Question 208: Centre of Mass and Rotational Motion

Question

Moment of inertia (M. I.) of four bodies, having same mass and radius, are reported as; I1 = M.I. of thin circular ring about its diameter,I2 = M.I. of circular disc about an axis perpendicular to disc and going through the centre,I3 = M.I. of solid cylinder about its axis andI4 = 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

I1 = I2 = I3 > I4

Answer

I1 + I3 < I2 + I4

Answer

I1 = I2 = I3 < I4

Answer

I1 + I2 = I3 + $${5 \over 2}$$ I4

Centre of Mass and Rotational Motion

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