IIT JEE , AIEEE Forum - Physics , Mechanics

rakhiagrawal

4 thin metal rods each of mass m & length l,r welded to form a square ABCD.WHAT IS THE MOMENT OF INERTIA of the composite structure about a line which bisects rod AB & CD & PERPENDICULAR TO THE PLANE OF THE STRUCTURE?

bhuwanaroracorroded

how can d line bisecting AB CD b perpendicular 2 d plane

give d diagram

rakhiagrawal

Re:moment of inertia

bhuwanaroracorroded

I=2(ml^2/12)+2(m(l/2)^2+mr^2)

elessar_iitkgp

Without considering the radius at all, ie, assuming the radius to be very small compared to L
MI = 2((1/12)mL² + mL²/4 ) = 2mL²/3

However if we consider the radius, the problem becomes much more interesting ...
The MI of a disk of radius R about its symmetrical axis is (1/2)mR². Consider two perpendicular axis on the disk's surface. If the moment of inertia about these axes be I, then by perpendicular axes theorem,
2I = (1/2)mR²
I = (1/4)mR²
Now, consider a elementary disk of radius R on the rod AB at a distance x from the axis of rotation and of thickness dx. Then, the MI of this disk about an axis on its surface, parallel to the axis is given by (1/4)(m/

R²L)(

R²dx)R².
The moment of inertia of this element about the axis will be (1/4)(m/L)(dx)R² + (m/L)(dx)x² = (1/4)(m/L)(R² + 4x²)dx
Hence the MI of the rod about the axis is (1/4)(m/L)_-L/2

^L/2 (R² + 4x²)dx = (1/4)m (R² + L²/3)
MI of AB and DC = (1/4)m (R² + L²/3)

MI of the rods AD and BC = mR² + m(L/2)² =mR² + mL²/4 (Using parallel axes theorem)
MI = 2((1/4)m (R² + L²/3) + mL²/4 + mR²) = 2mL²/3 + 5mR²/2

If you consider the radius R the answer wont be reached ... neglecting that, ie, assuming that the rods are thin, R

0, and then the answer is 2mL²/3.

ramyani

MI of the rods AD and BC = mR² + m(L/2)² =mR² + mL²/4

MI of the rods AD = ( 1/ 2 ) mR² + ( 1/ 2 ) mR²

MI of the rods BC = ( 1/ 2 ) mR²

plz elaborate. I just started this chapter.

rakhiagrawal

ans is 2/3ml^2.

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