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Mechanics
27 Jan 2008 15:12:09 IST
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1946
The ultimate challenge in Mechanics from Feynmann
None
Two particles A & B start from positions ( 0, 0 ) & ( 0, -d ) and move with constant speeds v & u respectively . A moves along x - axis and B moves such that its velocity is always aimed at A . Let r be the distance between them and 
be the angle made by the velocity of B with X - axis , at some time t .
be the angle made by the velocity of B with X - axis , at some time t . Prove that ,
r ( 1 - cos )^ ( u/v )
)^ ( u/v ) --- = ------------------------------
d ( sin ) ^ ( u/v + 1 )
) ^ ( u/v + 1 )I have solved the problem & will publish it on next sunday .
Let me see ,by then , who among u can solve it !!!
All are welcome .
Comments (11)
BHUWAN ARORA
Scorching goIITian
Joined: 21 Sep 2007
Posts: 299
27 Jan 2008 17:44:03 IST
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the path of d particle will be such that at
any time tangent drawn to d curve wil pass
through d particle
let at any time t d coordinates of particle B
be b ai+bj
eq of d curve on which d particle is moving is
y=f(x)
eq. of tangent (x-a)f1(a)=(y-b)
this will satisfy (vt,0)
(vt-a)f1(a)=-b
f1(a)=-b/(vt-a)
also b=f(a)
r2=(vt-a)2+b2
f1(a)=tan
da/dt=ucos
d(d-b)/dt=usin
getting these equations
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30 Jan 2008 01:46:29 IST
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The problem is best solved in plane polar coordinates. (For an analysis of motion in plane polar coordinates please refer to "Fundamentals of Mechanics" by I E Irodov)
Velocity of B in the frame of A,
(dr/dt)er + (r(d
/dt)eth = [u - vcos]er - (- vsin)eth
(dr/dt) = u-vcos ------------------(1)
And
r(d/dt) = vsin ----------------(2)
Dividing (1) by (2) and rearranging,
(1/r)dr = (u/v)cosec d - cot d
d
r (1/r)dr = pi/2theta(u/v)cosec d - pi/2thetacot d
ln(r/d) = ln[(cosec - cot)u/v/sin]
r/d = (1 - cos)u/v]/(sin)(u/v)+1
Was quite an enjoyable problem ... can you tell me its source??
Velocity of B in the frame of A,
(dr/dt)er + (r(d
/dt)eth = [u - vcos]er - (- vsin)eth(dr/dt) = u-vcos
------------------(1)And
r(d
/dt) = vsin ----------------(2)Dividing (1) by (2) and rearranging,
(1/r)dr = (u/v)cosec
d - cot dd
r (1/r)dr = pi/2theta(u/v)cosec d - pi/2thetacot dln(r/d) = ln[(cosec
- cot)u/v/sin]r/d = (1 - cos
)u/v]/(sin)(u/v)+1Was quite an enjoyable problem ... can you tell me its source??
10 Feb 2008 18:45:20 IST
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Hey , I have solved Irodov three times !!!!!!!!!!!!!!!
Could u plz give me the problem no in Irodov where the same problem with the SAME ANS SOUGHT is posed ?
If u can ,hurry .
If u can't ,plz think twice before posting any comment .
Thank u !!!
11 Feb 2008 15:21:08 IST
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Anyway, I can provide u a solution in cartesian co-ordinates .
see, the controlling differential eqns are clearly.........
dx/dt = u cos ..................... ( 1 )
..................... ( 1 ) dy/dt = usin .......................( 2 )
.......................( 2 )y/( x - vt) = tan ....................... ( 3 )
....................... ( 3 )y = - r sin .................. ( 4 )
.................. ( 4 )Now the task is to solve these coupled differential eqn .
From ( 3 ) we have
x - vt = y cot
dwrt 't'
dx/dt -v = dy/dt cot - y cosec^2() d/dt
- y cosec^2() d/dtcombining ( 1 ) & ( 2 ) with this eqn we have
u cos - v = u sin cos /sin + r cosec d/dt
- v = u sin cos /sin + r cosec d/dt or , d/dt = -v/r sin .................. ( A )
/dt = -v/r sin .................. ( A ) differentiating ( 4 ) wrt 't ' , we get
u sin = - d/d ( r sin ) d/dt
= - d/d ( r sin ) d/dt combining with ( A ) we get
u/v = cos + sin 1/r dr/d
+ sin 1/r dr/dor , u/v cosec - cot = 1/r dr/d
- cot = 1/r dr/dNow integrating it and putting the boundary condition that r= d at = 
/2
= /2 we get
r/d = ( 1- cos ) ^u/v / ( sin ) ^ ( u/v + 1 )
) ^u/v / ( sin ) ^ ( u/v + 1 )
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