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 and
I4 = M.I. of solid sphere about its diameter.
Then :
Solution
Let M and R be the mass and radius of four bodies. Then, as per
question, their moment of inertia are
<br/><br/>I<sub>1</sub> = ${{M{R^2}} \over 2}$,
<br/><br/>I<sub>2</sub> = ${{M{R^2}} \over 2}$,
<br/><br/>I<sub>3</sub> = ${{M{R^2}} \over 2}$,
<br/><br/>I<sub>4</sub> = ${2 \over 5}M{R^2}$
<br/><br/>$\therefore$ I<sub>1</sub> = I<sub>2</sub> = I<sub>3</sub> > I<sub>4</sub>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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