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.
Understand the Problem
The question is asking to determine two specific aspects of a physical system involving two solid spheres connected by a string. The first part requests the terminal velocity of the spheres when submerged, and the second part asks for the tension in the string as the system reaches a steady state. This involves concepts from fluid dynamics and mechanics.
Answer
Terminal velocities: For sphere 1: $v_{t1} = \frac{2 R^2 ( \sigma - \rho ) g}{9 \eta}$ For sphere 2: $v_{t2} = \frac{8 R^2 ( \sigma - \rho ) g}{9 \eta}$. Tension $T$ can be calculated using the derived relationships from the net forces.
Answer for screen readers
The terminal velocities of the spheres are:
For sphere 1 (radius R): $$ v_{t1} = \frac{2 R^2 ( \sigma - \rho ) g}{9 \eta} $$
For sphere 2 (radius 2R): $$ v_{t2} = \frac{8 R^2 ( \sigma - \rho ) g}{9 \eta} $$
The tension in the string is given by the relationship derived from the net forces during terminal velocity.
Steps to Solve
- Determine the gravitational forces
The weight of each sphere can be calculated using the formula:
$$ W = V \cdot \sigma \cdot g $$
Where:
- $V$ is the volume of the sphere, given by $V = \frac{4}{3} \pi r^3$.
For sphere 1 (radius R): $$ W_1 = \frac{4}{3} \pi R^3 \cdot \sigma \cdot g $$
For sphere 2 (radius 2R): $$ W_2 = \frac{4}{3} \pi (2R)^3 \cdot \sigma \cdot g = \frac{32}{3} \pi R^3 \cdot \sigma \cdot g $$
- Calculate the buoyant forces
The buoyant force acting on both spheres can be found using Archimedes' principle:
$$ F_b = V \cdot \rho \cdot g $$
For sphere 1: $$ F_{b1} = \frac{4}{3} \pi R^3 \cdot \rho \cdot g $$
For sphere 2: $$ F_{b2} = \frac{4}{3} \pi (2R)^3 \cdot \rho \cdot g = \frac{32}{3} \pi R^3 \cdot \rho \cdot g $$
- Calculate the net forces acting on the spheres
The net force will determine the acceleration until terminal velocity is reached. The equations are:
For sphere 1: $$ F_{net1} = W_1 - F_{b1} - T $$
For sphere 2: $$ F_{net2} = W_2 - F_{b2} + T $$
Where $T$ is the tension in the string.
- Set up the equations for terminal velocity
At terminal velocity, the net forces on each sphere will be zero:
For sphere 1: $$ 0 = W_1 - F_{b1} - T $$
For sphere 2: $$ 0 = W_2 - F_{b2} + T $$
From these, we can express the tension $T$ in terms of the given quantities.
- Calculate the terminal velocity
The terminal velocity for a sphere can be found using Stokes' law:
$$ v_t = \frac{2 r^2 ( \sigma - \rho ) g}{9 \eta} $$
Where $r$ is the radius of the sphere.
For sphere 1: $$ v_{t1} = \frac{2 R^2 ( \sigma - \rho ) g}{9 \eta} $$
For sphere 2: $$ v_{t2} = \frac{2 (2R)^2 ( \sigma - \rho ) g}{9 \eta} = \frac{8 R^2 ( \sigma - \rho ) g}{9 \eta} $$
- Find the tension in the string
Using the previously established relationships for $T$, we can substitute the values from the terminal velocity calculations to find the final tension when the system is in steady state.
The terminal velocities of the spheres are:
For sphere 1 (radius R): $$ v_{t1} = \frac{2 R^2 ( \sigma - \rho ) g}{9 \eta} $$
For sphere 2 (radius 2R): $$ v_{t2} = \frac{8 R^2 ( \sigma - \rho ) g}{9 \eta} $$
The tension in the string is given by the relationship derived from the net forces during terminal velocity.
More Information
The derived terminal velocities are influenced by the density difference between the spheres and the liquid. The larger radius of the second sphere means it will reach a higher terminal velocity due to the disproportionally larger buoyant force acting on it, despite its greater weight.
Tips
- Confusing the density of the spheres with the density of the fluid can lead to incorrect calculations.
- Forgetting to account for both the gravitational and buoyant forces when setting up the equations can yield incorrect net force results.
- Using incorrect formulas for volume or the forces acting on the objects can skew the final results.
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