Two solid spheres made of steel (density σ) of radii R and 2R each are connected to ends of a long massless inextensible string and passed over a smooth fixed pulley. The whole sys... Two solid spheres made of steel (density σ) of radii R and 2R each are connected to ends of a long massless inextensible string and passed over a smooth fixed pulley. The whole system is submerged inside a liquid of density ρ and coefficient of viscosity η. (Assume that the string is long enough that the spheres are far away from each other) Find 1. Terminal velocity of the spheres 2. Tension in the string as the system attains its steady state. Read more

Steps to Solve

1. Identify the Forces on Each Sphere

Each sphere experiences three main forces: buoyancy ($F_b$), the weight of the sphere ($F_g$), and the drag force ($F_d$).

• For Sphere 1 (radius $R$):

  • Volume: $V_1 = \frac{4}{3} \pi R^3$
  • Weight: $F_{g1} = \sigma V_1 g = \sigma \frac{4}{3} \pi R^3 g$
  • Buoyancy: $F_{b1} = \rho V_1 g = \rho \frac{4}{3} \pi R^3 g$

• For Sphere 2 (radius $2R$):

  • Volume: $V_2 = \frac{4}{3} \pi (2R)^3 = \frac{32}{3} \pi R^3$
  • Weight: $F_{g2} = \sigma V_2 g = \sigma \frac{32}{3} \pi R^3 g$
  • Buoyancy: $F_{b2} = \rho V_2 g = \rho \frac{32}{3} \pi R^3 g$

2. Equation of Motion for Each Sphere

In the steady state, the net force acting on each sphere is zero. Thus, we can write for each sphere:

• For Sphere 1: $$ F_{g1} - F_{b1} - F_{d1} = 0 $$

• For Sphere 2: $$ F_{g2} - F_{b2} - F_{d2} = 0 $$

3. Calculate Drag Forces

For a sphere falling through a fluid, the drag force is given by Stokes' law: $$ F_d = 6 \pi \eta R v $$

• For Sphere 1: $$ F_{d1} = 6 \pi \eta R v_1 $$

• For Sphere 2: $$ F_{d2} = 6 \pi \eta (2R) v_2 $$

4. Setting Up the Terminal Velocity Equations

Substituting the forces into the equations of motion:

• For Sphere 1: $$ \sigma \frac{4}{3} \pi R^3 g - \rho \frac{4}{3} \pi R^3 g - 6 \pi \eta R v_1 = 0 $$

• For Sphere 2: $$ \sigma \frac{32}{3} \pi R^3 g - \rho \frac{32}{3} \pi R^3 g - 6 \pi \eta (2R) v_2 = 0 $$

5. Solving for Terminal Velocities

Rearranging the equations for terminal velocities:

For Sphere 1: $$ v_1 = \frac{(\sigma - \rho) 2R^2 g}{9\eta} $$

For Sphere 2: $$ v_2 = \frac{(\sigma - \rho) 8R^2 g}{9\eta} $$

6. Finding Tension in the String

When both spheres reach terminal velocity, the tension $T$ in the string can be calculated based on the forces on one sphere:

Using Sphere 1: $$ T = F_{g1} - F_{b1} - F_{d1} $$

Substituting in the known values: $$ T = \sigma \frac{4}{3} \pi R^3 g - \rho \frac{4}{3} \pi R^3 g - 6 \pi \eta R v_1 $$

7. Final Substitution for Tension

Now, inserting the expression for $v_1$ derived earlier into the tension equation to find $T$ in terms of known quantities.