This disc of mass M, Radius R rolls down from a height H without slipping.
The floor below the hill has sticky tape of linear mass density P..and it winds around the disc as it rolls (its sticky side faces up). Finally the disc stops!!!!!
What is the total distance traveled by the disc apart from that traveled down the hill??
assume the tape to be really thin..also assume that the disc travels a sufficiently large distance.
buck up people..i need an answer here...
ReplyDeleteis it M/P
ReplyDeletetell if i shud post my attempt
Anuj sir please give the solution completely of the temple problem of hinge.
ReplyDeleteplease pity on me.....
ReplyDeleteno chhavi..but you can post the attempt!
ReplyDeletebhaiya is it
ReplyDeleteM(h/R - 1)/P
dude..can't the radius be less than 1?????
ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDeletechalo bhai...well done man!!i never solved this one myself...
ReplyDeletei dare you to this one!! come on guys...
ReplyDeleteis it MH/PR ?
ReplyDeletenearly..just a bit wrong..
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteM(H-R)/PR
ReplyDeleteI was conerving the potential energy. As mass has increased therefore, COM will reach a less height.
ReplyDeleteBhaiya, I want to ask if I am correct.
This type of similar question was in module too where the carpet was unrolling itself.
is the answer is MH/2PR.AS THE CENTER OF MASS OF TAPE GAINS A HIGHT OF 2R AT THE COST OF KINETIC ENERGY OF THE DISK WHICH IN TURN EQUAL TO MgH
ReplyDeleteAkash's last post gave the correct ans..
ReplyDeleteRohan, the center of mass of the wound-up tape gains a height of 'R'..and not '2R'..
also, the COM of the disc falls thru a height 'H-R', and not 'H'...
This question was a cooler version of the already quite cool carpet question in the module.. again, i was simply stuck up as to why the disc would stop...
But it turns out it would..or else them carpet companies would immediately fire some people involved in rolling carpets!!
is it due to symmetry of wounding the tape around the disk??????????
ReplyDeletebhaiya how did we get that centre of mass gains a height by R.
ReplyDeleteI was using the integration so as to get the rise in height by equating the gain in mass
PL=mass wound around the disc with limits from R to R+h
It gave me a complex thing. Please tell where I am wrong and how did we got the answer finally.
By the way bhaiya that was a good joke....
Amit,
ReplyDeletesuppose it moves distance L
so PL=mass wound around disc
now L/2piR=[L/2pir] + {L/2pir)
the greatest integer part gives no of COMPLETE LOOPS of tape around disc. obviously if integral loops are there, CM will be at center of disc ( due to symmetry)
now bunk the remaining part, as it is given assume it covers a very large distance, it will hardly have any effect on CM
take an example and think urself
tell if not clear
sry fr checkin late(its gettin difficult these days!)
ReplyDeleteanuj,yaar i think there is going 2 be some loss in energy..first up,as seen from the disc the particles of the tape will appear to make an inelastic collision with the ball..second,the "sticking"of particles takes place due to friction(consider the case when on a block moving with velocity v,another small block is"gently"placed..whaan bhi there is loss of KE-so i think the 2 cases are in a sense analogous..)
bta?
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ReplyDeleteAmit mg to pehle hi uski PE hai, we are equating LHS loss in PE and RHS as gain
ReplyDeletewhy u adding mg in RHS
?
yes akash I had got it myself
ReplyDeletei have come her just to remove it
anyway,thanx