A problem of Irodov---component method

Blazing goIITian

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19 Jan 2008 01:40:52 IST
Posts: 2537
19 Jan 2008 01:40:52 IST
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A problem of Irodov---component method

A problem of Irodov

If the expression for co-ordinates of position of particle is given, then the problem be best tackled by component method.
Question :

A point moves in the plane xy such that x = a sin b t  and y = a ( 1- cos b t  ) ,
where  a and  b are +ve constants. Find :
(a) the distance S traversed by the point during the time T ;
(b) the angle between the point's velocity and acceleration vector.

solution :
we shall solve the problem by component method.

the the distance S traversed by the point during the time T  is given by :
S = [ ][ ]  | V | dt    ( plz note that only speed is enough to calculate distance )
To calculate | V |  we have to find V  ( bold letters mean vector )
To calculate V  , we proceed remembering V = i Vx + j Vy
Where, i and j are unit vectors along X and Y axis.
Now how to calculate  Vx and Vy   ?
It is given,   x = a sin b t =>    Vx = dx / dt =  a  b  cos b t
also,  y = a ( 1- cos b t )  =>   Vy = dy / dt = a  b  sin b t
so,  V = i Vx + j Vy  = i ( a  b  cos b t ) + j ( a  b  sin b t )
so, | V | = [ ]  [ ( a  b  cos b t ) 2 + ( a  b  sin b t ) 2 ]  = a  b 
instantaneous speed    V  = a  b 
so,  S =    [ 0][T ] | V | dt  =  [0 ][ T]   a  b  dt  = a  b  T
S = a  b  T 

(b )
In the same component method, the acceleration of the point is calculated. The x component of acceleration is :
w= dVx / dt = d { a  b  cos b t }  / dt   =  -a b 2 sin b t
The y component of acceleration is :
w= dVy / dt = d { a  b  sin b t }  / dt   =  + a b 2 cos b t
so that acceleration vector is w = i wx + j wy
or, w = i ( -a b 2 sin b t ) + j ( a b 2 cos b t )

Now , we have velocity vector = i ( a  b  cos b t ) + j ( a  b  sin b t )
acceleration vector   w = i ( -a b 2 sin b t ) + j ( a b 2 cos b t )
Let C be the angle these 2 vectors. Then
w.V = [ i ( -a b 2 sin b t ) + j ( a b 2 cos b t ) ] . [ i ( a  b  cos b t )
+ j ( a b  sin b t ) ]

wV cos C = - a  2  b 3  sin b t cos b t  + a  2  b 3  sin b t cos b t = 0
cos C  = 0
= >   C  = 90 degree.

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Comments (2)


Blazing goIITian

Joined: 22 Apr 2007 11:21:45 IST
Posts: 2537
19 Jan 2008 01:58:27 IST
0 people liked this

In the same method u can solve another Q.:
A particle moves in the plane XY with constant acceleration w directed along the negative Y axis. The equn of motion of the particle has the form y = ax - bx^2 where a and b are positive constants. Find the velocity of the particle at the origin of the co-ordinates.

Blazing goIITian

Joined: 31 Jan 2008 16:40:48 IST
Posts: 334
8 Feb 2008 11:35:05 IST
0 people liked this

gud work....
dis methud will be useful in quite a lot of prob. specially in which vectrs are involved



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