Loss in energy of a ball dropped in a fluid

[Titan97]

Homework Statement

Two identical balls A and B are dropped from rest into a tall column of viscous liquid. Ball A is dropped from the surface of the liquid while ball B is dropped from a small height $h$ above the liquid. Let $E_A$ and $E_B$ be the loss in mechanical energy of bal A and B untillnterminal velocity is reached. Is $E_A>E_B$?

Homework Equations

None

The Attempt at a Solution

The only force that can reduce the total energy of the ball is the viscous force since it's non conservative.

Since ball B is dropped at some height above the liquid, I think it will travel more distance inside the liquid and work done by viscous force will be more. Hence $E_B>E_A$.

Is this correct?

 

[drvrm]

Two identical balls A and B are dropped from rest into a tall column of viscous liquid. Ball A is dropped from the surface of the liquid while ball B is dropped from a small height hhh above the liquid. Let EAEAE_A and EBEBE_B be the loss in mechanical energy of bal A and B untillnterminal velocity is reached. Is EA>EBEA>EBE_A>E_B?

Homework Equations

None

The Attempt at a Solution

The only force that can reduce the total energy of the is the viscous force since it's non conservative.

Since ball B is dropped at some height above the liquid, I think it will travel more distance inside the liquid and work done by viscous force will be more. Hence EB>EAEB>EAE_B>E_A.

Is this correct?

i can say that the energy loss of ball B will be greater as it started with higher kinetic energy and finally they are having same kinetic energy.

how can you say that the ball B has traveled larger distance than A... one has to do some calculation and check
as you are having velocity dependent resistive force and a larger velocity may not mean larger displacement!

 

[Titan97]

@drvrm let $\sigma$ be density of ball and $\rho$ be the density of liquid.
$$F=6\pi\eta Rv+V\rho g-V\sigma g=-m\frac{dv}{dt}$$
$$6\pi\eta Rv+V(\rho-\sigma)g=-mv\frac{dv}{dx}$$
$$\frac{vdv}{6\pi\eta Rv+V(\rho-\sigma)g}=\frac{-dx}{m}$$
Let $v$ be the terminal velocity and $u$ be the initial velocity. Let $s$ be the total distance traveled till terminal velocity.
$$(v-u)+6\pi\eta R\ln\left(\frac{v-\frac{V(\sigma-\rho)g}{6\pi\eta R}}{u-\frac{V(\sigma-\rho)g}{6\pi\eta R}}\right)=-\frac{6\pi\eta Rs}{m}$$

 

[Titan97]

@drvrm see the edited post.

 

[drvrm]

well, i was thinking on other lines;

the gravitational and buyancy forces are same for the two balls.
the viscous drag is say k.v... so if something starts with higher velocity will be facing a larger resistive force therefore its displacement will be less in a time slot...

 

[Titan97]

If force is more on B, then work done on B will also increase.

 

[drvrm]

If force is more on B, then work done on B will also increase.

this i agree so the energy loss will be more.
i was just thinking aloud logically if B can travel a larger distance to attain v-terminal than A

 

[Titan97]

But the answer given says $E_A>E_B$

 

[drvrm]

But the answer given says EA>EBEA>EBE_A>E_B

is there an experimental check of the answer? if its experimentally verified i.e. the answer than perhaps potential energy considerations may have been taken into consideration- that's a guess!

 

[drvrm]

But the answer given says EA>EBEA>EBE_A>E_B

actually the body B which fell with initial speed will travel a distance smaller than the the body A to to attain terminal velocity and that's why the loss of energy can/ will be larger for the body A than body B .

Actually the viscous drag is of the type F =- kv and in such condition the speed - time curve takes an exponential shape .
speed(t) = constant(V-terminal) x (1- exp (-K1. time)) K1 can be related to (massxeffective acceleration)/6.pi. eta.radius

the body B enters the liquid with initial velocity therefore faces larger resistance and its distance traversed ,which can be estimated from the area under speed time curve till it attains terminal velocity will be less than bodyA.
The loss in kinetic energy will be adjusted by the change in potential energy and net energy loss will be less than A. Though detail graph drawing has to be done.

just see
http://oregonstate.edu/instruct/mth252h/Bogley/w02/resist.html for a similar exercise of a body falling under air resistance.

 

[Titan97]

@haruspex does loss in mechanical energy mean $\Delta KE$ or the change in the total energy including the potential energy?

 

[haruspex]

@haruspex does loss in mechanical energy mean $\Delta KE$ or the change in the total energy including the potential energy?

It's the sum. https://en.m.wikipedia.org/wiki/Mechanical_energy

 

[Titan97]

The loss in energy for A is $-mgy_A+\frac{1}{2}mv^2$
The loss in energy for B is $\frac{1}{2}mu^2-mgy_B+\frac{1}{2}mv^2$

 

[haruspex]

The loss in energy for A is $-mgy_A+\frac{1}{2}mv^2$
The loss in energy for B is $\frac{1}{2}mu^2-mgy_B+\frac{1}{2}mv^2$

I think you have some signs wrong there. But I see a flaw in my earlier post... reworking it.

 

[haruspex]

A key statement here is that B is dropped from a small height h. This allows us to assume that the velocity on hitting the water is still less than the terminal velocity. Thus, B has a head start on attaining terminal velocity, and will therefore do so in less distance.
That shows B loses less GPE after hitting the water, but mechanical energy includes that impact KE (or, equivalently, the GPE lost in reaching the water).

Consider this: suppose A reaches terminal velocity at depth d. If B hits the water at speed u, there is some depth x at which A reached speed u. How does x compare with h? What does this tell us about total GPE lost from their respective initial positions?