Particles of masses 1 g, 2 g, 3 g, ....., 100 g are kept at the marks 1 cm, 2 cm, 3 cm, ....., 100 cm respectively on a metre scale. Find the moment of inertia of the system of the particles about a perpendicular bisector of the metre scale.
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Solve :- it is given that , Particles of masses 1 g, 2 g, 3 g, ....., 100 g are kept at the marks 1 cm, 2 cm, 3 cm, ....., 100 cm respectively on a metre scale.
you know, here it is clear that the metre scale is a system of 100 particles.
We can assume metre scale is rod of length 1m and mass of this rod , M= 1g + 2g + 3g + 4g + .... + 100g = 100(101)/2 = 5050g = 5.050g
and we have to find moment of rod about an axis normal to its plane and passing through mid point of it.
e.g., MI = ML²/12
= (5.050) × 1²/12
= 5.050/12 = 0.43 kgm²
Hence, the moment of inertia of the system of the particles about a perpendicular bisector of the metre scale is 0.43 kgm²
you know, here it is clear that the metre scale is a system of 100 particles.
We can assume metre scale is rod of length 1m and mass of this rod , M= 1g + 2g + 3g + 4g + .... + 100g = 100(101)/2 = 5050g = 5.050g
and we have to find moment of rod about an axis normal to its plane and passing through mid point of it.
e.g., MI = ML²/12
= (5.050) × 1²/12
= 5.050/12 = 0.43 kgm²
Hence, the moment of inertia of the system of the particles about a perpendicular bisector of the metre scale is 0.43 kgm²
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