33Most probably they say in FIITJEE... this question was cancelled...
As far as this question is concerned The length of rod cannot increase thats the trick...
a rod of length 2m moves in horizontal x-y plane. at any instant end 'a' of the rod is at the origin and has velocity V1 (see fig.). the other end B at the same instant is moving with velocity V2 (see fig.). the rod makes an angle of 30 deg with the x-axis at this instant.
(a) magnitude of angular velocity of the rod ???
>> 1 rad/s.
>> √3 rad/s.
>> √3/2 rad/s.
>> 1/2 rad/s.
N.B. : mujhe yeh question hi samajh me nahi aya :(
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12 Answers
1this is an easy one(but in the exam due to different data in question and in figure i didnt take the risk)
first of all the velocity along the rod's length is zero(as its length remains constant)
hence let the rod be r->
and both the velocities be a-> and b->
then just apply
r->.a-> and r->.b->
and from there find the velocities along the rod (which is r->.a->/2 and r->.b->/2)
equating them
2√3+Vy=3√3+6
Vy=6-√3
now find the perpendicular velocities at both the ends
and then find the instantaneous centre of rotation and apply wx=Va or Vb
33
9vel of end2 wrt 1 is 1i + 6-V j
now its perpendicular to rod
ωLsin30 = 1
ω=1
1can anyone expain to me celestines working
i think its some kind of shortcut
21yeah i guess it was cancelled.....tats why i guess everyone got more marks then expected in PT1 last year.....those 12 marks were given as a gift[3]
