A sphere is released from rest in a viscous liquid and obeys Stokes' law. Its initial acceleration, after accounting for buoyancy, is a₀ and its terminal speed is vₜ. What is its acceleration when its speed becomes vₜ/2?
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Question

A sphere is released from rest in a viscous liquid and obeys Stokes' law. Its initial acceleration, after accounting for buoyancy, is a₀ and its terminal speed is vₜ. What is its acceleration when its speed becomes vₜ/2?

Options

G1 Physics · Accepted Answer

Let the constant effective downward driving force be F₀. Stokes drag can be written as F(viscous) = kv, where k = 6πηr. At terminal speed, F₀ = kvₜ. At a general speed v, ma = F₀−kv = k(vₜ−v). At release, v = 0, so ma₀ = kvₜ. Therefore, a₀ = kvₜ/m. At v = vₜ/2, a = [k/m](vₜ−vₜ/2) = kvₜ/(2m) = a₀/2. Correct option: C.