A solid sphere of mass and radius having moment of inertia about its diameter is recast into a solid disc of radius and thickness . The moment of inertia of the disc about an axis passing the edge and perpendicular to the plane remains . Then and ar

Question

A solid sphere of mass $M$ and radius $R$ having moment of inertia $I$ about its diameter is recast into a solid disc of radius $r$ and thickness $t$ . The moment of inertia of the disc about an axis passing the edge and perpendicular to the plane remains $I$ . Then $R$ and $r$ are related as [2006]

Smart Achievers · Accepted Answer

(2)

For solid sphere, moment of inertia about diameter

$I_{A B} = \frac{2}{5} M R^{2} = I$

For solid disc

$I_{A' B'} = I_{y y'} + M r^{2} = \frac{1}{2} M r^{2} + M r^{2} = \frac{3}{2} M r^{2}$

$I_{A B} = I_{A' B'} (given)$

$∴$ $\frac{2}{5} M R^{2} = \frac{3}{2} M r^{2} \Rightarrow r = \frac{2}{\sqrt{15}} R$

Solution

1 $r = \sqrt{\frac{2}{15}} R$

2 $r = \frac{2}{\sqrt{15}} R$

3 $r = \frac{2}{15} R$

4 $r = \frac{\sqrt{2}}{15} R$

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Solution

Q. A solid sphere of mass M and radius R having moment of inertia I about its diameter is recast into a solid disc of radius r and thickness t. The moment of inertia of the disc about an axis passing the edge and perpendicular to the plane remains I. Then R and r are related as [2006]

1 r=215R

2 r=215R

3 r=215R

4 r=215R

Ans. (2) For solid sphere, moment of inertia about diameter IAB=25MR2=I For solid disc IA'B'=Iyy'+Mr2=12Mr2+Mr2=32Mr2 IAB=IA'B' (given) ∴ 25MR2=32Mr2 ⇒ r=215R

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1 $r = \sqrt{\frac{2}{15}} R$

2 $r = \frac{2}{\sqrt{15}} R$

3 $r = \frac{2}{15} R$

4 $r = \frac{\sqrt{2}}{15} R$