Question 204: Centre of Mass and Rotational Motion

Question

Moment of inertia (M. I.) of four bodies, having same mass and radius, are reported as;
I
1
= M.I. of thin circular ring about its diameter,
I
2
= M.I. of circular disc about an axis perpendicular to disc and going through the centre,
I
3
= M.I. of solid cylinder about its axis and
I
4
= M.I. of solid sphere about its diameter.
Then :

Accepted Answer

Let M and R be the mass and radius of four bodies. Then, as per
question, their moment of inertia are
I
1
= $${{M{R^2}} \over 2}$$,
I
2
= $${{M{R^2}} \over 2}$$,
I
3
= $${{M{R^2}} \over 2}$$,
I
4
= $${2 \over 5}M{R^2}$$
$$ 	herefore $$ I
1
= I
2
= I
3
> I
4

Answer

I
1
= I
2
= I
3
> I
4

Answer

I
1
+ I
3
< I
2
+ I
4

Answer

I
1
= I
2
= I
3
< I
4

Answer

I
1
+ I
2
= I
3
+ $${5 \over 2}$$ I
4

Centre of Mass and Rotational Motion

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