91Ó°ÊÓ

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 \(\dot{V}\) is \(\rho V a\) . (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 \(a_{n e 1}=\left[\left(\rho / \rho_{\text { bel }}\right)-1\right] a\) (f) Use the expression for \(a_{\mathrm{rel}}\) in part (e) to explain the movement of the balloons.

Step by step solution

01

Apply Newton's Second Law to the Volume Element

Consider a small air element in the tube with thickness \(dx\). The difference in force due to pressure is given by \((p + dp)A - pA = A dp\). According to Newton's Second Law, this difference equals the mass of the element times its acceleration, \( \rho A dx \cdot a \). So, \(A dp = \rho A dx \cdot a\). Simplifying, we get \( dp = \rho a dx \).

02

Integrate to Find Pressure at Front Surface

For the horizontal tube extending back from the windshield, integrate the expression \(dp = \rho a dx\) from \(0\) to \(x\) to find pressure as a function of \(x\). The integral gives \(p(x) - p_0 = \int_0^x \rho a dx = \rho a x\). Hence, \(p(x) = p_0 + \rho a x\).

03

Calculate Pressure Difference for Given Conditions

Using the formula \(p(x) = p_0 + \rho a x\) for a distance of \(2.5 \text{ m}\) and acceleration \(5.0 \text{ m/s}^2\), calculate the pressure difference. Assuming \(\rho = 1.225 \text{ kg/m}^3\) (density of air), the difference is \(\Delta p = \rho a x = 1.225 \times 5 \times 2.5 = 15.3125 \text{ Pa}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Pressure Difference

The concept of pressure difference is crucial when examining the dynamics within a moving car with a sealed environment. Pressure difference refers to the variation in pressure exerted between two points. In this scenario, the car's acceleration affects the air pressure across its interior. This concept is applied to a tube of air that stretches from the windshield to the back of the car.

The pressure on the rear end of a small volume element in the tube is slightly higher than the pressure on the front end. This results from the car's constant acceleration. According to Newton's Second Law, the change in pressure (\( dp \)) is proportional to the density of air (\( \rho \)), the acceleration (\( a \)) and the thickness of the air element (\( dx \)). This relationship is illustrated by the equation:\[ dp = \rho a \, dx \]Understanding this relationship helps us perceive how the car's motion changes the air pressure within, creating a measurable pressure difference.

Balloon Dynamics in Motion

The intriguing behavior of balloons in a moving car boils down to balloon dynamics. When the car accelerates, helium balloons move in the direction of the car’s acceleration, whilst air-filled balloons move in the opposite direction. This is due to the different densities of helium and air.

Since helium is less dense than the surrounding air, helium balloons move to areas of lower pressure. In a car accelerating forward, the region closer to the windshield experiences lower pressure due to the effects of acceleration on the air distribution.
On the other hand, air-filled balloons have the same density as the surrounding air but are still heavier than the surrounding air and move backwards towards areas of higher pressure when the car accelerates. This creates an apparent movement opposite to the car's acceleration for air-filled balloons.

Role of Acceleration

Acceleration is a measure of how quickly the velocity of an object changes. In this scenario, it plays a fundamental role in determining how forces are distributed within the car. Newton's Second Law relates force, mass, and acceleration, forming the basis for analyzing the situation.

When the car accelerates at a rate \( a \), it causes a shift in the internal pressure distribution. The equation \( dp = \rho a \, dx \) illustrates how acceleration directly contributes to the pressure difference along the car. The whole system, including balloons, responds to this acceleration-induced pressure difference.
Moreover, the acceleration defines the net force experienced by any object inside the car. A balloon, for instance, when not restrained, will have an effective acceleration (\( a_{rel} \)) relative to the car that depends on its density compared to the air density, as derived from the related equations. This explains how objects with different densities move differently in response to the car's acceleration.