IIT JEE , AIEEE Forum - Physics , Mechanics

Mr.IITIAN007

A sphere of mass m and radius r is projected in a viscous medium of coefficient of viscosity n (eta) with a speed v.Find the distance travelled and time taken before it stops when it is projected in :-
i)gravity free space;
ii)presence of gravity

cvramana

Case 1: work done by viscous force = change in kinetic energy.

Case 2: work done by viscous force + work done by gravity = change in kinetic energy.

Mr.IITIAN007

Ya I know that sir but I want the final answers.That is why I have posted it here.

Aatish

Respected experts pls answer this question with detailed solution Plssssssssssssssss

Thankxxxxxxxx

Aatish

Actually we tried for the answers but could not get satisfied with our answers thats why we have posted it here.......

cvramana

Consider it to be projected vertically upward. Then the forces acting on it or mg + viscous force.

Work done by gravity = - mgh

work done by viscous force = -

6

r v ds

Consider forces alone, then (mg + 6

r v) = - m dv / dt, find v versus t relation and then put v = ds / dt and find s versus t relation. Then do the work energy relation.

thevyzz

consider the ball to be projected in two dimensions

when there is no gravity ,

dv/dt = acc=( 6pi *eta* a* v) / m

intergrating and solving

we get

t = (log V * m) / (6* pi* eta* a)

for distance use the formula s = Vt - ((6pi* eta* a* V) / m)*t² / 2

thevyzz

in the above case, i've taken a as the radius ( tradition in TN state board :) )

since there is no gravity the ball willmove only in the direction in which it is projected and will stop at some point and sorry for taking a as the radius

thevyzz

anyone got it for under gravity , seems really tough, soln is running more than 2 pages and i'm still nowhere

somebody help

itz easy for vertical projection , buts damn diffficult for angular projection

pcdagr8

ok in presence of gravity lets resolve the velocity
the horizontal component has no force supporting but viscosity aginst it so viscosity decreses it to 0 after some time which is similar to the case of absence of gravity
in the vertical direction
theball first goes up due to its velocity
Weight-bouyantforce is acting don and viscosity down this slows down the ball and decreses velocity to 0
now W-B is acting down and viscosity is zero as v=0
as ball goes down w-b remains constant but as v increases viscosity in upward direction increases
now as w-b equals viscosity
the ball travels down with terminal velocity
no force is present so......it will jus go on moving..infinitely
time is infinite so is distance...
until theres some dimensions of container given..n u want time till it hits the walls

thevyzz

now if there is gravity, oh boy! its getting interesting

we need to split up into two dimentsions and

lets see the horizontal component

the ball is projected with angle Z ( Sorry, cant use the formula editor in ff)

initial horizonatal velocity = v cos Z

acceleration acting in horizonatal direction = (6 pi eta a v cosZ/m)

initial vertical velocity = v sin Z

first of all we need to find the time period, remember that the time periods will be different for different components of motion ( t is the time for the velocity in that dimension to become zero)

t for horizonatal component ( done by intergrating and solving)

t for horizonatal component = log(V)*m/ (6*pi*eta*a)

in the below eq, t need not neccessarily be the time taken for horizontal component, as it may also be the time for vertical component ( whichever is the smallest must be taken as t)

distance travelled = s = V*cosZ* t - ((6pi* eta* a* V *cosZ) / m)*t2 / 2

till now we have ignored buoyancy as there is no gravity component

still gotta find time taken, we cannot ignore buoyancy here

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