Incline Data That Do Not Give g
A physics class uses a 1.20 m aluminum dynamics track, a 0.250 kg cart with low-friction wheels, and a motion sensor at the top of the incline. Students set the track at angles from 5° to 25° and release the cart from rest. For each angle they fit the linear portion of the velocity-time graph and record acceleration. Their plot of acceleration versus sin θ is strikingly linear (R² = 0.998) with slope 7.8 m/s² and vertical intercept −0.35 m/s². The accepted value of g is 9.80 m/s². One student argues that the track is defective because the slope is not 9.80; another says the negative intercept proves friction, so the slope should still be 9.80. The teacher asks the group to decide what the data actually support and to propose a better experiment before writing a conclusion.
1. What do the slope and intercept imply physically? Evaluate both students' claims.
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With kinetic friction f = μmg cos θ, Newton's second law along the incline gives a = g sin θ − μg cos θ. Over 5° to 25°, cos θ changes from 0.996 to 0.906, so a plot against sin θ is not exactly straight; the negative intercept is consistent with a friction-like term, but its magnitude is not simply μg unless the cos θ variation is ignored. If friction were the only effect, the slope would still be g = 9.80 m/s², so the observed slope 7.8 is 20% low. That is too large to dismiss as random scatter, but it does not show that g is wrong or the track is defective; it shows the one-term friction model is incomplete. Rotational inertia of the cart wheels, track flex, or a misaligned sensor could reduce the slope. The first student overinterprets; the second correctly notes the sign of friction but incorrectly assumes the slope must be 9.80.
2. Using the simple constant-friction model, estimate μ from the intercept. Then calculate the predicted acceleration at θ = 20.0° if the slope were forced to 9.80 and compare it with the fitted line. What does the discrepancy reveal?
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Using the crude estimate μ ≈ 0.35/9.80 = 0.036, at 20.0° the model predicts a = 9.80(0.342) − 0.036(9.80)(0.940) = 3.35 − 0.33 = 3.02 m/s². The fitted line predicts a = 7.8(0.342) − 0.35 = 2.32 m/s². The 0.70 m/s² gap is about 23%, much larger than the likely scatter. If the entire gap were attributed to wheel rotational inertia, a = g sin θ / (1 + I/(mR²)), the effective factor is 7.8/9.80 = 0.796, so I/(mR²) ≈ 0.26, a plausible order for cart wheels. Thus the intercept suggests friction is present, but the low slope points to an additional inertial or systematic effect; the data cannot isolate g without a better model.
3. Design a follow-up procedure that would separate friction from rotational inertia and obtain a credible value of g. Include at least two measurements and how you would analyze them.
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Separate the effects rather than fit one line. First, measure friction directly: incline the track until the cart moves at constant velocity after a push; then tan θ = μ, independent of g. Second, test rotational inertia: repeat at a fixed angle with extra mass fixed securely to the cart body. Adding mass increases m but changes I/(mR²) little for the wheels, so if rotational inertia matters, the acceleration should increase toward g sin θ. Third, fit a = g sin θ − μg cos θ by multiple regression or plot a/cos θ versus tan θ; the slope is g and the intercept is −μg. Also level the track, align the sensor with the motion, and use an air track or low-friction cart. A credible g requires agreement between the free-fit slope and the independent friction measurement; otherwise report the model discrepancy, not a new value of g.