Problem 88 - University Physics with Modern Physics

Question

Untethered helium balloons, floating in a car that has all the windows rolled up and outside air vents closed, move in the direction of the car's acceleration, but loose balloons filled with air move in the opposite direction. To show why, consider only the horizontal forces acting on the balloons. Let \(a\) be the magnitude of the car's forward acceleration. Consider a horizontal tube of air with a cross-sectional area \(A\) that extends from the windshield, where \(x=0\) and \(p=p_{0},\) back along the \(x\) -axis. Now consider a volume element of thickness \(d x\) in this tube. The pressure on its front surface is \(p\) and the pressure on its rear surface is \(p+d p .\) Assume the air has a constant density \(\rho .\) (a) Apply Newton's second law to the volume element to show that \(d p=\rho a d x .\) (b) Integrate the result of part (a) to find the pressure at the front surface in terms of \(a\) and \(x\) . (c) To show that considering \(\rho\) constant is reasonable, calculate the pressure difference in atm for a distance as long as 2.5 \(\mathrm{m}\) and a large acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2}\) . (d) Show that the net horizontal force on a balloon of volume \(V\) is oVa. (e) For negligible friction forces, show that the acceleration of the balloon (average density \(\rho_{\text { bal }}\) is \(\left(\rho / \rho_{\text { bal }}\right) a,\) so that the acceleration relative to the car is \(a_{\mathrm{rcl}}=\left[\left(\rho / \rho_{\mathrm{bal}}\right)-1\right] a\) (f) Use the expression for \(a_{\mathrm{rcl}}\) in part (e) to explain the movement of the balloons.

Step-by-Step Solution

Answer

Helium balloons move forward due to positive relative acceleration, while air balloons don't and seem to move backward.

Step 1: Set up Newton's Second Law for the Volume Element

Consider an air volume element of thickness \(dx\) within the horizontal tube in the car. The force on the front face is \(-p \cdot A\) and at the back face is \(-(p + dp) \cdot A\). Apply Newton's second law: \( F = ma = \rho A dx \cdot a\). Solve for \(dp\): \(-dp \cdot A = \rho A dx \cdot a \, \Rightarrow dp = -\rho a dx \).

Step 2: Integrate dp to Find the Pressure

From step 1, we have \(dp = -\rho a dx\). Integrate both sides with respect to \(x\) from 0 to \(x\), considering the pressure changes from \(p_0\) to \(p(x)\): \[\int_{p_0}^{p(x)} dp = \int_{0}^{x} -\rho a dx\] resulting in \(p(x) - p_0 = -\rho ax\). Therefore, \(p(x) = p_0 - \rho ax\).

Step 3: Calculate the Pressure Difference Over 2.5 m

The pressure difference \(\Delta p\) over a distance \(L = 2.5 \text{ m}\) and acceleration of \(a = 5 \text{ m/s}^2\) is \(\Delta p = \rho a L\). Using typical air density \(\rho = 1.2 \text{ kg/m}^3\): \[\Delta p = 1.2 \cdot 5 \cdot 2.5 = 15 \text{ Pa}\]. Converting to atm: \(1 \text{ atm} = 101325 \text{ Pa}\), \(\Delta p \approx 0.00015 \text{ atm}\).

Step 4: Determine Net Force on the Balloon

The net horizontal force on a helium balloon of volume \(V\) is due to pressure difference: \(F_{net} = V \cdot \Delta p = V \rho a\). This gives the horizontal force as \(F_{net} = V \rho a\).

Step 5: Find the Acceleration of the Balloon

Consider the balloon's mass \(m = \rho_{\text{bal}} V\). Newton's second law gives \(ma_{\text{balloon}} = V \rho a\): \[\rho_{\text{bal}} V \cdot a_{\text{balloon}} = V \rho a\] simplifying to \(a_{\text{balloon}} = \left(\frac{\rho}{\rho_{\text{bal}}}\right)a\).

Step 6: Relative Acceleration of Balloon to the Car

The acceleration of the balloon relative to the car is \(a_{rcl} = a_{\text{balloon}} - a = \left(\frac{\rho}{\rho_{\text{bal}}} - 1\right)a\).

Step 7: Explain Balloon Movement

For helium, \(\rho_{\text{bal}}\) is less than \(\rho\), so \(a_{rcl}\) is positive, meaning it moves forward. For an air-filled balloon, \(\rho_{\text{bal}} = \rho\) giving \(a_{rcl} = 0\); thus, it appears to move backward as the car accelerates forward.

Key Concepts

Pressure Gradient in Fluids

When a car accelerates forward, the air inside creates a pressure gradient along its length. Pressure gradients are differences in pressure across a particular distance in a fluid. This concept is crucial in understanding how forces act within the system. In our scenario, the pressure at the front end, such as the windshield, is higher than at the back due to the car's acceleration. This creates a horizontal pressure gradient.

Applying Newton's Second Law, this gradient causes a net force on a fluid element.

• The pressure difference (\( dp \)) between two points is proportional to the acceleration \( a \) and the distance \( dx \), so \( dp = \rho a dx \).
• Here, \( \rho \) is the fluid's density.

In essence, the faster the car accelerates, the larger the pressure gradient within the fluid. Integrating this change gives us the overall pressure difference along any length, allowing us to calculate forces acting on objects within this fluid system.

Balloon Dynamics

Balloon dynamics in a moving car involve understanding how per unit pressure affects the movement of either helium or air-filled balloons. As the car moves, each type of balloon experiences different forces due to the surrounding air pressure and the fluid's density.

• For a helium balloon, it is less dense than the surrounding air. As the car accelerates, the pressure gradient drives the lighter helium balloon forward.
• In contrast, an air-filled balloon, having the same density as the surrounding air, doesn't experience this differing force and seems to move backward as the car moves forward.

This dynamic is a practical application of the principle of buoyancy, where the distribution of forces across a density gradient determines the direction of motion.

Acceleration in a Non-Inertial Frame

When discussing acceleration in a non-inertial frame, like inside a car, it's important to understand relative motion. A non-inertial frame is a reference frame that is accelerating. Here, the car itself is the non-inertial frame. The acceleration that we measure inside the car, such as that of balloons, differs from the actual external acceleration due to the car's movement.

• For a helium balloon, the acceleration it experiences relative to the car (\( a_{rcl} \)) can be calculated as \( a_{rcl} = \left(\frac{\rho}{\rho_{\text{bal}}} - 1\right)a \)
• This formula shows how much faster or slower the balloon appears to move in relation to the car's acceleration.

Understanding this helps us predict and explain why certain objects behave the way they do in accelerating frames, offering valuable insights into their apparent motion.

Physics of Gases

The physics of gases plays a foundational role in understanding how balloons behave under differing conditions of pressure and acceleration. Ideal gas laws, though simplified, offer insights into real gas behavior when pressure and temperature changes occur.

• Helium, being lighter, is easily displaced by air under pressure gradients.
• The density of gases affects how they respond to external forces such as those experienced by balloons in a moving vehicle.

The behavior of gases within moving systems illustrates how molecular motion and collisions can yield detectable effects like balloon displacement. This principle allows us to leverage the physics of gases for applications ranging from understanding everyday phenomena like balloon dynamics to designing systems in aerospace and engineering.