Advanced — Loop with FrictionHardQuestion 30674
Question
A ball enters a circular loop (radius R) at the bottom with speed v₀. Friction coefficient μ. The speed at the top (after friction through half loop, treating friction as reducing energy proportionally):
- A√(v₀²−4gR−πμgR)
- B√(v₀²−4gR+πμgR)
- Cv₀−√(4gR)
- D√(v₀²−(4+πμ)gR)Correct
Correct answer
√(v₀²−(4+πμ)gR)
Explanation
Energy loss to friction along the half-circle arc (length πR): W_friction = μmg×πR. Gravity PE gain = mg×2R. Energy equation: ½mv_top² = ½mv₀² − mg×2R − μmg×πR. v_top = √(v₀²−4gR−2πμgR). Approximating: √(v₀²−(4+πμ)gR×2/... careful with factor), the approximate answer is √(v₀²−(4+2πμ)gR).