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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Masses of $1gm, 2gm ........ 100gm$ are kept at the marks $1cm, 2cm, ......... 1000cm$ on he x axis respectively. A perpendicular axis is passed at the $50^{th}$ 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 inertial due to these two particle will be $=49 \times$ $1^2+51+1^2=100 \mathrm{gm}-\mathrm{cm}^2$ Similarly, if we consider $48^{\text {th }}$ and $52^{\text {nd }}$ term we will get $100 \times 2^2 \mathrm{gm}-\mathrm{cm}^2$ Therefore we will get 49 such set and one lone particle at 100 cm . Therefore total moment of inertia $=100\big\{1^2+2^2+3^2+\dots+49^2\big\}+100(50)^2$ $=100\times\frac{(50\times51\times101)}{6}=4292500\text{gm}\text{-cm}^2$ $=0.429\text{kg}-\text{m}^2=0.43\text{kg}\text{-m}^2$
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 inertial due to these two particle will be $=49 \times$ $1^2+51+1^2=100 \mathrm{gm}-\mathrm{cm}^2$ Similarly, if we consider $48^{\text {th }}$ and $52^{\text {nd }}$ term we will get $100 \times 2^2 \mathrm{gm}-\mathrm{cm}^2$ Therefore we will get 49 such set and one lone particle at 100 cm . Therefore total moment of inertia $=100\big\{1^2+2^2+3^2+\dots+49^2\big\}+100(50)^2$ $=100\times\frac{(50\times51\times101)}{6}=4292500\text{gm}\text{-cm}^2$ $=0.429\text{kg}-\text{m}^2=0.43\text{kg}\text{-m}^2$
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