Untethered helium balloons, floating in a car that has all

Chapter 12, Problem 12.88

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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 dx in this tube. The pressure on its front surface is p and the pressure on its rear surface is p + dp. Assume the air has a constant density \(\rho\). (a) Apply Newton’s second law to the volume element to show that dp = ra dx. (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 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 \(\rho V a\). (e) For negligible friction forces, show that the acceleration of the balloon (average density \(\rho_{\text {bal }}\)) is \(\left(\rho / \rho_{\mathrm{bal}}\right) a\), so that the acceleration relative to the car is \(a_{\mathrm{rel}}=\left[\left(\rho / \rho_{\mathrm{bal}}\right)-1\right] a\). (f) Use the expression for \(a_{\mathrm{rel}}\) in part (e) to explain the movement of the balloons.

Text Transcription:

p = p_0

rho

5.0 m/s^2

rho V a

rho_bal

(rho/rho_bal)a

a_rel = [(rho/rho_/bal)-1]a

a_rel