Gate Times Complete Error Formulas t down and up (cm/s') a_down N a up N Error Calculations (sec) Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Trial 6 sec Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Trial 6 Trial 1 56.1167 6 71.56212 6.815 6,025 4.843 3.011 1.741 0,407 0.371 0.335 0,306 0.337 0.340 0,339 Trial 2 51 01344 a 76.19833 a g (cm/s ) u (--) 7.021 5.273 5.093 3 259 1.987 3.657 0.653 0.611 0.613 0.612 0.612 0.616 Trial 3 32.19308 cm/s ] 76.42342 (cm/s') 982.025 0.01468 7.193 6.391 5.209 3.379 2.107 0.779 0.910 0.860 0.862 0.868 0.869 0.873 Trial 4 50.76384 48.98674 82.37028 77.75799 7.385 6.577 5.339 3.565 2.301 0.963 1.099 1.046 1.047 1.048 1.049 1.048 Trial 5 52 15172 5 79.92488 (cm/s') 6 (cm/s ) 7.489 6.681 5.503 3.667 2.39 1.069 0.204 0.198 0.19 0.191 0.194 0.194 Trial 6 51 68169 (cm/s'] 80.0689 (cm/s 126.74473 28.77124 Gan 7.653 6.839 5.659 3.827 2.55 1.227 0.487 0.473 0.473 0.474 0.474 0.475 8.454687 3.844772 S. (cm/s'] S. (cm/s') 7.761 6.943 5.765 3.939 2.671 1.341 0.816 0.795 0.797 0.797 0.799 0.802 D (Rad) 7.895 7.075 5,89 4.067 2.79 1,469 1.290 1.258 1.264 1,261 1.269 1.276 0.064577 (cm/s') (cm/s ) 1 = ( --) fa ( --) 7.945 7.127 5.945 4.115 2.845 1,509 Sa (Rad) 3,451612 1.569622 0 8.089 7.263 6.085 4.251 2.981 1.651 Note: to = time it takes the front of the glider to travel from the 1" gate 0.001745 3.337 7.503 6.325 4.493 3.221 1.889 to the 2" gate therefore the difference of the gate times gives the time fe(Rad) FE FE Terms for Error Calculations 3.485 7.651 6.471 4.637 3.371 2.039 intervals: to = Gar - Gas and to = Gyr-Gas etc. 0.02703 0.0%% D.OM 3.05E-06 141 3.05E-06 8.541 7.703 6.52 1.685 3.421 2.087 0.D06+00 0.ODE+00 3.689 7.847 6.667 4.827 3,559 2.229 M = W/t 0.008+00 Ge 3.809 7.959 6.783 4.949 3.681 2.349 (cm/s) Trial 1 Trial 2 Trial 3 Trial 4 Trial 5 Trial 6 (cm/s (cm/s'] ( unitless 3.98 8.141 6.959 5.129 3.859 2.529 60.647 57.164 73.529 56.766 66.176 66.372 980.35 982 025 0.01468 9.109 3.257 7.081 5.247 3.979 2.653 58.913 73.650 73.409 73.529 73.529 73.052 S S 9.345 8.487 7.309 5.477 4.211 2.879 74.176 78.488 78.306 77.765 77.675 77.320 *diff (cm/s'] (unitless) Error: Error: 9.505 8.639 7.465 5.631 4.36 3.039 81.893 86.042 85.960 85.878 85.796 85.878 0.171% 0.000 0.00000 9.897 9.031 7.859 6.021 4.763 3,441 -110.29 -113.64 -113.64 -117.8 -115.98 115.98 92.402 -95.137 95.137 -94.937 -94.937 -94.737 Total Cup VG -82.721 84.906 84.693 84.693 84.481 84.165 2.79 1 25 1.54 Gate pos. y down and up angle 69.767 -71.542 -71.203 -71.372 -70.922 -70.533 Un 120 YA 27 5 deg 97.5 Ya 45 3.7 Note: Use 2"SLOPE for the y and t data for each trial Fa 75.0 YC 67.5 a_down 56.117 51.013 32.193 50.764 52.152 51.682 52 5 Yo 90 Sangle a up 71.562 76.198 76.423 82.37 79.925 80.069 30.0 -22.5 deg 45 0.1 Distances YG 67.5 in cm YH 90Questions 1. (5 points) Is g sensitive to a change in angle of 0.10? Recalculate your experimental value of g by first adding, and then subtracting 0.1 to the angle of the track. How much does your experimental value change? 2. (5 points) Use the accepted value of g and your experimental results for upward and downward acceleration to calculate the inclination of the track. Report your result in degrees with three decimal places. You do not have to calculate the error for this value. 3. (10 points) a. What is the speed of the glider as it enters the first gate for the first trial? Describe how to use the linearized kinematic equation ( = =gt + Vo ) to determine this value. b. Calculate the length of the glider used in the experiment. Use the initial velocity from part (a) (Hint: We know how fast it was going as it entered the first gate and can calculate how much time it took to pass through the gate.) 4. (10 points) a. Suppose the glider was released just before the first photogate, with no time to build up speed. How would that affect your measurements of time, and your final result for g ? b. Suppose instead you push the glider down the track, increasing its initial velocity before it enters the first gate. How would this affect your measurements of time and final result for g?We then separate our vector equation into x-directed and y-directed pieces. Immediately, we notice that the normal force must balance the x-component of weight since there is no motion into or out of the track. We then see that since friction pointing up the track it is negative y-directed, and the y-component of weight is positive y-directed. -Wx + N =0 Wy - f = map Turning to trigonometry, we calculate the components of weight, given the angle 0. (fig 4.) N DOWN ap cost = W Wx = Wcost W Wy sind =- Wy = Wsind Figure 4 Substituting the results for W, and W, into our x- and y-equations we find the following: N = Wcose = mg cose And Wsine - f = map mg sine - UN = map mg sine - umg cose = map gsino - ugcose = ap Then, solving for g: ap g = sine - ucose In order to calculate the constant of gravitational acceleration, g, we must measure the angle of incline, 0, and the downward acceleration, ap, (by measuring distances and times, as in the Kinematics of Freefall experiment), however, we do not have a method to measure the coefficient of friction, M.Suppose, then, that the glider changes direction, and is moving up the incline instead. How would this change the free body diagram? Since friction opposes the direction of motion, it would point in the positive y-direction, the component of weight would also point in the negative y-direction. These forces would combine to decelerate the glider as it approaches the top of the track. A bumper placed at the bottom of the track will cause this change of direction. X UP Wx N cose = W Wx = Wcose au Wy sine = W W Wy = Wsine Figure 5: Velocity and acceleration are in different directions. The glider moves up the ramp in the negative y-direction, and the acceleration vector is pointing downward in the positive y-direction. The glider slows as it travels up the ramp. Using a similar analysis as above (and completed in your Prelab for this experiment) we find the following: gsine + ugcose = au Using the results for both upward and downward accelerations: gsine + ugcose = au gsine - ugcose = ap Adding these equations together allows us to eliminate the friction term and derive an expression for g in terms of the upward, and downward accelerations: 2gsine = au + ap g = - au + ap 2sine Also, subtracting these two equations (as in the Prelab) allows us to find an expression for the coefficient of friction: H =. au ab tand au + apAnalysis: Angle . Calculate the height difference between photogates r, and 75, call this value Ah. Calculate the distance between photogates r, and 75, call this value AL. Calculate the angle of the track using trigonometry (use the =DEGREES(ASIN()) formula in Excel): 0 = sin-1 (-) Calculate the error associated with the angle of the track: So = 0 SL V2 + Sn V2 AL Ah Time and Velocity . We are interested in the time it takes the middle of the glider to pass from the first photogate to the other gates: tA = (-21 + G2b )(Gif + G1b) 2 2 tB = (G3f + (3b) (Gif + Gib) 2 2 7 Physics 181 - UMass Boston etc. . Measuring from the first photogate, calculate the distance (VA, VB, etc) between the first gate and each subsequent photogate: YA = 11 -12 VB = 11 -13 etc.0 Measuring from the first photogate, calculate the distance (31A, 3'3: etc) between the rst gate and each subsequent photogate: 36121172 3J3=T1T'3 etc. 0 Calculate the velocity of the glider as it enters each photogate: \"A = J's/ta 0 As in our previous Freefall experiment, acceleration can be determined by using the slope of the velocity versus time graph. a,- = 2 * sweeter) . For each trial use the set of velocities and times and the excel SLOPEO formula to find acceleration down, and then up. 0 Average the calculated downward acceleration, and the average upward accelerations using the lDStats macro, then calculate g and ,u. 0 Calculate the error associated with g and p. 0 Compare your experimental result with the accepted value of980.35 cmfsz. Remember: Results are not complete without an estimate of the error. The only new step in calculating error for these equations is the error in the trigonometric formulas. Since our track is elevated to such a small angle [9 9: - Sln(69.7 169.4 Error associated with the angle of the track: Seat/(Erica 0.1 2 013/? 2 :> : g 33 3333\\/(0_1) (3.3) E>SQZO.67O After calculating the angle 91" the air tracks relative to the table surface using the Ah formula 6 : sin (E) and the error associated with the angle of the track, the result is not as expected because (Sing 7 33mg) and converted radiant. However two results on the grain satises the reguirement of being less than 1 radiant. The velocity of the glider as it is going down each photogate: 9A V : A tA 23.7 cm r> 26.93 VA 0.880 3 According to vector diagram because the glider going down to the air tracks so the According to vector diagram because the glider going down to the air tracks so the displacement will be negative and the velocity and it just going down because of the gravity force act on glider and it going opposite to the air pump. The velocity of the glider as it going up each photogate: VE:yE in 23. E> VE : 077 : 33.86 cm/s According to the vector diagram, the glider has been pushed by the air pump and it is going the same way as the air pump so the displacement and velocity will be positive. T113 unnolaw'h'n\" n? r1113 "HAQV who\" "min" Annm- 23. 7 cm E> an : 2 ( 0.88 ) 3.903? The acceleration of the glider when going up: a,t'v 21L 0 Solve for a,\"