Consider a rectangular block of wood moving with a velocity v0 in a gas at temperature T and mass density ρ. Assume the velocity is along
Consider a rectangular block of wood moving with a velocity v0 in a gas at temperature T and mass density ρ. Assume the velocity is along x-axis and the area of cross-section of the block perpendicular to v0 is A. Show that the drag force on the block is 4ρAv0√(kT/m), where m is the mass of the gas molecule.
1 Answers
n = no. of molecules per unit volume
vrms = rms speed of gas molecules
When block is moving with speed vo, relative speed of molecules w.r.t.
front face = v + vo
Coming head on, momentum transferred to block per collission
= 2 m (v + vo), where m = mass of molecule.
No. of collission in time Δt = 1/2(v + vo)nΔtA, where A = area of cross section of block and factor of 1/2 appears due to particles moving towards block.
Momentum transferred in time Δt = m(v + vo)2AΔt from front surface
Similarly momentum transferred in time Δt = m(v -vo)2nAΔt from back surface
Net force (drag force) = mnA [(v + vo) - (v - vo)2] from front
= mnA (4vvo) = (4mnAv)vo)
= (4ρAv)vo
We also have (1/2)mv2 = (1/2)kT (v - is the velocity along c-axis)
Therefore, v = √kT/m
Thus drag = 4ρA√(kT/m)vo