UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical EngineeringApplied 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 1|Page 5. width and length ) 2. there is no tension involved in this case. Vee support is inserted into the hole of the wooden pendulum. Step 4 was repeated for swinging the pendulum from left. 3.is pivoted to oscillate as shown in the figure. Besides these physical ramifications. 4. and time for 10 complete cycle was taken. The dimension for the wooden pendulum is taken ( thickness. Step 2 to 5 was repeated for cylindrical support for another hole of wooden pendulum. There is no requirement of string. the working of compound pendulum is essentially same as that of simple pendulum except in two important aspects: 2|Page . The wooden pendulum was swing from right at 15o . The wooden pendulum is hanging at the testing apparatus to swing it. a rigid body – instead of point mass . THEORY: Physical pendulum In this case. PROCEDURE: 1. 6.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) The apparatus used for the experiment. As a result. The time period of compound pendulum. center of mass is at a distance “L/2” from the point of suspension : h= L/2 Now.e. Clearly. let us consider a uniform rigid rod. 3|Page . length of pendulum used in equation is equal to linear distance between pivot and center of mass (“h”). however.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) • Gravity acts through center of mass of the rigid body. Hence. required to evaluate MI of the rod about the point of suspension. For illustration. we can evaluate above expression of time period for the physical pendulum. “O”. is given by : • In case we know MI of the rigid body. pivoted from a frame as shown in the figure. Applying parallel axes theorem. MI of the rigid rod about its center is: We are. • The moment of inertia of the rigid body about point suspension is not equal to “mL2” * as in the case of simple pendulum. therefore. i. Let this point be at a linear distance “Lo “from the point of suspension. we have: The important thing to note about this relation is that time period is still independent of mass of the rigid body. from the point of suspension is called point of oscillation of the physical pendulum. we should note that physical pendulum is an effective device to measure “g”. in turn. 4|Page . will change time period. This.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) Putting in the equation of time period. For this. point of oscillation will change if point of suspension is changed. Here. 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. However. A change in shape or size or change in mass distribution will change MI of the rigid body about point of suspension. The point defined by the vertical distance. Squaring and rearranging. time period is not independent of mass distribution of the rigid body. this device is used extensively in gravity surveys around the world. Further. "Lo “. Clearly. We only need to determine time period or frequency to determine the value of “g”. As a matter of fact. Point of oscillation We can think of physical pendulum as if it were a simple pendulum. 0 14.53 / 10 = 1.07 1 period of oscillation .5 14.5246 m Point 2 T2 = 2π L2 g 5|Page .407 s T = 2π L1 g (1. T1 T2 Point 1 = 14.81 ) L1= 0.07 / 10 = 1.53 2 T2 14.453)2 = (2 Л)2 ( L1 / 9.65 14.453 s = 14.1 14.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) SAMPLE OF CALCULATIONS: Experimental Result Point 1 Point For Point 1 2 3 Average T1 14.1 14.5 14. 78 kg/m3 To obtain the mass for each component.6 kg Density.9cm3 So.1 × 10-6 = 1016.6 kg / 590. VT = V1 – ( V2 + V3 ) = 640 – ( 45 + 4.81 ) L2 = 0.9 ) = 590. 6|Page . m = 0. Total volume.459)2 = (2 Л)2 ( L2 / 9.1 cm3 Given mass. ϼ = m / vtotal = 0.529 m Component 1 Component 2 Component 3 Component 1 Length Height Width Volume 80cm 1cm 8cm 640cm3 Component 3 1cm 45cm 1cm 45cm3 Diameter Height Volume Component 2 2.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) (1.5cm 1cm 4. 1124 kg.529 – 0.78 × (640×10-6) = 0.4107 = 0.529 .0458 kg m 2= ϼ × v = 1016.3572 m to obtain I0 : T = 2π IO mgrg At point 1 (T1)2 = (2 Л)2 ( I01 / mgrG ) (1.46 ) = -0.453) 2 = (2 Л)2 ( I01 / 0.982 × 10-3 kg X= 73 cm = 15 rG = x (L2 – x) / (L1 + L2 .78 × (45×10-6) = 0.6507 kg m3=ϼ×v = 1016.m2 7|Page .1467 / -0.73 ( 0.6 × 9.5203 + 0.3572) I01 = 0.9×10-6) = 4.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) m1= ϼ×v = 1016.2x) = 0.1.81 × 0.78 × (4.73 ) / ( 0. m2 At point 2 IG2 = mrG (L2 – rG) = (0.6)(0.3572)(0.3572) = 0.3572) = 0.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) At point 2 (T2)2 = (2 Л)2 ( I02 / mgrG ) (1.035 kg.142(1.81 × 0.5203-0. IG = mrG (L – rG) At point 1 IG1 = mrG (L1 – rG) = (0.0368 kg.529-0.6)(0.25)2 = 4.m2 Theoretical Result Area (A1) = 73 × 8 = 584 cm2 Area (A2) = = 3.909cm2 Area (A3) = 45 × 1 = 45cm2 8|Page .3572)(0.3572 ) I02 = 0.m2 To obtain IG : L1 = (IG1 + mrG²) / mrG Thus.407) 2 = (2 Л)2 ( I02 / 0.1054 kg.6 × 9. 091 = 35.091 cm2 Hanging at point 1 8cm Y 73cm *Height = 1cm Y= = 36. At = A1 – ( A2 + A3 ) = 534.6507×0.36 / 534.25(A2) .644 cm To obtain Io.6507×(0.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) Total area.365) 2) 9|Page . Moment of Inertia : I0 = IG + md2 For component 1 I01 = 1/12 ( mh12 ) + md12 = 1/12 (0.50.5 (A1) – 15(A3).82) + (0. At = 19037. total = IG + md2 Thus.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) = 0.7305×10-7 .356442) = 0.md2 = 0.total = I01 + I02 + I03 = 0.9.0.01245 = 0.01245 kg.7305×10-7 kg.1214 .total .m2 10 | P a g e .982 × 10-3 ×0.m3 For component 2 I02 = 1/4 mr22 ) + md22 = 1/4 (4.m3 For component 3 I03 = 1/12 ( mh32 ) + md32 = 1/12 ( 0. Io.0458×0.01252) + (4.m2 Io.m3 So.0125 2) = 9.45 2 ) + ( 0.0458×0.10895 – (0.10895 kg. IG can be obtain IG = Io.1214 kg.505 2 ) = 0.6)(0.982 × 10-3 ×0.03272 kg. 6507×0.091 = 37.3652) 11 | P a g e .356 cm To obtain Io.5(A3) At = 19951.6507×0.5 (A1) – 71. Moment of Inertia: I0 = IG + md2 For component 1 I01 = 1/12 ( mh12 ) + md12 = 1/12 (0.75(A2) – 22.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) Hanging at point 2 8cm Y 73cm *Height = 1cm Y= = 36.82) + (0.28 / 534. 982 × 10-3 ×0.7175 2) = 2.5649×10-3 .0458×0.m2 Io.2252 ) = 3.total = I01 + I02 + I03 = 0.1214 . Io.m2 Percentage Error % At point 1 : 12 | P a g e .md2 = 0.2. IG can be obtain IG = Io.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) = 0.m3 For component 3 I03 = 1/12 ( mh32 ) + md32 = 1/12 ( 0.0458×0.0915 × 10-3 kg So.45 2 ) + ( 0.982 × 10-3 ×0.1157 kg.3.5649×10-3 kg.01252) + (4.0915 × 10-3 = 0.total .m3 For component 2 I02 = 1/4 mr22 ) + md22 = 1/4 (4.1157 – (0.03202 kg.total = IG + md2 Thus.1214 kg.37356)2 = 0.6)(0. 10895 x 100% 0.287% IG1 = 0.0368 = 12. Percentage error = 0.514% At point 2: I02 = 0.m2 (experiment) = 0. Percentage error = 0.1115 = 2.03202 kg.m2 (experiment) = 0.10895 kg.035 – 0.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) I01 = 0.m2 (theory) So. Percentage error = 0.03272 x 100% 0.0368 kg.m2 (theory) So.m2 (experiment) = 0. Percentage error = 0.035 kg.m2 (experiment) = 0.061% IG2 = 0.1157 kg.m3 (theory) So.0368 – 0.03272 kg.989% 13 | P a g e .1115 kg.1157 x 100% 0.1134 kg.1115 – 0.035 = 6.1134 – 0.1116 = 2.m2 (theory)So.03202 x 100% 0. we found out there are errors identified that affects the results. The experiment been ran out in a very conducive laboratory but possibilities of the present of blowing wind must be considered. a new set of apparatus must be replaced with the unreliable one. 14 | P a g e . Since the percentage error is less than 15%. The experiment is best conducted in a vacuum. Minimize the parallax error (reading been taken parallel to the eyes). • Environment factors –This factor slightly can be taken as a minor cause of the error.UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) DISCUSSION: During the experiment. To obtain a better experiment result. we found out the reading is not accurate. • 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. CONCLUSION: From this experiment. we obtained the experimental value is slightly differs to the theoretical value. • 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. Precising the value by taking readings more than 2 or 3 times (during handling the stopwatch) and get the average. we can conclude the experiment is succeed and the objective of the experiment is achieved. From the discussion above. The value k is different compare to the theoretical value because of errors as below. the handling of stopwatch timing is not accurate. org/content/m15585/latest/ 15 | P a g e .UNIVERSITI TEKNOLOGI MARA Faculty Of Mechanical Engineering Applied Mechanics Lab (MEC424) REFERENCES • Engineering Mechanics Dynamics.C Hibbeler Publisher : Pearson Prentice Hall • http://cnx. 11th Edition In SI Units by R.