The physics of the chewing gum and helium experiment JosĂŠ Gonçalvesi In a YouTubeii video we see an experiment involving gum and helium gas, creating a balloon that can lift a person. Using the law of hydrostatics, the driving force, impulse I, must be at least equal (as can be seen in a certain part of the video) to the weight of the man with the balloon, W. So, W = I Replacing the weights, W, by m g and knowing that đ?‘š = đ?œŒđ?‘‰ , we have
4 3 đ?œ‹đ?‘… Ă— đ?œŒđ??ťđ?‘’ + đ?‘šđ?‘€ đ?‘” = đ?œŒđ??´đ?‘–đ?‘&#x; đ?‘‰đ??´đ?‘–đ?‘&#x; đ?‘” 3 so the mass of man, mM, will be 4 đ?‘šđ?‘€ = đ?œ‹đ?‘… 3 đ?œŒđ??´đ?‘–đ?‘&#x; − đ?œŒđ??ťđ?‘’ 3 Replacing the values of đ?œŒđ??´đ?‘–đ?‘&#x; (15ÂşC) = 1,225 kg m−3 and đ?œŒđ??ťđ?‘’ = 0,20 kg m−3 we obtain đ?‘šđ?‘€ = 4,294 đ?‘…3 That equation represents the dependency of the mass of the man with the ballon’s radius On the following chart of mM = mM(R)
mH / kg
mH / kg 140 120 100 80 60 40 20 0 0
0,5
1
1,5
2
2,5
3
3,5
R/m
We have, for a mass exceeding 65 kg, the diameter of the balloon will be greater than 5 m. Therefore, we can easily see that in the video is not the case. i
JosÊ Gonçalves, Physics teacher with Master Science degree in Physics Teaching. Email: josegoncalves@eufisica.com ii http://www.youtube.com/watch?v=pTnwz6MqMl8