A small spherical lead shot of radius 3.0 \times 10^{-3}\text{ m} and density 8000\text{ k… — JAMB UTME

A small spherical lead shot of radius $3.0 \times 10^{-3}\text{ m}$ and density $8000\text{ kg/m}^3$ falls vertically through a viscous oil of density $800\text{ kg/m}^3$ and dynamic viscosity $0.18\text{ Pa}\cdot\text{s}$ . What is the magnitude of the terminal velocity of the sphere in $\text{m/s}$ ? (Take $g = 10\text{ m/s}^2$ )

Cevap: 0.8 m/s

Cevap

The terminal velocity of the lead shot is

0.8\text{ m/s}

At terminal velocity, acceleration is zero because the upward forces (viscous drag

6\pi \eta r v_t

plus buoyant upthrust

\frac{4}{3}\pi r^3 \rho_f g

) completely balance the downward weight of the sphere (

\frac{4}{3}\pi r^3 \rho_s g

). Solving for velocity gives

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta} = 0.8\text{ m/s}

Adım Adım Çözüm

Formulate the force balance equation at terminal velocity.

Weight (

W

) = Upthrust (

U

) + Viscous drag (

F_v

), which simplifies to

v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta}

When terminal velocity is attained, the net acceleration of the sphere is zero, so upward forces balance downward force.

Substitute the physical parameters into the terminal velocity formula.

v_t = \frac{2 \times (3.0 \times 10^{-3})^2 \times (8000 - 800) \times 10}{9 \times 0.18} = 0.8\text{ m/s}

Direct calculation using Stokes' law and Archimedes' principle.

Anahtar Kavram

Terminal Velocity and Stokes' Law in a Viscous Medium

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