1kya centre of mass pe ???????????????
hey guys i am not getting the answer, plzz help
a rod of length L is lying on a smooth roller of height `h` as shown. find the coefficient of friction between the lower end of the rod and ground if the minimum angle i can make with ground is θ.
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23 Answers
39yes......................................... plzzzz that will be a great help
1y ??
y not...
any point apply it to any point you 'd get same thing "maybe for iit"
1y u have applied torque balance...about the point on ground
1due to this site i have regained confidence in rotational it seems
i jus hope same happens with all the physics chapters i think i was (yes was :( )good in 11th ...
1did u get till
N = mgL sinθcosθ/2h
N is for the roller...
just apply torque balance...about the point on ground
mgL cosθ/2 = N h/sinθ...
so N = mgL sinθcosθ/2h
now break it into components say Nx and Ny
Nx = mgL sin2θcosθ/2h
Ny = mgL sinθcos2θ/2h
now Nx = friction
friction = μ .(mg-Ny)
this should be all right..
hence the answer
now substitute the values ...and we are done [1]
1ok i got it after seeing but u tell me how am i to know that this
infinitesimal length going ahead of that "h" line was not a mistake
i mean it was not as prominent as in HCV figure
maybe i should have been more attentive sorry mathie..
doing it..
62philip neither is the figure wrong nor the question..
Think for just a couple more seconds :)
39hey see rotation topic in hc verma. am i mad to give wrong figure?
rotational mechanics question 45
39this i also wrote dudes.. plzzz solve i am not able to do math further ....
1why is θ and h both given [7]
i can't understand...
isn't h = L sinθ
3
just do basic eq. conditions.....and for min angle we take Ff max= meu N.....
