Timeline for Getting acceleration due to gravity from dropping ball experiment [closed]
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Sep 7, 2017 at 10:02 | history | closed |
sammy gerbil Bill N Jon Custer By Symmetry Kyle Kanos |
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| Sep 5, 2017 at 21:03 | vote | accept | Peter | ||
| Sep 5, 2017 at 21:03 | vote | accept | Peter | ||
| Sep 5, 2017 at 21:03 | |||||
| Sep 5, 2017 at 21:03 | vote | accept | Peter | ||
| Sep 5, 2017 at 21:03 | |||||
| Sep 5, 2017 at 17:48 | comment | added | Bill N | Why use Mathematica? Set up a system of 3 equations and 3 unknowns: $v_1^2=2gL_1$, $v_2^2=2gL_2$, and $v_2=v_1+g\Delta t$. Solve for $g$. | |
| Sep 5, 2017 at 11:53 | history | edited | Mitchell | CC BY-SA 3.0 |
deleted 8 characters in body
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| Sep 5, 2017 at 11:48 | review | Close votes | |||
| Sep 7, 2017 at 10:02 | |||||
| S Sep 5, 2017 at 10:44 | history | suggested | Plexus | CC BY-SA 3.0 |
improved formatting
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| Sep 5, 2017 at 10:14 | review | Suggested edits | |||
| S Sep 5, 2017 at 10:44 | |||||
| Sep 5, 2017 at 8:51 | answer | added | Farcher | timeline score: 1 | |
| Sep 5, 2017 at 6:21 | history | edited | Qmechanic♦ |
edited tags
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| Sep 5, 2017 at 5:46 | answer | added | kpv | timeline score: 2 | |
| Sep 5, 2017 at 5:06 | comment | added | user93237 | .. the (unknown) drop time. You then have two equations with two unknowns (i.e., 't0' and $g$). The unraveling of the solutions doesn't appear easy (I used Mathematica), but it is possible to get expressions for the actual start time 't0' and the acceleration $g$ from the measured quantities of L1, L2, t1, and t2. You just need to measure the distances L1 and L2 very accurately and hope that the gate timers are accurate and have little jitter. | |
| Sep 5, 2017 at 5:02 | comment | added | user93237 | A big problem with your setup is that, as you are aware, you have no way of knowing what the true t=0 start time of the ball drop is. Putting the first timer gate as close as possible to the drop point doesn't seem like a good option to me because the ball will not have v=0 when it triggers the first timer gate and a small timing error can lead to a large error in $g$ (as Mitchell below showed). A work-around may be to instead put the timer gates at distances L1 and L2 down from the drop point and measure their trigger times t1 and t2. The L1=g (t1-t0)^2/2 and L2=g (t2-t0)^2/2, where t0 is .. | |
| Sep 5, 2017 at 4:28 | comment | added | user93237 | Your methods #2 and #3 can't be used because you have no way of measuring the instantaneous final velocity with your setup. Also, your calculation for $v_f$ is really a calculation of the average velocity. As for Method #1, that can be applied but I'm not sure that putting a timer gate near the initial v=0 drop point is the best position to get the highest measurement accuracy of g. Seems to me that you might get more accurate g measurement by locating the first timer gate a bit further down (and, of course, accurately measuring its position). You should so an error analysis of your setup. | |
| Sep 5, 2017 at 4:10 | answer | added | Mitchell | timeline score: 3 | |
| Sep 5, 2017 at 3:54 | review | First posts | |||
| Sep 5, 2017 at 4:05 | |||||
| Sep 5, 2017 at 3:53 | history | asked | Peter | CC BY-SA 3.0 |