Problem No.2
I got this rod with balls on it and i want to hinge it at some point so that it does not... you know wobble like a see-saw.. or fall down...
The balls and the rod have the following property:
At '1', mass of the ball is 1kg;
At '2', the mass is 2kg..at '100' the mass is 100 kg.
Assume the balls to be really small and the rod to be massless.
Tell me the point of hinging.
70+ 40/71
ReplyDelete66 from the left end?(assuming the rod is a meter scale sort of thing)
ReplyDeleteYO!! Digvijay got it right. Lets keep the soln. a secret for another day..
ReplyDeleteI got 67 from the left end.
ReplyDeletei mean the hinge should be placed on the 67th ball. :)
ReplyDeletesorry mates.. correct ans. is 67.
ReplyDeletegot a bit excited seeing 66.... thanks kunal
by 66 i meant 66 unit distance..not 66th ball.
ReplyDeleteThis comment has been removed by the author.
ReplyDeletethat is my way.
ReplyDeletei just balanced torques.
n got the answer.pls check anuj bhaiya.
http://img213.imageshack.us/img213/4305/mysolutn.jpg
anmol, you could have done it a bit more simply...
ReplyDeleteanyways.. talk abt it tomorrow.
ans is 67th ball
ReplyDeleteanmol the legth will not be 100-x but 99-x as scale is starting from 1 and not 0
ReplyDeletei also did the same way
thnx anurag.
ReplyDeletei'll check it again.
67th ball :)
ReplyDelete67th ball. find centre of mass assuming size remains same and only density increases. similar question in hcv.
ReplyDeletesum of squares of natural numbers till 100 divided by the sum of natural numbers till hundred
ReplyDeletejust formed Fnet=0 and Torque(net)=0 equations
taking moments about the end of the rod was easier than taking them about the hinge..
so I had to include normal reaction
66 did what vaibhav said just solved fr COM
ReplyDeletethe hinge should be the ball with mass 67kg
ReplyDeletesince the balls are in translational as well as rotational equilibrium,net torque about any point is 0.i considered it about one the sides
ya balancing torques is a good approach.. but you may also like to calculate the rod's center of mass...
ReplyDeleteWhy go into torques and all when you can just find the c.o.m. and the weight on both sides will become equal. If you hinge it at the c.o.m. it HAS to stay in equilibrium :)
ReplyDeleteWhy Akash's method is wrong?
ReplyDeletend plz post the solution by torque method also .. !! :)
This comment has been removed by the author.
ReplyDeletesum of all weights is balanced by the normal reaction from the hinge
ReplyDelete[summation (n)]=N
also,
taking moments about the free end
(1.1 + 2.2 + 3.3.......100.100)=N*x
=>summation (n^2)=N*x
find x by substituting values....
vahi kia hai jo COM nikalne k lie kia
also, moments about hinge lekar hume summation(n)=N vali equation ki zaroorat nhi padti...kynki second eq main N hota he nhi...yeh karna generally questions ko asan bana deta hai..(see liquids sse 3 or ose 18)
par is case main mushkil ho jaega kynki direct summation ka formula nhi lag paega
PS..what is akash's method?
Akash and even me did lik dis in first atempt .. !!
ReplyDeletefindin the total mass = 100 * 101 / 2
to be in equi .. dono side same mass hona chiye ..
100*101 / 4 (taken haf mass) = x(x+1)/2 (X from left)
hop u got . wat i did ...
whats rong in this ??
ur wrong because by no means does a balnced rod imply mass on both sides of the rod is same..
ReplyDeletewat ur lukin for is torques(rotational equilibrium)
the only reaason y this method works for the centre of mass of the system as well is because torque about the centre of mass is 0!
Do note that even while calculating the centre of mass, ur summing the product of mass AND length....
in ur case, u have completely ignored length...
ur method will only prove a success if the number of balls are equal on both sides... which is not the case over here
Hey people!! the mass of a body may not be equally distributed abt. its COM!!!!!
ReplyDeleteI realised it while trying to calculate the COM of a solid cone..Check it out using the normal approach of integration..and by equating the masses on the 2 side.
isn't it 3/4 times the heigth/length taken from the tip?
ReplyDelete