Free Counters
The last question drew little response, so this is something better (with a counter, which sucks):
This is a fixed vertical cylinder (top view), and it is connected by a thread (of length equal to the circumference of cylinder) to a ball.
Now, the ball is wound around the cylinder, keeping the string taut at all times.
The path traced by the ball is called an involute...simply speaking: find the length of the involute.
cool...the counter is working!!!
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
ReplyDelete2*pi*pi*r
ReplyDeleter being the radius of the cylinder
same as sambhav.....2*pi*pi*r
ReplyDeleteyes same as sambhav
ReplyDeletehey anuj....
ReplyDeleteThis is mohit.. Georgian...
Ans is d same....
Jee went gud...
Rslts cming in few days...
Keep in touch..
yes...same
ReplyDeletenever thought we could use integration for things like this also
hey guys post your approaches...
ReplyDeletemy differential equation
[integral(2piR - RQ)dQ] from 0 to 2pi
bingo...same to same for me too!!!
ReplyDeleteOye Mohit!! Best of luck yaar (hope to see you in IITD..me ragging my classmates!! what fun!)
ReplyDeleteaccha...people got it right. its a cool problem naa??
but the real problem (Ref. IE Irodov) had a wee bit of modifications:
1)the thread was of length 'L', instead of the perimeter..
2)you had to calculate the time taken by the ball to wind up completely, given that it is projected with an initial velocity 'V'..
1)pi times L
ReplyDelete2)pi times L times the reciprocal of V...
someone help me with it, i m blank here
ReplyDeleteconsider an instant when the particle has swept an angle theta,
ReplyDeleteat that point, it can move in an infinitesimally small arc over which it has the same length of string to move
total length of string is 2piR
length of string that has wound around the cylinder is RQ
so the lenght of string availabe/free=(2piR-RQ)
the length of arc in which the particle sweeps round this radius is (2piR-RQ)dQ
integrate from 0 to 2pi!
sahi yaar Sambhav...i was just drawing the soln on Paint when this happened: i am not being able to attach images!!!
ReplyDeleteand so, for the cycloid question, we have no diagram!
i perhaps understood from sambhav's expression, but got ans as L^2 / 2R for 2nd ques
ReplyDeleteintegral (L-rQ)dQ , Q from 0 to L/R
thanks
This comment has been removed by the author.
ReplyDeleteL^2/2R=(4*pi*pi*R^2)/2r=(2*pi*pi*R)=(pi*L)
ReplyDeleteyaar L=2 pi r yaha kese valid ho sakta hai when its given not equal to circumference
ReplyDeleteabe yaar..!
ReplyDeletehadd hai meri bhi..
yea you are right...(maine soch bhi kaise lia itna simple hoga??)
yes first part have been done but what about the time taken as the velocity is not constant.
ReplyDeletethere is no reason for the velocity to change
ReplyDeletelength of string is l-rQ
ReplyDeletedx=(l-rQ)dQ
dx=(2pir-rQ)dQ
integrating from 0 to 2 pi yields the answer