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    How is terminal velocity of a spherical particle proportional to (radius)^2?

    thanks a lot.
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    m\ddot{x}= mg - \lambda\dot{x}

    Is the equation for a particle falling under gravity. i.e. the force on the particle equals the force due to gravity minus the frictional drag.

    At terminal velocity, the force on the particle is zero as the velocity is constant. So we have:

     \dot{x}=\frac{m}{\lambda}g

    (((So the terminal velocity is inversely proportional to the frictional constant \lambda . This constant is proportional to the area (i.e. radius squared) because what is causing the friction is particles in the air bumping into the spherical particle as it falls downwards.

    The bigger the cross-sectional area of the spherical particle, the more frequently air particles collide with it and so the greater the frictional constant. Since terminal velocity is inversely proportional to the frictional constant, terminal velocity is inversely proportional to the area of the spherical particle. Ignore this it's wrong )))
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    (Original post by 3nTr0pY)
    m\ddot{x}= mg - \lambda\dot{x}

    Is the equation for a particle falling under gravity. i.e. the force on the particle equals the force due to gravity minus the frictional drag.

    At terminal velocity, the force on the particle is zero as the velocity is constant. So we have:

     \dot{x}=\frac{m}{\lambda}g

    So the terminal velocity is inversely proportional to the frictional constant \lambda . This constant is proportional to the area (i.e. radius squared) because what is causing the friction is particles in the air bumping into the spherical particle as it falls downwards.

    The bigger the cross-sectional area of the spherical particle, the more frequently air particles collide with it and so the greater the frictional constant. Since terminal velocity is inversely proportional to the frictional constant, terminal velocity is inversely proportional to the area of the spherical particle.
    Can you help give me a simpler explanation please
    Thanks
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    (Original post by krisshP)
    How is terminal velocity of a spherical particle proportional to (radius)^2?

    thanks a lot.
    Do you just want the formula for terminal velocity or do you want to see the derivation?

    It's derived from Stokes' Law which states that the viscous force on the sphere is given by
    F = 6 \pi \eta vr
    r is its radius, so the force depends on r
    v is the velocity of the sphere

    In a simple treatment you can say that at terminal velocity, this force is balanced by (equal to) the weight of the sphere. The weight of the sphere depends on its density and mass, and mass depends on r3

    So you finish up with an equation where on one side you have
    F = 6 \pi \eta vr
    and on the other a term with r3

    The r in the Stokes' formula cancels with one of the r in the r3 on the other side to give
    v = constant x r2

    If you want a fuller proof I can post it.

    The actual formula for terminal velocity is
    v_t=\frac{2r^2(\rho_s-\rho_l)g}{9\eta}
    ρs is density of sphere
    ρl is density of liquid
    η is viscosity of the liquid
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    (Original post by Stonebridge)
    Do you just want the formula for terminal velocity or do you want to see the derivation?

    It's derived from Stokes' Law which states that the viscous force on the sphere is given by
    F = 6 \pi \eta vr
    r is its radius, so the force depends on r
    v is the velocity of the sphere

    In a simple treatment you can say that at terminal velocity, this force is balanced by (equal to) the weight of the sphere. The weight of the sphere depends on its density and mass, and mass depends on r3

    So you finish up with an equation where on one side you have
    F = 6 \pi \eta vr
    and on the other a term with r3

    The r in the Stokes' formula cancels with one of the r in the r3 on the other side to give
    v = constant x r2

    If you want a fuller proof I can post it.

    The actual formula for terminal velocity is
    v_t=\frac{2r^2(\rho_s-\rho_l)g}{9\eta}
    ρs is density of sphere
    ρl is density of liquid
    η is viscosity of the liquid
    Well said. My post is wrong as I stupidly forgot m is proportional to r^3....duhhhh. And forgetting Stokes law...embarassing.

    Still, I've never found Stokes law intuitively obvious. To me a proportional dependence on r^2 for lambda seems natural. Shame my intuition is wrong! Is there an obvious reason why it should be r?
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    (Original post by 3nTr0pY)
    W

    Still, I've never found Stokes law intuitively obvious. To me a proportional dependence on r^2 for lambda seems natural. Shame my intuition is wrong! Is there an obvious reason why it should be r?
    I can't think of one. It obviously comes out in the maths but I agree that r2 feels more intuitive for the dependence of the force.
    I've never actually seen a rigorous proof of Stokes' formula, only the classic A-Level dimensional proof which obviously doesn't give the 6 Pi constant.
    (And of course all this only applies to lamina flow.)
    I'll have a search around on Google as it's got me thinking now...
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    (Original post by Stonebridge)
    I can't think of one. It obviously comes out in the maths but I agree that r2 feels more intuitive for the dependence of the force.
    I've never actually seen a rigorous proof of Stokes' formula, only the classic A-Level dimensional proof which obviously doesn't give the 6 Pi constant.
    (And of course all this only applies to lamina flow.)
    I'll have a search around on Google as it's got me thinking now...
    I saw a more rigorous one than that at Uni, although even then they still had to fudge the constant. (I think it gave something like 4*pi*viscosity*density instead of 6*pi*viscosity*density). Completely forgot the proof though.

    I'm thinking that perhaps the proportional to r^2 bit comes from the acceleration being proportional to v/r. So you get (v/r)*r^2 = vr...with the extra constants. But that's just a guess.
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    (Original post by Stonebridge)
    Do you just want the formula for terminal velocity or do you want to see the derivation?

    It's derived from Stokes' Law which states that the viscous force on the sphere is given by
    F = 6 \pi \eta vr
    r is its radius, so the force depends on r
    v is the velocity of the sphere

    In a simple treatment you can say that at terminal velocity, this force is balanced by (equal to) the weight of the sphere. The weight of the sphere depends on its density and mass, and mass depends on r3

    So you finish up with an equation where on one side you have
    F = 6 \pi \eta vr
    and on the other a term with r3

    The r in the Stokes' formula cancels with one of the r in the r3 on the other side to give
    v = constant x r2

    If you want a fuller proof I can post it.

    The actual formula for terminal velocity is
    v_t=\frac{2r^2(\rho_s-\rho_l)g}{9\eta}
    ρs is density of sphere
    ρl is density of liquid
    η is viscosity of the liquid
    Thanks
 
 
 
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