Lab Pendulum 2

UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

Applied Mechanics Lab (MEC424)

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

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UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

Applied Mechanics Lab (MEC424)

The apparatus used for the experiment.

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:

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UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

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Applied Mechanics Lab (MEC424)

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”).

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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 :

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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,

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UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

Applied Mechanics Lab (MEC424)

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 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|Page

Applied Mechanics Lab (MEC424)

UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

SAMPLE OF CALCULATIONS:

Experimental Result Point 1

Point 2

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

T = 2π

L1 g

(1.453)2 = (2 Л)2 ( L1 / 9.81 ) L1= 0.5246 m Point 2

T2 = 2π

L2 g

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Applied Mechanics Lab (MEC424)

UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

(1.459)2 = (2 Л)2 ( L2 / 9.81 ) L2 = 0.529 m

Component 1

Component 2

Component 3

Component 1

Component 3

Length

80cm

1cm

Diameter

Height

1cm

45cm

Height

1cm

Width

8cm

1cm

Volume

4.9cm3

640cm3

45cm3

Volume

Component 2 2.5cm

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, 6|Page

Applied Mechanics Lab (MEC424)

UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

m1= ϼ×v

m3=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

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 :

T = 2π

IO mgrg

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.m2 7|Page

Applied Mechanics Lab (MEC424)

UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

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.m2

To obtain IG :

L1 = (IG1 + mrG²) / mrG Thus,

IG = mrG (L – rG)

At point 1 IG1 = mrG (L1 – rG) = (0.6)(0.3572)(0.5203-0.3572) = 0.035 kg.m2 At point 2 IG2 = mrG (L2 – rG) = (0.6)(0.3572)(0.529-0.3572) = 0.0368 kg.m2

Theoretical Result

Area (A1) = 73 × 8 = 584 cm2

Area (A2) = = 3.142(1.25)2

Area (A3) = 45 × 1 = 45cm2

= 4.909cm2

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Applied Mechanics Lab (MEC424)

UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

Total area, At = A1 – ( A2 + A3 ) = 534.091 cm2

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 :

I0 = IG + md2 For component 1 I01 = 1/12 ( mh12 ) + md12 = 1/12 (0.6507×0.82) + (0.6507×(0.365) 2) 9|Page

UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

Applied Mechanics Lab (MEC424)

= 0.1214 kg.m3

For component 2 I02 = 1/4 mr22 ) + md22 = 1/4 (4.982 × 10-3 ×0.01252) + (4.982 × 10-3 ×0.0125 2) = 9.7305×10-7 kg.m3 For component 3 I03 = 1/12 ( mh32 ) + md32 = 1/12 ( 0.0458×0.45 2 ) + ( 0.0458×0.505 2 ) = 0.01245 kg.m3

So, Io,total

= I01 + I02 + I03 = 0.1214 - 9.7305×10-7 - 0.01245 = 0.10895 kg.m2

Io,total = IG + md2 Thus, IG can be obtain IG = Io,total

- md2

= 0.10895 – (0.6)(0.356442) = 0.03272 kg.m2

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UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

Applied Mechanics Lab (MEC424)

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:

I0 = IG + md2 For component 1 I01 = 1/12 ( mh12 ) + md12 = 1/12 (0.6507×0.82) + (0.6507×0.3652) 11 | P a g e

UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

Applied Mechanics Lab (MEC424)

= 0.1214 kg.m3

For component 2 I02 = 1/4 mr22 ) + md22 = 1/4 (4.982 × 10-3 ×0.01252) + (4.982 × 10-3 ×0.7175 2) = 2.5649×10-3 kg.m3 For component 3 I03 = 1/12 ( mh32 ) + md32 = 1/12 ( 0.0458×0.45 2 ) + ( 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.m2

Io,total = IG + md2

Thus, IG can be obtain IG = Io,total - md2 = 0.1157 – (0.6)(0.37356)2 = 0.03202 kg.m2

Percentage Error

%

At point 1 : 12 | P a g e

UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

Applied Mechanics Lab (MEC424)

I01 = 0.1115 kg.m2 (experiment) = 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, Percentage error = 0.035 – 0.03272 x 100% 0.035 = 6.514% At point 2: I02 = 0.1134 kg.m2 (experiment) = 0.1157 kg.m2 (theory)So, Percentage error = 0.1134 – 0.1157 x 100% 0.1116 = 2.061%

IG2 = 0.0368 kg.m2

(experiment)

= 0.03202 kg.m2 (theory) So, Percentage error = 0.0368 – 0.03202 x 100% 0.0368 = 12.989%

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UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

Applied Mechanics Lab (MEC424)

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.

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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.

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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. 14 | P a g e

UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering

Applied Mechanics Lab (MEC424)

REFERENCES •

Engineering Mechanics Dynamics, 11th Edition In SI Units by R.C Hibbeler Publisher : Pearson Prentice Hall

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

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