Question

particles of masses 1g,2g,3g....., 100g are kept at the marks 1cm,2cm3cm......., 100cm respectly on a mitre scale.find the Momentum of inertia of system of prticles about a perpendicular bisector of the metre scale. give me answer with full solution help.

Answer 1

perpendicular bisector will be at the mark 50 cm on the scale. 49 masses is placed at one side of the scale at (1,2,3,4 ....49)

and 50 masses is placed at the other side of the scale at ( 51,52,53,54,......100) ,and one mass is placed at the 50th mark.

so moment of inertia I = 2* [m*12 + m*22 + m* 32 + .....m*492  ] + m* 502

                                =2m[49(49+1)(98+1)/6  + m *2500

                                =83350m    g cm2

Answer 2

Answer:

$0.4295 \: kg \: {m}^{2}$

Explanation:

There are two ways to do this:

1. Although the scale is not of uniform mass per unit length, you can sort of take it to be that.

You find mass per unit length by:

$\frac{m}{l} = \frac{n \times (n + 1)}{100 \times 2}$

Then integrate it with proper limits. I am not doing it here as it gives an approximate answer.

2. This method is more accurate.

Here you are basically adding the moment of inertia of all the particles. The pictures above are for the second method.

Hope it helps.

EDIT: Definitely don't go with the first method. The nature of the system is discrete, but we do integration only for contiguous systems, so don't do it.