1.

Particles of masses 1g, 2g, 3g, …….. 100 g, are kept at the marks 1 cm, 2cm, 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 meter scale.

Answer»

Solution :`[1xx49^(2)+2xx48^(2)+3xx47^(2)+……49xx1^(2)+52xx2^(2)+…….98xx48^(2)+99xx49^(2)+100xx50^(2)]10^(-7)`
`= 100[50^(2)+49^(2)+48^(2)+…..+1^(2)]xx10^(-7)`
`= 10^(-5)XX(1)/(6)xx50xx51xx101=0.43 KGM^(2)`


Discussion

No Comment Found