1. mailtoarko

2. mailtoarko

if u know how to solve then pls help otherwise please don't interfere unnecessarily

3. theyatin

@morganFREEDDOM if muslim doesn't means terrorist . . .

4. mailtoarko

wats that to do with the question

5. theyatin

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

6. mailtoarko

can u explain me the process

7. theyatin

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

8. mailtoarko

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)

9. theyatin

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

10. Vincent-Lyon.Fr

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.

11. srjmukherjee2

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

12. srjmukherjee2

Its me sourajit

13. Vincent-Lyon.Fr

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.

14. srjmukherjee2

actually the original mass is 9M which I inputted at last

15. srjmukherjee2

can u plz post what is wrng in my solution so that it can be corrected

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(R-R/3)^2=MR^2/2 NOW SUBTRACT BOTH THE MI AND U WILL GET 4MR^2 AS THE ANS.....

17. Dhawal

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