Question

A \(1.0-g\) balloon is filled with helium gas. When a mass of \(4.0 \mathrm{~g}\) is attached to the balloon, the combined mass hangs in static equilibrium in midair. Assuming that the balloon is spherical, what is its diameter?

Short answer

Answer: The diameter of the helium-filled balloon is 16.4 cm.

Step-by-step solution

Determine the buoyant force

We know that the balloon and the attached mass are in static equilibrium, which means that the gravitational force acting on them is equal to the buoyant force acting on them due to the displaced air. The combined mass of the balloon and the attached mass is 1.0 g + 4.0 g = 5.0 g. Convert this mass to kilograms: 5.0 g = 0.005 kg. The gravitational force acting on them is given by weight (W) = mass (m) × gravity (g): W = 0.005 kg × 9.81 m/s^2 = 0.04905 N. This is the buoyant force as well.

Find the volume of the helium

Now that we have the buoyant force acting on the balloon, we can find the volume of the helium being displaced. Buoyant force (F) = mass of the displaced air (m) × gravity (g). We know that mass = density (ρ) × volume (V), so F = ρ_air × V × g. We want to find the volume (V). Rearrange the formula to solve for V: V = F / (ρ_air × g). The density of air is approximately 1.2 kg/m^3, and the buoyant force we found earlier is 0.04905 N. So, V = 0.04905 N / (1.2 kg/m^3 × 9.81 m/s^2) = 0.00416 m^3.

Calculate the diameter of the balloon

We have the volume of the helium. Now we can use the formula for the volume of a sphere to find the diameter of the balloon. The volume of a sphere (V) is given by V = (4/3) × π × (d/2)^3, where d is the diameter. Rearrange the formula to solve for the diameter: d = 2 × (3V / (4π))^1/3. Substitute the volume of the helium that we calculated earlier, V = 0.00416 m^3: d = 2 × (3 × 0.00416 m^3 / (4π))^1/3 = 0.164 m. Therefore, the diameter of the balloon is 0.164 m, or 16.4 cm.

Key concepts

Understanding Static Equilibrium

Static equilibrium is a fundamental concept in physics that occurs when an object is not moving or rotating. This means all the forces and moments (torques) acting on the object are balanced. In the case of the helium balloon exercise, the balloon, when in static equilibrium, has the buoyant force perfectly balancing the gravitational force acting down due to its own weight and the weight of the attached mass.

The condition for static equilibrium is mathematically expressed as the sum of all forces equals zero (\( \textbf{F}_{total} = 0 \) ), where the buoyant force is equal in magnitude but opposite in direction to the weight of the balloon and the additional mass. Similarly, for an object to remain in static equilibrium, the sum of all torques must also equal zero (\( \textbf{τ}_{total} = 0 \) ), ensuring the object does not start to spin.

The Physics of Helium Balloons

Helium balloons float because helium is less dense than air. When a balloon is filled with helium, it displaces a volume of air that weighs more than the helium inside the balloon. This creates an upward buoyant force. Applied physics tells us that if the buoyant force is greater than the combined weight of the balloon and any load it carries, the balloon will ascend until reaching a point in the atmosphere where the surrounding air is as dense as the helium.

In the helium balloon exercise, the system reaches static equilibrium when the upward buoyant force exactly offsets the downward gravitational force on the system. One of the remarkable properties of helium gas responsible for this phenomenon is its low density (approximately 0.1785 kg/m³) compared to the density of air (approximately 1.2 kg/m³). This significant difference in density is what allows helium balloons to float and be used effectively for lifting.

Calculating Spherical Volume

Formula and Calculation

A spherical object's volume can be calculated using the formula \( V = \frac{4}{3} \pi \left( \frac{d}{2} \right)^3 \), where \(V\) is the volume and \(d\) is the diameter of the sphere. The diameter is twice the radius, so we often see the formula written with radius as \( V = \frac{4}{3} \pi r^3 \).

To find the diameter from a known volume, as in the helium balloon exercise, the formula needs rearranging: \(d = 2 \times \left( \frac{3V}{4\pi} \right)^{\frac{1}{3}} \).

Practical Use in the Exercise

In our particular problem, determining the volume of helium gas using buoyant force allows us to solve for the diameter of the balloon using this spherical volume formula. This step is crucial in linking the balloon’s volume with its physical dimensions, which in turn can be used to calculate or predict many other properties and behaviors of spherical objects or systems.