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Question

# Particles of masses 1g,2g,3g,.......100g are kept at the marks 1cm,2cm,3cm,.......100cm respectively on a metre scale. Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.

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Solution

## As we know,Masses of 1 gm,2 gm....100 gm are kept at the marks 1 cm,2 cm,....1000 cm on the x axis respectively. A perpendicular axis is passed at the 50th particle.Therefore on the L.H.S. side of the axis, there will be 49 particles and on the R.H.S. side, there are 50 particles.Consider the two particles at the position 49 cm and 51 cm.Moment inertia due to these two-particle will be =49×12+51+12=100 gm−cm2Similarly if we consider 48th and 52nd tern will get 100×22 gm−cm2Therefore we will get 49 such set and one lone particle at 100 cm. Therefore total moment of inertia =100{12+22+32+...+492}+100(50)2.=100×(50×51×101)/6=4292500 gm−cm2=0.429 kg−m2=0.43 kg−m2.

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