Marble Run Loop: Friction, Contact and Braking
A school physics club is building a marble run for its annual exhibition. A steel marble of mass 20 g is released from rest at height h above the bottom of a vertical circular loop of radius R = 0.40 m. Apart from a horizontal rough patch 0.50 m long just before the loop, where the coefficient of kinetic friction is 0.25, the track is frictionless. The marble is treated as a particle that slides without rolling, and g = 10 m/s². When the marble is released from h = 1.10 m, the club finds that it loses contact with the track before reaching the top of the loop. They want a release height that guarantees the marble stays in contact all the way round, and they also plan to fit a braking strip, made of the same rough surface, after the loop so that the marble stops within 0.30 m. The club adviser suspects that no ordinary rough strip can be that short and asks for numbers before the design is finalised.
1. Find the minimum release height h_min that keeps the marble in contact with the track at the top of the loop.
Read the worked answer
Contact condition at the top. At the top of the loop the weight and the normal reaction both point towards the centre, so
mg + N = m v_t² / R.
Contact is just maintained when N = 0, giving v_t = √(gR) = √(10 × 0.40) = 2.0 m/s.
Energy needed at the top. ½mv_t² + mg(2R) = ½(0.020)(4.0) + (0.020)(10)(0.80) = 0.040 + 0.160 = 0.200 J.
Friction loss before the loop. W_f = μmgL = 0.25 × 0.020 × 10 × 0.50 = 0.025 J.
Release height. mgh_min = 0.200 + 0.025 = 0.225 J, so h_min = 0.225 / (0.020 × 10) = 1.125 m ≈ 1.13 m.
The observation is consistent: 1.10 m < 1.125 m, so the marble is short of the required speed at the top even though it reaches the top of the track.
2. If the marble is released from h = 1.30 m, what is its speed at the top of the loop and the normal force on it there?
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Initial energy: mgh = 0.020 × 10 × 1.30 = 0.260 J. Friction removes 0.025 J, and the marble must still climb 2R = 0.80 m, costing mg(2R) = 0.160 J.
KE at the top = 0.260 − 0.025 − 0.160 = 0.075 J.
v² = 2(0.075)/0.020 = 7.5 m² s⁻², so v ≈ 2.74 m/s (above the 2.0 m/s threshold, so contact is maintained).
N = m v²/R − mg = (0.020)(7.5)/0.40 − (0.020)(10) = 0.375 − 0.200 = 0.175 N.
The track pushes the marble vertically downward, i.e. towards the centre of the loop; a positive value confirms it is still on the track.
3. Can the same rough surface stop the marble within 0.30 m after the loop? Support your answer with numbers.
Read the worked answer
Energy at the bottom after the loop (release from 1.30 m) = 0.260 − 0.025 = 0.235 J, so
v = √(2 × 0.235 / 0.020) = √23.5 ≈ 4.85 m/s.
Stopping distance for the same surface (μ = 0.25): s = v²/(2μg) = 23.5/(2 × 0.25 × 10) = 4.7 m, which is about 16 times the 0.30 m target.
For s = 0.30 m the required coefficient would be μ = v²/(2gs) = 23.5/(2 × 10 × 0.30) ≈ 3.9, far beyond any ordinary track surface (typical μ < 1).
Conclusion: the adviser is right - the design must instead lower the release height, lengthen the braking strip, or add a separate energy absorber such as a soft block or a spring.