A particle, initially at a distance d from the origin on the X axis takes off on a peculiar trajectory, driven by an external agent dealing with which is none of our bussiness. The trajectory is such that at any position of the particle (r, Q) in polar coodinates, its velocity is [(-VocosQ icap) + (Vo(1-sinQ) jcap)], where Vo is a constant
We are to find its angular velocity about the origin, its radial velocity about the origin, and its distance from origin as it strikes the Y axis, in terms of the bold data
Free question-> a particle at (3,4) is moving with velocity (1,2,3), find x+y+z, if its angular velocity about origin is (x,y,z)
-Sambhav
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ReplyDeleteyou aren't getting what 'radial' velocity means shaan, it isn't a vector, it is the rate at which the distance 'r' from the origin varies
ReplyDeleteand you are almost there for the distance from the origin at Y axis
good one!
ReplyDeletedistance from origin k =sin-1(1-e*-d) angular velocity = -v.cosk/k radial velocity = v.e*-d
ReplyDeleteshivam: whats' e*?
ReplyDeletesambhav: then, is radial acc also not a vector?..i think it is..
it is e raised to the power -d.
ReplyDeletefree ques
ReplyDeletei think it is 1/5
is it
ReplyDeleteVradial= Vo(1-sinQ)
Vang= V/r * cosQ
dist = droot2
1/5 ???
plz... reply if i am wrong so that i can think about it more......
ReplyDeleteno shivam, ur answer is wrong
ReplyDeleteanonymous and shaan have answered most of it correctly, but no one has yet come up with the distance on Y axis correctly.
it is simple to give the answers in that format what i did is that dy/dt=v.(1-sinq)
ReplyDeleteand dx/dt=-v.cosq.
and divided the above two eqns and get relation b/w x and y (i mean r and q) so we dont need coordinates.is it right method because we have same method in ncert for a diff exercise
in that format should it not be q.
ReplyDelete1)i didn't get what you meant in ur last comment...(try to use as little as pronouns as possible while taking part in discussions)
ReplyDelete2)answer is asked in terms of quantities written in bold only, x and y are completely out of picture
3)definitely, dy/dt=v.(1-sinq)and dx/dt=-v.cosq, but x and y are themselves functions of q
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ReplyDeletehi, is the dist from origin on y axis= d/2?
ReplyDeleteps: it took me over 3 days for this to strike!!! :)
congos shaan! :-)
ReplyDeletethe soln easy not as easy to strike, you have to assume coordinates at(r,Q) as (rcosQ, rsinQ) and velocity as (-VocosQ,Vo(1-sinQ)), the derivative of position with time is velocity, and neither r nor Q is constant
ReplyDeletefind dr/dt and dQ/dt
divide to get dr/dQ, solve the DE to get the motion in polar coordinates, at Q=0, r=d, find r at Q=pi/2
can anyone give the curve in cartesian coordinates?
for the free question, you have to use the formula---->angular velocity=(rxv)/|r|^2...(you can use this to get the angular velocity in the above problem also)
ReplyDeleteremember that v(perpendicular)/r is only the magnitude, u have to assign direction as k cap if it is anticlockwise
curve in cartesian coordinates:
ReplyDeletex^2+y^2= (d-y)^2
thanks a ton for this( & many other) wonderful questions
ReplyDelete