IIT JEE , AIEEE Forum - Physics , Mechanics

arun5678

Four point masses, each of value m, are placed at the corners of a square ABCD of side 1. The moment of inertia through A and parallel to BD is

1) ml²

2) 2 ml²

3) 2l

4) zero

catch_arnnie

wat's " l " here ?

rphy

just draw the fig acc to ques.

moment of inertia of pt masses = mr^2
r is the dist from axis of rotation.

MOI of A abt the axis is zero since it is on axis.

MOI of B & D abt axis is m(l/

2 )^2

MOI of C abt axis is m(

2 l )^2

total is 3ml^2

LastMinuteGenius

1st, consider the MI about centre of mass perpendicular to the plane of the square:

4*(mR²)

now, R = L/

2 , where L is side of square

so, MI about CM is 4*(mL²/2)

so, MI = 2mL²

now, use parallel axis theorem to find MI passing thro A and perpendicular to plane of square:

so, new MI is 2mL² + (4m)D²

now, D is nothing but R = L/

2 (just draw and think of the figure...)

so, new MI is 2mL² + 4mL²/2

so New MI is 4mL²

now, use perpendicular axis theorem

take 2 axes,mutually perpendicular, one parallel to BD and other perpendicular to BD intersecting at A.

Now, MI's about these 2 axes is NOT equal (cause the mass distribution is not symmetric about both axes)

so, I_{perpendicular to BD} + I_{parallel to BD} = 4mL²

=> 2*(m*L²/2) + I_{parallel to BD} = 4mL²

=> I_{parallel to BD} = 4mL²- mL²

=> I_{parallel to BD}= 3mL² (answer)

None of the given answers are correct!!!!

Do RATE me If this is CoRReCt.....

nitin62225

ans= 3ml²

soln:

2m(l/

2)²+m(

2l)²=3ml²

^{-----------------l= side of the square.}

pottermania1990

i think ntin is correct

even i got 3mI^2

LastMinuteGenius

yes the answer is 3mL^2
look at me soln...

edison

Q) Four point masses, each of value m, are placed at the corners of a square ABCD of side 1. The moment of inertia through A and parallel to BD is

1) ml²

2) 2 ml²

3) 2l

4) zero

Sol) Here the axis is passing through A and parallel to diagonal BD.

The perpendicular distance of point B from the axis = diagonal / 2

=

2/2 = 1/

2

similarly the perepndicular distance of D from axis = 1/

2

Also perpendicular distance of point C from teh axis = diagonal =

2

so teh moment of inertia = m (1/

2)² + m(1/

2)² + m(

2)²

or I = m/2 + m/2 + 2m = 3m

so moment of inertia = 3m

But in the options the answer is in terms of l which is not mentioned in teh question as you have given side of square to be 1(one)

edison

If side of square is taken to be L then moment of inertia follwing above approach comes out to be

I = 3mL^2

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