Question

Calculate moment of inertia of a ring of mass 500 g and radius 0.5 m about an axis of rotation coinciding with its . diameter and tangent perpendicular to its plane.
(Ans: 6.25 x 10⁻² kgm²,0.25 kgm²)

Answer 1

Answer:

$I_{D} = 6.25\times10^{-2} kg m^{2}$

$I_{T} = 0.25 kg m^{2}$

Explanation:

As per the question,

Given data is,

mass = 500 g = 0.5 kg

( because 1 kg = 100 gm )

radius = 0.5 m

Now moment of inertia of a rigid body is defined as angular mass of rotating body.

According to question,

The fomula for moment of inertia of ring about its diameter :

$I_{D} = \frac{MR^{2} }{2}$

$I_{D} = \frac{0.5\times \left ( 0.5 \right )^{2} }{2}$  

$I_{D} = 0.0625 kg m^{2}$  

$I_{D} = 6.25\times10^{-2} kg m^{2}$

The fomula for moment of inertia of ring about its tangent :

$I_{T} = MR^{2}$

$I_{T} = 0.5\times \left ( 0.5 \right )^{2}$  

$I_{T} = 0.25 kg m^{2}$  

Hence,  $I_{D} = 6.25\times10^{-2} kg m^{2}$

             $I_{T} = 0.25 kg m^{2}$

Answer 2

Answer:

Moment of inertia of ring about an axis of rotation coinciding with its diameter and tangent perpendicular to the plane

I =3/2 mr^2

Given: m=500 gm =0.5 kg ,r=0.5 m .  I=3x0.5x0.5^2 =0.1875 kg m^2