A sphere of radius 5.2 mm is moving vertically through air at 8.3 m/s. The viscosity of air is 1.8 x 10^-5 Pas. Calculate the instantaneous acceleration produced on the sphere if its mass was 4.8g.

Answer 1

To calculate the instantaneous acceleration of the sphere, we use Stokes' Law to determine the drag force acting on the sphere and then apply Newton’s Second Law to find the acceleration.

Given Data:

1. Radius of the sphere, r = 5.2 mm = 5.2 × 10³ m
2. Velocity of the sphere, v = 8.3 m/s
3. Viscosity of air, η = 1.8 × 10⁻⁵ Pa·s
4. Mass of the sphere, m = 4.8 g = 4.8 × 10⁻³ kg

Step 1: Calculate Drag Force using Stokes’ Law

Stokes' Law states that the drag force on a sphere moving through a viscous fluid is given by:

F_d = 6πηrv

Substituting the values:

F_d = 6 × (1.8 × 10⁻¹) × (5.2 × 10⁻³) × (8.3)

F_d = 4.6596 × 10³ N

Step 2: Apply Newton’s Second Law

The acceleration is given by Newton’s Second Law:

a = F_d / m

Substituting values:

a = 4.6596 × 10⁻³ / (4.8 × 10⁻³)

a = 0.971 m/s²

Final Answer:

The instantaneous acceleration produced on the sphere is 0.971 m/s2.

Answer 2

Solution and Explanation

We need to calculate the instantaneous acceleration of a sphere moving through the air using StokesLaw, which describes the drag force acting on a small sphere moving through a viscous fluid.

Step 1: Given Data

Radius of the sphere = 5.2 mm = 5.2 x 10^-3 m

Velocity of the sphere = 8.3 m/s

Viscosity of air = 1.8 x 10^-5 Pa·s

Mass of the sphere = 4.8 g = 4.8 x 10^-3 kg

Step 2: Apply Stokes Law for Drag Force

Stokes Law states that the drag force Fd acting on a small sphere moving through a viscous fluid is given by:

Fd = 6 * pi * viscosity * radius * velocity

where:

viscosity = 1.8 x 10^-5 Pa·s

radius = 5.2 x 10^-3 m

velocity = 8.3 m/s

pi is approximately 3.1416

Substituting the values:

Fd = 6 * 3.1416 * (1.8 x 10^-5) * (5.2 x 10^-3) * 8.3

Fd is approximately 1.4644 x 10^-5 N

Step 3: Apply Newton’s Second Law

Newton’s Second Law states:

F = ma

Solving for acceleration:

a = F / m

Substituting the values:

a = (1.4644 x 10^-5) / (4.8 x 10^-3)

a is approximately 0.00305 m/s^2

Final Answer:

The instantaneous acceleration of the sphere due to the drag force is 0.00305 m/s^2.

Explanation

The acceleration is very small because the drag force exerted by the air on the sphere is relatively weak compared to its mass. This is expected in high-speed motion through a low-viscosity fluid like air. Over time, this drag force will slow the sphere down, reducing its velocity until equilibrium is reached, known as terminal velocity.