Lab 8 Circular Motion Part 1

Lab 8 Circular Motion Part 1 - Lab 8 Circular Motion Part 1...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lab 8 Circular Motion Part 1 / <1; Lab Report (100 pts) ' ,fie-«u PHYS 1114 2014 Section- V 9/ Day and Time: 2 J}; '» ;_‘ ‘ "v" 7. "37 x} :9 J, Station Check: Table 1. Hanging Mass and Trolley Angular Speed Data (47 pts) ‘ l m 4 (kg) R = radius = Lgn) M = mass of trolley = _____’________‘ Combined Mass 1na (kg) Combined Weight mag (N) (0 ave (rad/ sec) 83 Lab 8 Circular Motion Part1 Exercises and Questions 1. Table 2. Forces Exerted 0n the Trolle lit 2. Identify all of the forces exerted on the trolley at the instant the trolley gleaves the sthel pin and is pulled inward by the string (Fp = 0N). To identify a force onft‘liineyumea‘fifito identify the physical object that exerts the force, to identify the type of force that the object exerts (e.g., tension, gravity, normal, static friction, kinetic friction, etc), and to label the f 1:“ , 13k,etc.) (12 pts) _. —o F 5, force with an appropriate vector symbol (e.g., T, F g’n W13 Apply Newton’s Second Law for uniform circular motion (Equations 5 and 7) to the trolley M and Newton’s Second Law PM = m a to the hanging mass ma. (12 pts) ZFy = ma ay ZFmd 2 M032 R 2Fy = May Show that the tension T in the string is given by the weight mag of the hanging mass ma, that is, T =ma g. ,9. Quay PVC. ‘ff'qflzv : ma {3 ‘ , ’: W??? * i r I w 3 r 2:4. rm Ema l 1 f L (Lu a . K, "“W" “ "ii C _ I Show that the magnitude of the centripetal force Fc on the trolley is given by Fc =mag-fS =MRw2, (8) where fs is the magnitude of the net radial static frictional force f5 exerted by the track on the trolley in Exercise 1. TC, A M3,th 1?) i» G5 » , ~. “9 "" M“; T: Whig! 0, E J 84 if .1" Lab 8 Circular Motion Part 1 3. Transfer your average angular speed wave vs. force ma g data to the provided Excel ‘ Worksheet. 4. Follow the graphical analysis procedures outlined in the Functional Relationship Lab Report to determine the slope and y-intercept pf the line of best fit. (12 pts) _ z, ' — w 1' Slope — .2043 Y'lntercept A» 5. The theoretical results in Exercise 3 predict that the slope of the linear graph of wz vs. ma g is equal to l/ (M R). (13 pts) I Calculate the theoretical value of the slope from the measurements of R and M. Show all work and units. ‘ ‘ ‘ _./‘ m .3 WW 2 Kg“; . , \ M 1“ /3\ ' As a measure of how well the theoretical relationship between angular speed a) and weight mag of the hanging mass ma expressed in Equation 8, l f (02 = —— rn - —-S—, (9) MR( ag) MR agrees with the experimentally determined relationship wfve = (slopeof line of best fit).(ma g) + (vertical intercept), calculate the percent error between the theoretical value of the slope l/ (M R) and the slope of the line of best fit to the measured wjve vs. ma g data. Show all work. 2,. ~~ } s 1,1. ’ 90.5 a // V K 2 4413.? f”!- (9:; 2:50“ f" (V. ,..- .1 r '7 ' Estimate the magnitude of the maximum net static friction force that the track exerts on the trolley in the radial direction. (Refer to Question 4.) Show all work. {so ~—-§ ,Lijgm “I”; r \S 2*? Ts: ifli‘JON ‘ 1 ea. ~§ ‘. 041 1—5 Fog 'I’r ; "‘ i 85 Lab 8 Circular Motion Part1 6. The weight mg of the hanging mass ma alone is not equal to magnitude of the centripetal force Fc, yet the percent differences between M Ra)2 and mag decrease with larger hanging masses. Why do the percent differences improve? (NOTE: Refer to Equation 8 to explain) (4 pts) T: ‘ l or“ a". x We a ‘ "\ is C 275.165 » / W? a; v A» ‘fcpgp ,r» (“mgé'e ; x A g ‘4 w. r V “HQ war/“Ive N ’ ' ‘ x .J ,, 86 Data with Force as the Independent Variable M (kg) ‘- R (m): ma 03a ma 9 (ma/92 M (03m)2 R Percent (kg) (rad/ (N) (rad/sec)2 (N) Difference ve sec) ”’ ” 11““ 9.000 0.204 18.9% ' 16.810 0.382 25.000 0.568 12.1% 0.836 33.640 0.764 1.032 49.000 1.112 0N9 23 H 3mm. 006 00.0 00.0 0V0 . txfiwéw 2 A 2 NU; YESWJD C 5 mm 0 z wdm U r :u/Ji r99»va .fl. . « 0N0 m 43x. x308 .. 3mg 355. 9:0ch D5 3925 .m> veg—ow cwonw 3.30:4. ¢m§w>< 0.0m- 0.0V- 0N.0- 0.0V 0.0m 0.00 0.0V 0.00 0.8 N35 :3 ...
View Full Document

This note was uploaded on 01/10/2012 for the course PHYS 2014 taught by Professor Nandi during the Spring '08 term at Oklahoma State.

Page1 / 6

Lab 8 Circular Motion Part 1 - Lab 8 Circular Motion Part 1...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online