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A bead of mass m slides on a fixed horizontal circular wire of radius R. At time t=0, the bead is given a speed v0 along the tangent to the wire. The coefficient of kinetic friction between the bead and wire is muₖ. What is the magnitude of the tangential acceleration of the bead at t=0?

1. muₖ * g
2. muₖ * v0² / R
3. (muₖ / R) * sqrt(v0⁴ + g² * R²)
4. muₖ * (g + v0² / R)

Correct answer: (muₖ / R) * sqrt(v0⁴ + g² * R²)

Solution

The wire exerts a normal force with horizontal (centripetal) component m*v0²/R and vertical component mg. The magnitude of the net normal force is sqrt((m*v0²/R)² + (mg)²). Kinetic friction equals muₖ times this normal force, giving tangential deceleration = (muₖ/R)*sqrt(v0⁴ + g²*R²).

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