Using Real Data, and Analysing Errors( continued)
YEARS 7 – 12 IDEAS ARTICLES FOR THE CLASSROOM
Using Real Data, and Analysing Errors( continued)
The results using average times to calculate the accelerations show excellent fits to a straight line for both parts of the data, but they are not the same line. The results are almost symmetrical because the zero of the horizontal( h 0 in the table above) was adjusted by 0.151 mm, corresponding to about 30˚ rotation of the thread. This is less than half of the interval( 90˚) used in leveling the air track, so physically reasonable and symmetry makes the remaining analysis easier.
The slopes of the lines give estimates of gravitational acceleration. Both values are a little high. From earlier work with dropping
balls, three timing errors have been identified and all are mixed together in the data. The timing errors occur starting the timer, stopping the timer, and releasing the ball or glider. Adding 0.084 second to the timings brings the results in both directions to about 9.8 ± 0.05 m. s – 2. This small timing error is plausible; it is well within human reaction time and it has been consistently observed with many groups of students. The value is consistent across experiments dropping balls, timing on the linear air track and using Atwood’ s machine.
Figure 3: The adjusted timings are consistent with the expected value of gravity.
In Table 1, results for 0 and-1 turns are conspicuously absent. We observed that the glider at both ends of the track did not accelerate. In fact, when moved out a small distance it drifted towards the end. This suggests our track was slightly bowed upwards in the centre. This invalidates the assumption of uniform acceleration when the height in the centre is large compared to the slope we have from the screw thread.
A mathematical model of the glider accelerating on a slightly curved track( parabolic, higher in the centre by h) was constructed. The end-to-end slope is θ radian. The model should be valid for small θ. Using energy conservation, kinetic energy gain = ½. m. v 2 = potential energy loss = m. g. ∆y where ∆y is the vertical distance through which the glider has fallen.
v = dx / dt = √( 2g. ∆y) Lz = x
&
! dddd = TT = LL 4
') 2ggh! dddd
= LL
') zz( nn − 4 + 4zz)) 2ggh arctanh( 2
√nn) e of h through which the end has been lowered. n is the multiple of h through which the end has been lowered.
This model is simple enough to be examined in detail with a spreadsheet calculation.
56 SCIENCE EDUCATIONAL NEWS VOL 67 NO 3