Spool Pulled by a String on Its Inner Axle
A spool of mass 2.0 kg rests on a horizontal table. Its moment of inertia about the central axis is 0.030 kg·m². The spool has an outer radius of 0.20 m that contacts the table and an inner axle of radius 0.10 m. A light string is wound around the inner axle and leaves from the lowest point of the axle, so the string is tangent to the bottom of the inner circle. The string is pulled with a constant force of 12 N directed 37° above the horizontal. The coefficients of friction between the spool and the table are μs = 0.40 and μk = 0.30. Take g = 9.8 m/s², sin 37° = 0.60, and cos 37° = 0.80. The spool is released from rest. The string remains taut and does not slip on the axle, and the spool stays in contact with the table. Air resistance and rolling resistance are negligible.
1. Determine the direction and magnitude of the static friction force on the spool, assuming it rolls without slipping. Then test whether this assumption is consistent with the available friction.
Read the worked answer
Let a be the acceleration of the centre to the right and α the clockwise angular acceleration. For rolling without slipping, a = αR.
The horizontal component of the pull is F cosθ = 12×0.80 = 9.6 N to the right. Let friction f act to the right; a negative result would mean it acts left.
Translation: 2.0a = 9.6 + f.
Rotation about the centre: the pull acts at the bottom of the inner axle and gives clockwise torque rF cosθ = 0.10×9.6 = 0.960 N·m. Friction at the table contact gives clockwise torque −Rf. Hence 0.030α = 0.960 − 0.20f.
Using a = αR gives α = a/0.20 = 5a, so 0.030×5a = 0.960 − 0.20f, i.e. 0.15a = 0.960 − 0.20f.
From translation a = (9.6 + f)/2.0. Substituting: 0.15(9.6 + f)/2 = 0.960 − 0.20f, so 0.72 + 0.075f = 0.960 − 0.20f. Thus 0.275f = 0.24 and f = 0.873 N to the right.
The normal force is N = mg − F sinθ = 19.6 − 7.2 = 12.4 N. Maximum static friction is 0.40×12.4 = 4.96 N. Since 0.873 N < 4.96 N, the no-slip assumption is self-consistent.
2. The applied force is now gradually increased while its direction is kept at 37°. Find the largest force for which the spool still rolls without slipping, and state what happens immediately beyond it.
Read the worked answer
Keep f to the right and write the rolling solution as a function of F.
Translation: M a = F cosθ + f.
Rotation: I a/R = rF cosθ − R f.
Eliminating a gives f = F cosθ [r/R − I(1 + r/R)/(R²(M + I/R²))]. Numerically, I/R² = 0.030/0.04 = 0.75 kg, M + I/R² = 2.75 kg, and the bracket is 0.10/0.20 − 0.030×1.5/(0.04×2.75) = 0.50 − 0.409 = 0.0909. Thus f = 0.0909 F cos37° = 0.0727F, to the right.
The normal force is N = Mg − F sinθ = 19.6 − 0.60F.
Slipping begins when f = μsN: 0.0727F = 0.40(19.6 − 0.60F) 0.0727F = 7.84 − 0.24F 0.3127F = 7.84 F ≈ 25 N.
At this force N ≈ 4.6 N > 0, so contact is maintained. Just beyond this force the required static friction exceeds μsN; kinetic friction takes over, the contact point slips backwards, and the spool no longer rolls without slipping.