Question

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

Answer:

Explanation:

Masses of 1gm, 2gm …………100 gm are kept at themarks 1cm, 2cm…………..100 cm on the xis resectively. A perpendicular axis is passed at the 50th particle.

Therefore on the L.H.S. of the axis there will be 49 particles and on the R.H.S there are 50 particles.

Consider the two particles at the position 49 cm and 51 cm.

Moment of inertia due to these two particles will be:

=49×(1)2+51×(1)2

=100×1=100gm−cm2

similarly if we consider 48th and 52nd term we will get 100×(2)2gm−cm2.

Threfore we wil get 49 such set and one lone particle at 100 cm. Threfoee totl moment of inertia

=100(12+22+32+........+492)+100(50)2

=100(1^2+2^2+3^2+..........+50^2)=100xx(50xx51xx101)/6=100x25xx17xx101=100xx101xx425=4292500gm-cm^2=429kg-m^2=0.43kg-m^2`