2. Μηχανικό Μαύρο Κουτί: κύλινδρος με μια μπάλα μέσα σε αυτόν.
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1 Experiental Copetition: 14 July 011 Proble Page 1 of. Μηχανικό Μαύρο Κουτί: κύλινδρος με μια μπάλα μέσα σε αυτόν. Ένα μικρό σωματίδιο μάζας (μπάλα) βρίσκεται σε σταθερή απόσταση z από το πάνω μέρος ενός κυλίνδρου μεγάλου μήκους μάζας. Ο κύλινδρος έχει μια σειρά από μικρές τρύπες κατά μήκος του άξονα του. Αυτές οι τρύπες χρησιμεύουν για τη στήριξή του ώστε να κρέμεται κατακόρυφος. Θα πρέπει να πάρετε τις αναγκαίες πειραματικές μετρήσεις έτσι ώστε να προσδιορίσετε τις αριθμητικές τιμές των πιο κάτω φυσικών μεγεθών μαζί με το αντίστοιχο σφάλμα μέτρησης: i. Τη θέση του κέντρου μάζας του κυλίνδρου μαζί με τη μπάλα. Επίσης να σχεδιάσετε το σχήμα της διάταξης του απλού πειράματος που κάνατε για τον προσδιορισμό του κέντρου μάζας. [1.0 points] ii. Την απόσταση z. iii. Τον λόγο. [3.5 points] [3.5 points] iv. Την επιτάχυνση λόγω της βαρύτητας, g. [.0 points] Υλικά και Όργανα: Ένας κύλινδρος με τρύπες μαζί με μια μπάλα κολημένη στο εσωτερικό του, μια ορθογώνια βάση μαζί με ένα λεπτό καρφί, πλαστικό καπάκι σε σχήμα καρφιού, χάρακας, χρονόμετρο, νήμα, μολύβι και κολλητική ταινία. z O στήριξη x R Βάση Στερεωμένη με ταινία στην ακρη του τραπεζιού L Λεπτό καρφί για στήριξη x η απόσταση από το πάνω άκρο του κυλίνδρου μέχρι το κέντρο μάζας. R η απόσταση από το σημείο στήριξης μέχρι το κέντρο μάζας.
2 Experiental Copetition: 14 July 011 Proble Page of Κολλητική ταινία Κύλινδρος με τρύπες και με μπάλα μέσα Καπάκι καρφιού χρονό μετρο Νήμα (για ισορροπία) Βάση κανόνας Προσοχή: Το λεπτό καρφί είναι πολύ αιχμηρό. Όταν δεν χρησιμοποιείται, θα πρέπει να προστατεύεται με το πλαστικό καπάκι. Χρήσιμες πληροφορίες: d q 1. Για ένα τέτοιο φυσικό εκκρεμές, {( + ) R + I }» - g ( + ) Rq, όπου I dt είναι η ροπή αδράνειας του κυλίνδρου με τη μπάλλα ως προς το κέντρο μάζας και θ είναι η γωνιακή μετατόπιση.. Για κύλινδρο μεγάλου μήκους L και μάζας, η ροπή αδράνειας ως προς άξονα που περνά από το κέντρο μάζας του κυλίνδρου και είναι κάθετος σ αυτόν κατά προσέγγιση είναι 1 L ( ) Το θεώρημα των παράλληλων αξόνων: I = I centre of ass + x, όπου x είναι η απόσταση του σημείου περιστροφής από το κέντρο μάζας του συστήματος, και η συνολική μάζα του συστήματος. 4. Η μπάλλα μπορεί να θεωρηθεί ως υλικό σημείο που βρίσκεται στον κεντρικό άξονα του κυλίνδρου. 5. Υποθέστε ότι ο κύλινδρος είναι ομογενής και η μάζα των καλυμάτων στα άκρα του είναι αμελητέα.
3 ODIFIED Q_EXPERIENT_SOLUTION_14JULY.DOCX Experiental Copetition: 14 July 011 Question Page 1 of 9 Solution:. echanical Blackbox: a cylinder with a ball inside z O pivot x R L In order to be able to calculate the required values in i, ii, iii, we need to know: a. the position of the centre of ass of the tubing plus particle (object) which depends on z,, b. the oent of inertia of the above. The position of the ay be found by balancing. The I can be calculated fro the period of oscillation of the tubing plus object. Analytical steps to select paraeters for plotting L z I. x L is readily obtainable with a ruler. x is detered by balancing the tubing and object. (1) 1
4 ODIFIED Q_EXPERIENT_SOLUTION_14JULY.DOCX Experiental Copetition: 14 July 011 Question Page of 9 II. For sall-aplitude oscillation about any point O the period T is given by considering the equation: sin R I g R g R. () T I R g R where I x z x Note that. (3) 1 L L 3 1 L x Lx z x 3. (4) g I. (5) 4 R T R ethod (a): (linear graph ethod) The equation (5) ay be put in the for: 4 4 I T R R. (6) g g Hence the plot of TRv.s. R will yield the straight line whose 4 Slope. (7) g and y-intercept 4 I g I Hence,. (8). (9) The value of g is fro equation (7): g 4. (10)
5 ethod (b): iu point curve ethod The equation (5) iplies that T has a iu value at ODIFIED Q_EXPERIENT_SOLUTION_14JULY.DOCX Experiental Copetition: 14 July 011 Question Page 3 of 9 RR I. (11) Hence R can be obtained fro the graph Tv.s. R. And therefore I R. (1) This equation (1) together with equation (1) will allow us to calculate the required values z and. g At the value R R equation (5) becoes T R R 4 R 8 R g 4. (13) T T fro which g can be calculated. 3
6 ODIFIED Q_EXPERIENT_SOLUTION_14JULY.DOCX Experiental Copetition: 14 July 011 Question Page 4 of 9 Results L 30.0 c 0.1 c x 17.8 c 0.1 c (fro top) x R (c) tie (s) for 0 cycles T (s) R (c) R (c ) TR(s c) Notes: at x R15.1,16.1 c, ties for 10 cycles. 4
7 ODIFIED Q_EXPERIENT_SOLUTION_14JULY.DOCX Experiental Copetition: 14 July 011 Question Page 5 of 9 ethod (a) 16 TR(sc) R (c ) Calculation fro straight line graph: slope s /c, y-intercept s c g 4 giving g (961 0) c/s c.5c I L L 3 Fro equation (4): I x z x 5
8 Then z 17.8 ODIFIED Q_EXPERIENT_SOLUTION_14JULY.DOCX Experiental Copetition: 14 July 011 Question Page 6 of z (14) The centre of ass position gives: z Fro equations (14) and (15): z17.8. (15) z z z And z c Error Estiation Find error for g : Fro (10), g g g 16.3 c/s 0 c/s i) Find error for z : 3.10 First, find error for r c r ( ) r.5 c r L xc Since error fro r contributes ost ( ~ while, ~ ), we estiate error r L xc propagation fro r only to siplify the analysis by substituting the and ax values into equation (4). Now, we use r ax r r The corresponding quadratic equation is z z The corresponding solution is ( z 17.8) ax 7.55 c 6
9 ODIFIED Q_EXPERIENT_SOLUTION_14JULY.DOCX Experiental Copetition: 14 July 011 Question Page 7 of 9 If we use r r r , the corresponding quadratic equation is z z The corresponding solution is ( z 17.8) 6.96 c So ( z 17.8) 0.3 c ( z 17.8) Note that ~ So, we still ignore the error propagation due to L, xc z 17.8 The error z can be estiated fro z ( z17.8) 0.3 c ii) Find error for : z17.8 We know that.8 ( z17.8)
10 ODIFIED Q_EXPERIENT_SOLUTION_14JULY.DOCX Experiental Copetition: 14 July 011 Question Page 8 of 9 ethod (b) Calculation fro T-R plot: T(s) R(c) Using the iu position: T T at I R and g 8 R T Fro graph: R c and T s g c/s I. (16) 8
11 ODIFIED Q_EXPERIENT_SOLUTION_14JULY.DOCX Experiental Copetition: 14 July 011 Question Page 9 of 9 Fro equations (14), (15), (16): z z 17.8 x x z And z c Error estiation i) Find error for g : 8 R Using the iu position: g T R T g R T ii) Find error for z : g c/s, we have First, find error for r R 79.1 c. r R R 3.56 c This r is equivalent to r in part 1. So, one can follow the sae error analysis. As a result, we have z c z 0.8 c i) Find error for : Following the sae analysis as in part I, we found that.96 ; ( ) NOTE: This iu curve ethod is not as accurate as the usual straight line graph. 9
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