33The worst ever answer which i had got ever.. is definitely wrong as in phys answers are always simple.. :(
s=(R/2μ).ln((√v0+√R2g2+v0)/Rg)
A small collar of massm is given an initial velocity of magnitude v0 on the horizontal circular track fabricated from a slender rod.If the coefficient of kinetic friction is μk, determine the distance travelled before the collar comes to rest.
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33okie ...
surely tommorow..
feeling sleepy .. and going to sleep ...
33normal reaction between collar and track is
√(mg)2+(mv2/R)2
so frictional force acting will be μk√(mg)2+(mv2/R)2
so mdv/dt=-μk√(mg)2+(mv2/R)2
dv/dt=-μk√(g)2+(v2/R)2
dv/ds*ds/dt=-μk√(g)2+(v2/R)2
vdv/dt=-μk√(g)2+(v2/R)2
.....(correction with a minus sign)
Solve it to get 's'
Haven't taken gravity free space as in that case only normal reaction will be mv2/R and as μmv2/R will decrease v and hence normal reaction will also so when v→0 the force also→0 so it will take infinite time to stop...
33
3v2μk/R = -vdv/dx!!!!!!!!!!!!!
μkx/R = ln (0)
x = R /μk.................
defently rong!!!!!!!!!!!!!!!!!! rite im half asleep [13]
3horizontal usuall means to neglect gravity ........ !!! it cant be soo complex!!!!!!!!