find velocity of COM [gud ques]

btw i learnt the following way of finding IAOR....

take 2 convenient points A and B of a rigid body ........
draw their velocity vectors..........
now draw lines _|_ to the vectors v_A and v_B

such that the _|_ passes thru the tails of d vectors

the point where these 2 _|_ intersect is the IAOR....

if they dont intersect ...the point is at infinity ....

IS DIS METHOD CORRECT??? ALSO IS DER ANE ODER APPROACH????

a book named vector mechanics and dynamics says that this IAOR cannot be used to fidn acceleration ...y so ?

not as challenging physics as maths in your question arshad... but I assure you its doable

I have already done that so .... I will let someone else do it
however if no one does I'll just post the answer (and probably some equations)

hey karna pls explain how did u get that answer .......

i guess u took an axis about wich the COM performs pure rotation...so how do we come to kno wer is dat axis ...secondly ..the axis u hav taken ...is the distance of COM always constant from the axis ?how ?? diagram dekh ke lag to nahi raha ???do we need the distance of COM frm axis to be constant ??????

and i m able to understand nothing frm ur one line answers ..pls elaborate [2]

i agree with philip.

help me out with this one ................. pls some one post a solution that i can understand!!!! PLZ!!! [2]

qwerty,d axis xyz hs taken is known as d "instantaneous axis of rotation"...u will get to read of it in most d buks...

.....the normal force at that instant.............

i learnt frm some where that IAOR can be used to find only velocities ..and not accelerations ...is it true??

aah !
qwerty is actually correct (partially) :

"i learnt frm some where that IAOR can be used to find only velocities ..and not accelerations ...is it true"

However the method applied (finding α using torque about IAOR) by xyz is also correct

how can both not be wrong ?? ;)

solved it .....
using constraint hoping u know the methods
first draw a line parallel to the perpendicular and from the com got it..
now consider the groud as a reference for x1 and x2(u know it)
since x1/x2=1/2 midpnt theorem ...
v1/v2=1/2 or vcom in y direction is v/2

for vel in x axis............
first lets find the vel of the other pnt.....
y^2+x^2=l^2
differentiating it
2xv_x=-2yv_y
solving it we get vel in x direction as vtan@

now for com do the same thing but now take x1 and x2 from aong the perpendicualr since x1/x2=1/2 similarly vx =vtan@/2
now v=v^2_x+v^2_y
using it we get v^2(1+tan^2)/4 solving it vel is vsec@/2

qwerty....no need to worry about this "IAOR"...its just a problem solving technique......but i wud like to add few more things to your post #34

for example......

tnx yaar ......

its zero.

as far as i remember ....the ans is v/2 cosecA
or v/2 sec A ..... dont remember exactly

nishant sir plz tell wer i am wrong...

find that distance(b/w axis and com) using geometry
v_cm*d=vcosA*l