A thin ring of radius 2m is spun with an angular velocity of 10rad/sec and kept to roll (like a tyre) on a rough floor with coeff. of friction .5;
Its subsequent motion is analysed for the next 1 minute;
Find the time duration for which:
1) the friction acting is kinetic in nature;
2) the friction acting is static in nature;
why wud static friction act when the ring stops????
ReplyDeleteit wud be in equilibrium,there is no force which friction shud oppose
anyways...here's my an
kinetic - 4sec
static - 0sec??
i am with john there would be no slipping tendency after coming to rest after 4s
ReplyDeletei should hav been more clear:
ReplyDeletethe ring is kept such that it rolls;
it is not kept face down on the ground..as i guess you have assumed...
silly mistake that one
ReplyDelete3s kinetic friction and static does not come in this q
i think i am making some mistake
ReplyDeletei got 2s kinetic friction, and then static friction (i.e R.W.S.)
i m also getting 2s kinetic and for d rest of tym, no friction at all
ReplyDeletesorry i too got no friction,
ReplyDeleteyeah 2sec kinetic then Vcm becomes wR so no friction
ReplyDeleteyes 2s silly mistake really very silly
ReplyDeleteya..2s kinetic and then no friction...
ReplyDeleteimp. fact: static friction never acts!
2 sec for kinetic and 0 for static
ReplyDeleteplz. note:
ReplyDeletethe time after which slipping stops:
rw/(1+k^2/R^2)