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Vertical motion

Viscosity of water

The vertical velocities v in cm/s versus the instant of time after releasing (at t=0) a metal ball bearing with a radius of r=0.24 cm in water. Data are taken from J.P. Owen and W.S. Ryu, Eur. J. Phys. 26, 1085 (2005). The curve is given by v(t)=α×tanh(t/T), for α=1.00 m/s and T=0.117 s.

For the velocity as a function of time we deduced previously the expression

v(t) = -α×tanh(t/T)     where     v_term = -α   and   T=α/g

In their article on the subject Julia P Owen and William S Ryu give measured values for the velocities at certain instants of time for a metal ball bearing with a radius of r=0.24 cm which is vertically falling in water. Those values can be compared with the above theoretical expression.
At the previous page we find that the terminal velocity for a metal ball bearing with a radius of r=0.24 cm (¹⁰log(0.24)=-0.62) equals 1.00 m/s (¹⁰log(100)=2.0).

Furthermore, also counting with the buoyant force of water, the relation between T and g reads

T = αρ_s / g(ρ_s-ρ_water)

where ρ_s=8.02×10³ kg/m³ represents the density of the metal of the ball bearing and ρ_water=1.00×10³ kg/m³ the density of water. Substituting the values for α, g=9.8 m/s² and the densities, we find T=0.117 s.

Note that the curve in the figure fits very well the measured velocities.

buoyant force