Physics, asked by ANIKETJADHAV08, 28 days ago

A metal ring has moment of inertia 2kg m- about a transverse axis through it's centre. It is melted and recast in to a thin uniform disc of the same radius. What will be the disc's moment of inertia about its diameter?​

Answers

Answered by Anonymous
14

mass of ring and disc is M = 1 kg

Moment of inertia of ring at diameter (Ir)d

= 1 kg m²

Rr = Rd

What we have to find out:

Moment of inertia of disc about own axis =

Id =?

Using theorem of perpendicular axes, for a ring M.I about its axis passing through C.M and perpendicular to its plane is twice the M.I about its any diameter, which is given

by,

(Ir)c = 2 (Ir)d

= 2 x 1

MRr² = 2 kg m²

Rr² = Rd² = 2 meter

Hence,

Moment of inertia of disc about own axis is given by,

1 Id=- MRd² 1 = -x 1 x 2 2 Id = 1 kg m²

Answered by ᏚɑvɑgeᏀurL
12

In geometry, a diameter of a circle is any straight line segment that passes through the centre of the circle and whose endpoints lie on the circle. It can also be defined as the longest chord of the circle. ... In more modern usage, the length of a diameter is also called the diameter.

The equation for diameter of a circle from circumference is:

d = c / π d=c/\pi d=c/π

d = 2 r d = 2r d=2r.

a = π ( d / 2 ) 2 a = \pi (d/2)^2 a=π(d/2)2.

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