Trolley on an Incline: Is the Shortfall Due to Friction?
A student investigates whether a trolley rolling down a ramp obeys a = g sin θ. The ramp is set at θ = 8.0° ± 0.5°. The trolley has total mass 0.500 kg, including four identical solid-disc wheels, each of mass 0.012 kg and radius 6.0 mm. A card of width 50.0 mm on the trolley passes through two light gates separated by 0.400 m along the ramp. The gate times are 41.2 ms and 28.8 ms. The student calculates the speeds as 1.21 m/s and 1.74 m/s, then a = (v₂² − v₁²)/(2 × 0.400) = 1.10 m/s². The value of g sin θ is 1.37 m/s². The student writes: 'The shortfall is caused by friction. Taking the normal reaction as mg cos θ, the coefficient of friction is μ = (1.37 − 1.10)/(9.81 cos 8.0°) = 0.028. The wheels must be slipping.' A repeat gives 1.08 m/s².
1. Calculate the acceleration expected if the wheels roll without slipping and the only resistance is the rotational inertia of the wheels. Hence comment on whether the measured acceleration is fully explained by rotational inertia.
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Each wheel is a solid disc: I = ½ m r². For four wheels, total I = 4 × ½ × 0.012 × r² = 0.024 r². The translational mass is 0.500 kg, so the effective inertial mass is M + I/r² = 0.500 + 0.024 = 0.524 kg. The component of weight down the slope is M g sin θ = 0.500 × 9.81 × sin 8.0° = 0.683 N. Thus a = 0.683/0.524 = 1.30 m/s². Equivalently a = g sin θ/(1 + I/(M r²)) = 1.365/1.048 = 1.30 m/s². The measured 1.10 m/s² is about 0.20 m/s² lower, whereas rotational inertia accounts for only about 0.06 m/s² (1.365 − 1.303). Also the angle uncertainty gives δ(g sin θ) = g cos θ δθ ≈ 9.81 × 0.990 × 0.0087 = 0.084 m/s², so the expected value is roughly 1.30 ± 0.08 m/s². The measured value is still outside this range. Therefore rotational inertia alone does not explain the measured acceleration; another retarding effect is present.
2. Evaluate the student's conclusion that μ = 0.028 and that the wheels are slipping. Give two reasons why the calculation is not a valid determination of a coefficient of friction.
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The conclusion is not justified. First, the wheels roll rather than slide: the contact point is instantaneously at rest, so the relevant resistance is rolling resistance and bearing friction, not a sliding coefficient μ. Using μ = (g sin θ − a)/(g cos θ) treats the trolley as a sliding block and mislabels the result. Second, the calculation ignores the rotational inertia of the wheels, which reduces the acceleration even with no friction at all; including it changes the inferred resistance. Third, the light-gate speed is an average over the card, and the card width shifts the effective measurement points, so the acceleration has additional systematic uncertainty. The value 0.028 is therefore an effective resistance coefficient at best, not a valid coefficient of friction. The claim that the wheels 'must be slipping' is also unsupported: slipping would usually increase the retarding force and could produce a larger shortfall, but the evidence does not distinguish slipping from bearing friction, rolling resistance, or air resistance. A better conclusion is that the measured acceleration is lower than the rotational-inertia prediction by about 0.20 m/s², and further tests are needed to identify the cause.
3. Suggest one experimental change that would allow friction and rotational inertia to be separated, explaining what would be measured.
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Use an air track or a friction-compensated ramp so that the only significant retarding effect is the wheels' rotational inertia. Measure the acceleration for the same total mass but with wheels of different moment of inertia (for example, solid discs and hoops of the same mass and radius). If a plot of a against I/(M r²) is linear and extrapolates to g sin θ at I = 0, the rotational-inertia model is supported; any residual intercept or slope gives the friction term. Alternatively, lock the wheels and slide the trolley body down the same ramp to measure kinetic friction separately, then compare the rolling and sliding accelerations. In both cases, use several gate spacings and plot v² against displacement, so that the acceleration comes from a gradient rather than a single pair of timings.