If I1 is the moment of inertia of solid cylinder of radius R and length l about it geometrical axis. A concentric cylinder having radius R2 and lengthl2 is carved off from original cylinder and I2 is MOI of carved off cylinder. Find I1I2.

Thus I1I2=32MOl of solid cylinderI1=12mR2Density =ρ=mv=mπR2lI2=m2(R/2)22I2=m2(R/2)22I2=12mπR2l×π(R2)2(l2)(R2)2Where,m2=mπR2l×π(R2)2l2I2=12mR232I1I2=12mR212mR232=321

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If $I_{1}$ is the moment of inertia of solid cylinder of radius R and length $l$ about it geometrical axis. A concentric cylinder having radius $\frac{R}{2}$ and length $\frac{l}{2}$ is carved off from original cylinder and $I_{2}$ is MOI of carved off cylinder. Find $\frac{I_{1}}{I_{2}}$ .

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answer is 32.

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Detailed Solution

Thus $\frac{I_{1}}{I_{2}} =32$

MOl of solid cylinder

$I_{1} = \frac{1}{2} {mR}^{2}$

Density $= ρ = \frac{m}{v} = \frac{m}{π R^{2} l}$

$I_{2} = \frac{m_{2} {(R/2)}^{2}}{2}$

$I_{2} = \frac{m_{2} {(R / 2)}^{2}}{2}$

$I_{2} = \frac{1}{2} \frac{m}{π R^{2} l} \times π {(\frac{R}{2})}^{2} (\frac{l}{2}) {(\frac{R}{2})}^{2}$

Where, $m_{2} = \frac{m}{π R^{2} l} \times π {(\frac{R}{2})}^{2} \frac{l}{2}$

$I_{2} = \frac{1}{2} \frac{m R^{2}}{32}$

$\frac{I_{1}}{I_{2}} = \frac{\frac{1}{2} m R^{2}}{\frac{1}{2} \frac{m R^{2}}{32}} = \frac{32}{1}$

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