = mrG (L1 – rG)

= (0.6)(0.3572)(0.5203-0.3572)

= 0.035 kg.m 2

At point 2

IG2 = mrG (L2 – rG)

= (0.6)(0.3572)(0.529-0.3572)

= 0.0368 kg.m 2

Theoretical Result

Area (A1) = 73 × 8 Area (A2) = Area (A3) = 45 × 1

= 584 cm2 = 3.142(1.25)2 = 45cm2

= 4.909cm2

Total area, At = A1 – ( A2 + A3 )

= 534.091 cm2

8 | P a g e

L1 = (IG1 + mrG²) / mrG


Hanging at point 1

8cm

Y 73cm

*Height = 1cm

Y =

= 36.5 (A1) – 15(A3).25(A2) - 50.

At

= 19037.36 / 534.091

= 35.644 cm

To obtain Io,

Moment of Inertia :

For component 1

I01 = 1/12 ( mh12 ) + md1

2

= 1/12 (0.6507×0.82) + (0.6507×(0.365) 2)

= 0.1214 kg.m 3

For component 2

9 | P a g e

I0 = IG + md2


I02 = 1/4 mr22 ) + md2

2

= 1/4 (4.982 × 10-3 ×0.01252) + (4.982 × 10-3 ×0.0125 2)

= 9.7305×10 -7 kg.m 3

For component 3

I03 = 1/12 ( mh32 ) + md3

2

= 1/12 ( 0.0458×0.452 ) + ( 0.0458×0.505 2 )

= 0.01245 kg.m 3

So,

Io,total = I01 + I02 + I03

= 0.1214 - 9.7305×10-7 - 0.01245

= 0.10895 kg.m 2

Io,total = IG + md2

Thus, IG can be obtain

IG = Io,total - md2

= 0.10895 – (0.6)(0.356442)

= 0.03272 kg.m 2

10 | P a g e


Hanging at point 2

8cm

Y

73cm

*Height = 1cm

Y =

= 36.5 (A1) – 71.75(A2) – 22.5(A3)

At

= 19951.28 / 534.091

= 37.356 cm

To obtain Io,

Moment of Inertia:

For component 1

I01 = 1/12 ( mh12 ) + md1

2

= 1/12 (0.6507×0.82) + (0.6507×0.3652)

= 0.1214 kg.m 3

11 | P a g e

I0 = IG + md2


For component 2

I02 = 1/4 mr22 ) + md2

2

= 1/4 (4.982 × 10-3 ×0.01252) + (4.982 × 10-3 ×0.7175 2)

= 2.5649×10 -3 kg.m 3

For component 3

I03 = 1/12 ( mh32 ) + md3

2

= 1/12 ( 0.0458×0.452 ) + ( 0.0458×0.2252 )

= 3.0915 × 10 -3 kg

So,

Io,total = I01 + I02 + I03

= 0.1214 - 2.5649×10-3 - 3.0915 × 10-3

= 0.1157 kg.m 2

Io,total = IG + md2

Thus, IG can be obtain

IG = Io,total - md2

= 0.1157 – (0.6)(0.37356)2

= 0.03202 kg.m 2

Percentage Error

%

At point 1 :

I01 = 0.1115 kg.m2 (experiment)

12 | P a g e


= 0.10895 kg.m3 (theory)

So,

Percentage error = 0.1115 – 0.10895 x 100%

0.1115

= 2.287%

IG1 = 0.035 kg.m2 (experiment)

= 0.03272 kg.m2 (theory)

So,


0.035

= 6.514%

At point 2:

I02 = 0.1134 kg.m2 (experiment)

= 0.1157 kg.m2 (theory)So,


0.1116

= 2.061%

IG2 = 0.0368 kg.m2 (experiment)

= 0.03202 kg.m2 (theory)

So,


0.0368

= 12.989%

13 | P a g e


DISCUSSION:

During the experiment, we found out there are errors identified that affects the results. The

value k is different compare to the theoretical value because of errors as below;

Human error – This experiment is conducted by human so there must be some error

in terms of readings and procedure. Parallax error is one of the most common errors

in conducting the experiment and then, the handling of stopwatch timing is not

accurate.

Device error – The apparatus used is not reliable because one of the parts of the

apparatus (protractor) is gone but the angle still can be obtained by using the

protractor which is drawn by pencil.

Environment factors –This factor slightly can be taken as a minor cause of the error.

The experiment been ran out in a very conducive laboratory but possibilities of the

present of blowing wind must be considered.

CONCLUSION:

From this experiment, we obtained the experimental value is slightly differs to the theoretical

value. From the discussion above, we found out the reading is not accurate. Minimize the

parallax error (reading been taken parallel to the eyes). The experiment is best conducted in

a vacuum. Precising the value by taking readings more than 2 or 3 times (during handling the

stopwatch) and get the average. To obtain a better experiment result, a new set of apparatus

must be replaced with the unreliable one. Since the percentage error is less than 15%, we

can conclude the experiment is succeed and the objective of the experiment is achieved.

REFERENCES

Engineering Mechanics Dynamics, 11th Edition In SI Units by R.C Hibbeler

14 | P a g e


Publisher : Pearson Prentice Hall

http://cnx.org/content/m15585/latest/

15 | P a g e

lab pendulum 2

Documents

applied mechanics

time period

rigid body

wooden pendulum

height 1cm

vee support

mechanical

simple pendulum