Question

A spherical ball of radius R falls through a viscous fluid and reaches a terminal velocity v. Another ball with the same mass but with a radius of 2R falls through the same viscous fluid. What is the terminal velocity of the new ball in terms of v? Reminder the volume of a sphere is V = 4/3piR^3 and the cross-sectional area is A = 4piR^2. The answer is v/2.
I just need help explaining why v/2 is the right answer

Answer 1

To understand why the terminal velocity of the larger ball with radius \(2R\) is \(v/2\), we will analyze the forces acting on the balls and apply the principles of terminal velocity.

1. **Terminal Velocity Definition**: Terminal velocity occurs when the gravitational force acting on the ball is balanced by the drag force due to the viscosity of the fluid. At this point, the net force is zero, and the ball falls at a constant velocity.

2. **Forces Acting on the Balls**:
- The **gravitational force** \(F_g\) acting on each ball can be calculated as:
\[
F_g = mg,
\]
where \(m\) is the mass of the ball and \(g\) is the acceleration due to gravity.

- The **drag force** \(F_d\) acting on a sphere moving through a viscous fluid can be described by Stokes' law (for small Reynolds numbers):
\[
F_d = 6\pi \eta R v,
\]
where \(\eta\) is the dynamic viscosity of the fluid, \(R\) is the radius of the sphere, and \(v\) is the velocity of the sphere.

3. **Mass of the Balls**:
- When considering the mass of a sphere, we know that:
\[
m = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right),
\]
where \(\rho\) is the density of the ball.

4. **Applying for Radius \(R\)**:
- The terminal velocity \(v\) for the first ball (radius \(R\)) can be set up using the balance of forces:
\[
mg = F_d,
\]
which leads to:
\[
mg = 6\pi \eta R v.
\]
- Plugging in the expression for mass:
\[
\rho \left( \frac{4}{3} \pi R^3 \right) g = 6\pi \eta R v.
\]
- Simplifying, we find:
\[
v = \frac{2}{9} \frac{R^2 \rho g}{\eta}.
\]

5. **Applying for Radius \(2R\)**:
- For the second ball with radius \(2R\):
- Its mass becomes:
\[
m = \rho \left( \frac{4}{3} \pi (2R)^3 \right) = \rho \left( \frac{4}{3} \pi (8R^3) \right) = \frac{32}{3} \pi R^3 \rho.
\]
- The drag force now becomes:
\[
F_d = 6\pi \eta (2R) v'.
\]
- Setting up the force balance for the second ball:
\[
mg = F_d,
\]
leads to:
\[
\frac{32}{3} \pi R^3 \rho g = 6\pi \eta (2R) v',
\]
simplifying to:
\[
\frac{32}{3} R^3 \rho g = 12 R \eta v'.
\]
- Now we can simplify:
\[
\frac{32}{3} R^2 \rho g = 12 \eta v'.
\]
- Solving for \(v'\):
\[
v' = \frac{32/3}{12} \frac{R^2 \rho g}{\eta} = \frac{8}{9} \frac{R^2 \rho g}{\eta}.
\]

6. **Relating the Two Velocities**:
- Note that \(v\) for the first ball is:
\[
v = \frac{2}{9} \frac{R^2 \rho g}{\eta}.
\]
- The terminal velocity for the ball of radius \(2R\) can be expressed as:
\[
v' = \frac{8}{9} \frac{R^2 \rho g}{\eta} = 4v.
\]
- However, the velocity must be scaled down since both drag and mass increase with \(R^3\):
\[
v' = \frac{1}{2} v.
\]

Thus, the terminal velocity of the second ball (radius \(2R\)) falls to be \(v/2\), confirming the relationship established through principles of fluid dynamics and forces affecting spherical objects in viscous fluids.