Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of inertia about their diameter axis AB as shown in figure is $\sqrt{\frac{8}{x}}$ . The value of $x$ is:

Question

Smart Achievers · Accepted Answer

(67)

$I_{sphere} = \frac{2}{3} M R^{2} = M k_{1}^{2}$

$I_{cylinder} = \frac{1}{12} M (4 R^{2}) + \frac{1}{4} M R^{2} + M {(2 R)}^{2} = \frac{67}{12} M R^{2} = M k_{2}^{2}$

$\frac{k_{1}}{k_{2}} = \sqrt{\frac{2}{3} \times \frac{12}{67}} = \sqrt{\frac{8}{67}}$

Solution

Q.
Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of inertia about their diameter axis AB as shown in figure is $\sqrt{\frac{8}{x}}$ . The value of $x$ is:

Ans.
(67)

$I_{sphere} = \frac{2}{3} M R^{2} = M k_{1}^{2}$

$I_{cylinder} = \frac{1}{12} M (4 R^{2}) + \frac{1}{4} M R^{2} + M {(2 R)}^{2} = \frac{67}{12} M R^{2} = M k_{2}^{2}$

$\frac{k_{1}}{k_{2}} = \sqrt{\frac{2}{3} \times \frac{12}{67}} = \sqrt{\frac{8}{67}}$

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Solution

Q. Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of inertia about their diameter axis AB as shown in figure is 8x. The value of x is:

Ans. (67) Isphere=23MR2=Mk12 Icylinder=112M(4R2)+14MR2+M(2R)2=6712MR2=Mk22 k1k2=23×1267=867

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Q.
Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of inertia about their diameter axis AB as shown in figure is $\sqrt{\frac{8}{x}}$ . The value of $x$ is:

Ans.
(67)

$I_{sphere} = \frac{2}{3} M R^{2} = M k_{1}^{2}$

$I_{cylinder} = \frac{1}{12} M (4 R^{2}) + \frac{1}{4} M R^{2} + M {(2 R)}^{2} = \frac{67}{12} M R^{2} = M k_{2}^{2}$

$\frac{k_{1}}{k_{2}} = \sqrt{\frac{2}{3} \times \frac{12}{67}} = \sqrt{\frac{8}{67}}$