1thing gone completely over my head.
viscous force may be anything dependent(ie whatever ure saying, height, velocity etc..)
but that still does not prove why this eqn
mg-buoyant force= drag force fails to hold.
1. Q.A container filled with viscous liquid is moving vertically downwards with constant speed 3vo. A sphere of radius r is moving vertically downwards (in liquid) has speed vo. The coefficient of viscosity is eta . There is no relative motion between the liquid and the container. Then the magnitude of viscous force acting on sphere is
ans is given as 12∩ (eta) rvo
my doubt that is that the way we derive Vt for non accelerated system is:
mg= Drag force + buoyant force
so, mg-buoyant force= drag force
and so unless any of mg or buoyant force changes(which wont unless system is accelerated which in this case is not)
why should the expression for drag force change???
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8 Answers
1Dunno exactly, its a guess. Viscous force is non-constant. When you read the derivation of viscous force, you come to know that it is height dependent.
Similarly here, viscous force is radius dependent. The velocity is that of centre of mass...so there will be a variation of viscous force moving either side of centre
Hence, the answer.
dunno if i am toking crap!!
11Or perhaps it is possible that in this case, it is not in equilibrium at all.
NOTE: nowhere does it mention that Vo is terminal
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ho sakta hai , viscous force at any moment of time is given as
6∩ (eta) r v where v is the velocity at that time and not the terminal velocity.
Correct me if I am wrong.
1i think its true
this wiki article seems to prove it
http://en.wikipedia.org/wiki/Stokes'_law