A ball bearing falls through a viscous liquid (a) Using a labelled diagram of a tall vessel, show all the forces acting on it (b) When will it attain termin...

Answer Details

Ball bearing falling through a viscous liquid

(a) Labelled diagram showing the forces acting on the ball

Inside a tall vessel of viscous liquid, a spherical ball bearing falling vertically downwards experiences three forces:

• Weight, \(W = mg\), acting vertically downwards through the centre of the ball.
• Upthrust (buoyancy), \(U\), acting vertically upwards, equal to the weight of liquid displaced.
• Viscous drag, \(F\), a fluid-friction force given by Stokes' law \(F = 6\pi\eta r v\), acting vertically upwards, opposing the motion.

(b) When terminal velocity is attained

As the ball accelerates downwards its speed \(v\) increases, and because the viscous drag \(F = 6\pi\eta r v\) grows with speed, the total upward force \((U + F)\) increases too. The ball attains its terminal velocity at the instant the upward forces exactly balance the downward weight, so the resultant force and hence the acceleration become zero:

From that moment onwards there is no further acceleration, and the ball continues to fall through the liquid with a constant (maximum, uniform) velocity, which is the terminal velocity.