A massless rod has a 'mass' glued to it. The position of the mass is 1m from the left end, and 3m form the right end.
This rod-mass combo is placed on the frictionless inclined-planes shown. At equilibrium, it makes an angle 'B' with the horizontal. What is this angle?
PS: What is the number of cuboids in a 'Rubik's Revenge'? (Ignore the internal structure of the toy.)
To see a Rubik's Revenge: See This
is ans for 2nd part 1000?
ReplyDeleteis B=0 ??
ReplyDeletei m getting 100 for 2nd part
ReplyDeleteor maybe 480 after re-reading the question
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ReplyDeleteI think 30 degree. I have a reason but, I have not used that anywhere else or I couldn't feel it while using.
ReplyDeletebhaiya, gud question :)
30 for me too
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ReplyDelete19.11° ,i think so
ReplyDeleteThe 'rod-mass' question had a -ve solution (for B) as well. You would have obtained it had you solved it mathematically rather than intuitively.
ReplyDeleteI've changed the diagram to make it a bit easier.
For the 'Rubik's Revenge' question:
You have to find the number of cuboids in a 4*4*4 cube (visible/invisible). I had precisely stated to ignore the internal structure of the toy. As for the answer, Akash's first answer was right.
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ReplyDeleteWhen i solved it by eqns, i got B as -60 and 30
ReplyDeleteand for the Rubik's question, i thought maybe neglecting internal structure meant to neglect that there is 2*2*2 cube which is not visible, my bad :P
Hi Akash: 30* and -60* are very 'naive' answers. One expects better :P.
ReplyDeletethat was an epic mistake actually, two more 30 degrees encouraged me to ignore it !
ReplyDeleteis it tan inverse(-2/sqrt3) ?
I applied the logic that every system tries to minimize its energy(to get equilibrium) so, rod will move down to decrease it potential energy hence, 30 degree.
ReplyDeleteAmit, consider a block on the tip of a hemisphere, if u displace it a little, it moves down, it does minimise its potential energy, but not to get equilibrium!
ReplyDeleteF=-dU/dr
ReplyDeleteat equilibrium , potential energy is not minimum(can be sometimes but not general), but constant
block moves down because initially it was at unstable eq.(here it would be stable). And definitely p.e. is not the only energy, it tries to minimise total energy but, here only P.E. plays a role. there would be no K.E. or any type of nuclear energy.
ReplyDeleteequilibrium means the net force will be zero, just try to balance the forces and torques at this position and see if it comes 0
ReplyDeleteit should and it would. Also equilibrium position means energy should be min.
ReplyDeleteone of us is wrong, lets wait for Bhaiya before we build misconceptions :)
ReplyDelete-60* and 30* would be the answers for a 'stable' equilibrium.
ReplyDeleteNow its high time we started working on the unstable equilibrium scenario.
India is going to lose, so here's a hint:
Consider all the forces on the rod (they are 3 in number). Arrive at a condition for rotational equilibrium of the rod. (Open your third eye).
btw, Akash's answer matches with mine..so we're probably right.
Anuj bahiya that is not correct.My answer is also correct according to your previous diagram.But you dont give me credit for that.Instead of that you changed the figure
ReplyDeletemy previous answer was "19.11° ,i think so"
ReplyDeleteHi *%#@$||- (I can't make all those symbols).
ReplyDelete19.11* is a 'wrong' answer. In the previous diagram, a positive value of 'B' did no exist. The -ve value of 'B' was 49.106*, which is atan(2/root3).
atan means tan inverse.
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ReplyDeleteThere is a small mistake, in equation 4.
ReplyDeleteThank you bhaiya for this help.Next time i will do your problem carefully.
ReplyDeleteby considering the glue friction we get
ReplyDeleteatan(3root3) - 30
i.e. 49 degrees
bhaiya need a bit of help frm u
a dog and cat are at (0,0) and (0,d)
the cat runs parallel to x axis with a constant speed u
the dog starts chasing it with velocity v s.t. its velocity is always directed towards the cat
(u<v)
Find time after which dog catched it
thats an easy question yaar
ReplyDeletet=(vd)/(v^2-u^2)
Also friend don't ask your problems here,as bhaiya is against it. 5 months ago i also asked my problems in this blog,but bhaiya told me not post your personal problems here.
ReplyDelete