Trolley on an incline with a hanging mass
A laboratory trolley of mass M = 0.80 kg rolls on a straight track inclined at angle α = 12° to the horizontal. Friction in the axle is negligible, but the track's small wheels have total moment of inertia I_w = 4.0×10⁻⁵ kg·m² and radius r = 5.0 mm, so their rotation cannot be ignored when the trolley accelerates. A light string attached to the trolley passes over a frictionless pulley at the top of the incline and hangs vertically, carrying a load of mass m = 0.25 kg. The system is released from rest with the trolley 1.20 m from the pulley. Take g = 9.81 m/s² and sin 12° = 0.208. A student predicts the trolley will move up the incline because m is 'small but the incline is gentle'.
1. Determine the direction of motion and the acceleration of the trolley, treating the trolley plus its wheels as a single body but accounting for the wheels' rotational inertia.
Read the worked answer
Compare the driving force from the hanging mass with the downhill component of the trolley's weight (the wheels are part of the trolley and roll without slipping on the track, so their translational mass is already in M; their rotational inertia adds an effective mass).
Driving force from load: F_m = m g = 0.25 × 9.81 = 2.45 N. Downhill component along the track: F_∥ = M g sin α = 0.80 × 9.81 × 0.208 = 1.63 N.
Since F_m > F_∥, the trolley moves up the incline; the student is wrong.
For the acceleration, the wheels' rotational inertia adds an effective mass. For each wheel rolling without slipping, v = ωr, so the extra effective mass per wheel is I_w/r². With r = 5.0×10⁻³ m, I_w/r² = 4.0×10⁻⁵ / 2.5×10⁻⁵ = 1.6 kg per wheel - which is larger than M itself, so this is clearly a case where the model must be questioned (see next task). Taking the statement at face value for a single effective wheel term, the total effective mass of the trolley is M + I_w/r² = 0.80 + 1.6 = 2.4 kg.
Equation of motion (up the incline positive): (2.4 + 0.25) a = m g − M g sin α = 2.45 − 1.63 = 0.82 N a = 0.82 / 2.65 ≈ 0.31 m/s².
The trolley accelerates up the incline at about 0.31 m/s².
2. The value I_w/r² = 1.6 kg is physically suspicious for a trolley of mass 0.80 kg. Identify the inconsistency and explain what a careful experimenter should do before trusting the acceleration value.
Read the worked answer
The computed effective mass I_w/r² = 1.6 kg per wheel exceeds the entire trolley mass. That would require the wheels to store more rotational energy than the whole trolley has translational energy at the same speed - implausible for small steel wheels of radius 5 mm. The realistic moment of inertia of a small solid wheel of mass m_w and radius r is ½ m_w r², giving I_w/r² = ½ m_w, i.e. at most a few grams of effective mass, not 1.6 kg.
A careful experimenter should:
- Check the units and magnitude of the given I_w before substituting. A value of 4×10⁻⁵ kg·m² for a 5 mm wheel corresponds to a wheel mass of about 3.2 kg (from I = ½ m r²), which contradicts the trolley mass.
- Re-measure or re-estimate I_w from the wheel's mass and geometry, or determine it experimentally (e.g. rolling-time method).
- Report the sensitivity: with a corrected I_w/r² of order 10⁻³ kg, the effective mass is essentially M, and a = 0.82 / 1.05 ≈ 0.78 m/s² - more than double the earlier value.
Conclusion: the direction of motion is robust (F_m > F_∥), but the acceleration is highly sensitive to the wheel term, so the data must be validated before use.
3. Using the corrected estimate that the wheels contribute negligibly, find the speed of the trolley after it has moved 1.20 m up the incline, and state one assumption that limits the accuracy of this prediction.
Read the worked answer
With wheels neglected, the net force along the motion is constant: F_net = m g − M g sin α = 2.45 − 1.63 = 0.82 N. Total mass in motion: M + m = 1.05 kg (string mass negligible). a = 0.82 / 1.05 ≈ 0.78 m/s².
Using v² = 2 a s with s = 1.20 m: v² = 2 × 0.78 × 1.20 = 1.87 m²/s², so v ≈ 1.37 m/s.
Assumptions/limits: the string stays taut and does not stretch; the pulley is frictionless and massless; air resistance is neglected; the load does not swing (it moves vertically only). If the load swings, part of the driving force goes into transverse motion and the predicted speed is too high. Also, if the trolley reaches the pulley before 1.20 m, the model breaks down.