by bhuvankar we get a range of ws due to the fact that frictn can act both in the upwards and the downwords dirns..........
first let us assume that fn is acting in the upwards dirn...........then we have for vertical eqbm...
Ncos@ + KNsin$=mg.......(K=coeff of fn)
from here we have N=mg/(ksin$+cos$).............1)
then for horizontal eqbm.........we have,(seeing in pseudo frame we apply an outward centrifugal force.......)
kNcos$= Nsin$ + mw2 Rsin$..........
substituting for N from 1) we get the expn that.......
g(sin$-Kcos$)/(ksin$+cos$) = w2 Rsin$
on simplifying which we get the expn for w as..........
sqrt(g(sin$-Kcos$)/((Rsin$)(ksin$+cos$)))...............
similarly on reversing the dirn of frn.ie in dwnwrd dirn.......we will get our result as sqrt(g(sin$+Kcos$)/((Rsin$)(ksin$+cos$))).................
Moderation Team
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