E-StoreA 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 :
wx = dVx / dt = d { a b cos b t } / dt = -a b 2 sin b t
The y component of acceleration is :
wy = 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 V = 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.
Comments (2)
dis methud will be useful in quite a lot of prob. specially in which vectrs are involved
Preparing for IIT-JEE ?
Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor
@ INR 1,995/-
For Quick Info
|
|
|
|
|
| 35,229,755 Engineering Questions Attempted |
6,69,530 Visitors Monthly |
79 Engg. Exam Covered |
862 Students Currently Online |
Connect with Us  |
 |
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.