Cart on an Incline: Deciding What the Missing 15% Really Is
A lab group places a 0.500 kg dynamics cart on a track inclined at 8.0° ± 0.5° and ties it by a light string over a small pulley to a 0.200 kg hanging mass. A motion sensor records velocity, and a straight-line fit to the first 0.60 s of data - the cart is released 0.15 s after recording starts - gives a = 1.55 m/s². The point-mass prediction, ignoring friction and the pulley, is a = (m₂g − m₁g sin θ)/(m₁ + m₂) = 1.83 m/s². The group's draft write-up says only that friction caused a 15% error. Their instructor asks them to defend that claim with numbers before they revise the procedure. Apparatus details: the pulley is a solid disk of mass 0.020 kg and radius 0.030 m; the cart rides on four hard plastic wheels on an aluminum track; the incline is set with the track's built-in scale, readable to about ±0.5°; the string is light and does not slip. The cart reaches roughly 0.9 m/s by the end of the fit window.
1. Rank the three candidate explanations - pulley rotational inertia, wheel friction, and a ±0.5° error in the incline angle - by the size of the acceleration error each could produce, with a numerical estimate for each. State which one the group should investigate first.
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Work in terms of the force along the track. The point-mass prediction is a = (1.96 − 0.682)/0.700 = 1.83 m/s², so the driving net force is 1.278 N. To reach 1.55 m/s² the net force must be 0.700 × 1.55 = 1.085 N - a shortfall of 0.193 N, or 15%.
Pulley. I = ½MR² = ½(0.020)(0.030)² = 9.0×10⁻⁶ kg·m². The pulley behaves as an extra effective mass I/R² = ½M = 0.010 kg, so a = 1.278/(0.710) = 1.80 m/s². Effect ≈ 0.03 m/s², about 1.6%. Not enough.
Angle. sin 8.0° = 0.1392, sin 8.5° = 0.1478, sin 7.5° = 0.1305. Worst case shifts the driving force by 0.500 × 9.8 × 0.0086 ≈ 0.042 N, so a changes by ±0.06 m/s², about 3%. Not enough either.
Friction. A constant 0.193 N retarding force accounts for the whole gap. Relative to the cart's normal force (4.85 N) that is μ ≈ 0.04, entirely ordinary for plastic wheels on aluminum, and it would include pulley-bearing friction.
Only friction has the right order of magnitude, so it should be pursued first. The group should be careful to call 0.193 N an effective total, not proof that the wheels are responsible.
2. A classmate argues that if friction is the cause, the velocity–time graph must still be perfectly straight, so the group cannot be missing anything important. Evaluate that claim and describe what feature of the data would separate dry (constant-magnitude) friction from friction that grows with speed.
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The claim is half right and therefore misleading. Dry friction has a magnitude independent of speed, so the net force stays constant, v(t) is exactly straight, and the missing 0.28 m/s² hides entirely in the slope. In that case the linear fit is the correct model and no residual pattern will betray anything - the classmate is right about the shape but wrong to conclude that nothing is missing.
Speed-dependent drag behaves differently. If friction or air resistance grows with speed (∝ v or ∝ v²), the net force falls as the cart accelerates, so v(t) is concave down and the fitted slope drifts lower across the window. The cart reaches about 0.9 m/s, fast enough that a few percent of curvature should be visible. The diagnostic is a systematic residual pattern - points below the fitted line early and above it late, or a quadratic term whose coefficient exceeds its own uncertainty by a clear margin - as opposed to random scatter about a straight line.
Two further tell-tales: a kink near the release instant as static friction is broken, and a nonzero velocity intercept if the release is not clean. A single straight-line fit is a model choice, not a test, and the group should fit both models and compare residuals rather than assuming one.
3. Describe one change to the apparatus and one change to the analysis that would let the group separate pulley inertia from friction instead of lumping them together. Show how the analysis would extract the two quantities.
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Analysis change (the stronger option). Repeat the run with several hanging masses - 0.100, 0.150, 0.200, 0.250, 0.300 kg - keeping cart, incline and pulley fixed. The equation of motion is m₂g − m₁g sin θ − f = (m₁ + m₂ + I/R²)a. Plot the driving term (m₂g − m₁g sin θ) on the vertical axis against the measured a. The slope is the effective inertia M_eff = m₁ + m₂ + I/R², so any excess of the slope over m₁ + m₂ is the pulley's contribution I/R² (and any wheel rotational inertia). The intercept gives f directly, since a = 0 when the driving term equals f.
The logic is that pulley inertia changes the slope while friction changes the intercept; a single point confounds them, a line does not. A slope consistent with m₁ + m₂ within uncertainty is positive evidence that both the pulley and the wheels contribute negligible rotational inertia, which would shift suspicion onto plain bearing and wheel friction.
Apparatus change. Remove the string and hanging mass and let the cart roll freely down the same incline over the same distance. Comparing free-roll acceleration with the towed acceleration at matched speeds, with no string attached, isolates string-and-pulley losses from wheel friction.