GENERAL PROBLEM SOLVING TECHNIQUE

Any good IIT-JEE like problem , when analysed contains basically of three parts

1. Catchwords
2. Guessing simple Concept
3. Links

One good problem is linked with many simple concepts .This mix gives us is, an tricky looking question.Your first effort should be, to read(mark the catchwords) and somehow break the question so that you can identify(better word would be ‘guess’) the simple concepts involved .For identifying simple concepts, one must resort to drawing small diagrams of the problems and then a careful thought procedure should follow.Once you have identified the simpler concepts involved in it, then goto stage number two that is linking the concepts.Now this linking stage is where your earlier efforts and mental level come to the fore, in this area of linking no one can help you except, you, yourself.

In linking you use result of one portion and apply it to solve another.We will illustrate all these stages with an example:

IRODOV 1.241A uniform disc of radius R=20cm has a round cut as shown in the figure .The mass of the remaining portion of the disc equals m = 7.3kg.Find the moment of inertia of such a disc relative to the axis passing through its centre of inertia and perpendicular to the plane of the disc.

SOLUTION
JEE APPROACH: Remember , our aim is to show you method of attacking a problem by the above analysis , hence we will give an essay type description.

1.identifying catchwords:—-
    1.1. ‘uniform disc’ , ’round cut’,'passing through centre of inertia’….

2.guessing simple concepts:—-
    2.1. ‘ uniform disc ‘=> AREA proportional to MASS
    2.2. ’round cut’=> area removed=> proportional mass removed..leads to depletion of Moment of inertia…..also center of gravity of remaining disc will not be the at the same position as of full disc.
    2.3. ‘ passing through the centre of inertia’=>maybe parallel axis theorem ….!!!!???

3.linking and solving the problem:—-
Remember ,practice makes you a ‘better linker’.What you get by solving good number of problem is EXPOSURE which makes you a better linker.Back to the problem…

Taking clue from 2.1, suppose you had the full disc , if we remove the portion then the area removed is pi*(R/2)^2 => mass removed
=>(M/(pi*R^2))*pi*(R/2)^2=M/4.Mass of remaining portion is,m = M-M/4=3M/4

Taking clue from 2.2, we have removed a disc of radius R/2 from the orignal disc , so if we remove contribution to moment of inertia of that disc from full disc we will get the moment of inertia of the remaining

Next we try to find the center of gravity(CG) of the remaining portion .The CG of small disc is at the center of the small disc and that of full disc is at the center of full disc(our origin)….full disc=small disc + remaining portion!!! .So CG of small disc*its mass + CG of remaining portion*its mass = CG of full disc*total mass(equalising the mass moments) gives: (R/2)*(M/4) + (3M/4)*Xcm=M*0 => Xcm= - R/6

Taking clue from 2.3, moment of inertia(MI) of removed disc about ITS OWN CENTER is [(M/4)*(R/2)^2]/2 , and about the CG of remaining portion is [(M/4)*(R/2)^2]/2 +[(M/4)*(R/2 -Xcm)^2](parallel axis thm) =41(MR^2)/288
MI of full disc about CG of remaining disc is [M*R^2]/2 + [M*(0 -Xcm)^2]=19(MR^2)/36
=> MI of remainig disc = MI of full disc - MI of removed(small) disc=37(MR^2)/96=37(mR^2)/72 Ans.

try to see the’ linking’ carefully ,see how results of 2.1 is used in 2.2 and then in 2.3 to arrive at the solution.

In future , try to proceed by the above procedure and you will be able to win over lots of problem !!!!