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mailtoarko
Group Title
i am posting the picture of my question please help friends
 one year ago
 one year ago
mailtoarko Group Title
i am posting the picture of my question please help friends
 one year ago
 one year ago

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mailtoarko Group TitleBest ResponseYou've already chosen the best response.0
if u know how to solve then pls help otherwise please don't interfere unnecessarily
 one year ago

theyatin Group TitleBest ResponseYou've already chosen the best response.0
@morganFREEDDOM if muslim doesn't means terrorist . . .
 one year ago

mailtoarko Group TitleBest ResponseYou've already chosen the best response.0
wats that to do with the question
 one year ago

theyatin Group TitleBest ResponseYou've already chosen the best response.0
yes it was but not yours Q and answer of your q is that radius of disk varies inversely square to the M.I. so 1/9M.I of the disk must be subtracted from the actual m.I. of the disk
 one year ago

mailtoarko Group TitleBest ResponseYou've already chosen the best response.0
can u explain me the process
 one year ago

theyatin Group TitleBest ResponseYou've already chosen the best response.0
first find the actual M.I. of the disk with mass 9M and radius R then subtract M.I. of mass M and radius R/3
 one year ago

mailtoarko Group TitleBest ResponseYou've already chosen the best response.0
then it should be 9/2 mr^2  ( 1/2*mr^2/9) but in the solution its given 9/2 mr^2  ( 1/2*mr^2/9 + 1/2m*(2r/3)^2)
 one year ago

theyatin Group TitleBest ResponseYou've already chosen the best response.0
well boy i am confused too. . . sorry cant help any further. . . i'll give you a site you try post your question there it might help
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.0
Right answer is 4MR² You should use additivity of moments of inertia. 1st work out mass of small missing disc. 2nd work out its moment of inertia about its own axis, then use parallel axis theorem to find its moment about centre of big disc.
 one year ago

srjmukherjee2 Group TitleBest ResponseYou've already chosen the best response.0
See, MI = (MR^2)/2 (normal disc) Let σ be the Mass per unit area So, σ=M/πr^2 area is pi(r^2)  pi{(r/3)^2} = 8/9(pi)(r^2) So the mass of disk given becomes 8/9(M). So acc. to formula (MI = (MR^2)/2) [8/9(M)×r^2]/2=4/9Mr^2 NOW put M=9M and thus MOI=4Mr^2
 one year ago

srjmukherjee2 Group TitleBest ResponseYou've already chosen the best response.0
Its me sourajit
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.0
I'm afraid something is missing there. You need to apply parallel axes theorem somehow. Why do you write M=9M in the end? You should start with the right mass from the beginning.
 one year ago

srjmukherjee2 Group TitleBest ResponseYou've already chosen the best response.0
actually the original mass is 9M which I inputted at last
 one year ago

srjmukherjee2 Group TitleBest ResponseYou've already chosen the best response.0
can u plz post what is wrng in my solution so that it can be corrected
 one year ago

adityalakhe Group TitleBest ResponseYou've already chosen the best response.2
ans is 4 mr^2 first take the mi of actual disk i.e. 9mr^2divided by 2 then subtract the mi of small disk for that mass of small disk is m since mass distribution is uniform so mi of small disk abt its own axis is M(R/3)^2/2 AND THEN APPLYING PARALLEL AXIS THEOREM MI OF SMALL DISK ABT AXIS OF BIGGER DISK IS M(R/3)^2/2+M(RR/3)^2=MR^2/2 NOW SUBTRACT BOTH THE MI AND U WILL GET 4MR^2 AS THE ANS.....
 one year ago

Dhawal Group TitleBest ResponseYou've already chosen the best response.0
find mass of disc (cavity) then subtract [intertia of cavity+(mass of cavity) x(distance from cavity axis to actual rotation axis)squared] that gives u answer
 one year ago
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