In an elastic balloon the relation between the internal excess pressure and the tension of the wall can be readily calculated if we assume that the balloon is spherical and that the material is homogeneous and of negligible weight. If we suppose the balloon divided into two hemispheres by a plane horizontal partition, the area of this partition will be πr2 and the downward force on the upper surface due to the excess pressure p will be πr2p. The balloon wall meets the partition at right angles along a length 2πr. Hence if T is the tension in the wall, the upward force exerted by this tension on the partition is 2πrT. But as these two forces must be equal we have πr2p = 2πTr, so that p = 2 T/r. (1) When such a balloon is filled without stretching the wall the pressure inside is equal to the prevailing atmospheric, and the radius r0 may be termed the initial radius. If we assume that the balloon is perfectly obedient to Hooke's law, then T1 = K (r1 — r0)/r0; but from (1) we learn that p1 = 2 T1/r1; hence, by substitution, p1 = 2 K/r0—2 K/r1, or r1 (2K/r0 — p1) = 2 K. (2)
No takes yet. Share an insight, caveat, or question.
Osborne et al. (1909) studied this question.