A circular disc has radius R_ 1 and thickness T_ 1 . Another circular disc made of the same material has radius R_ 2 and thickness T_ 2 . If the moment of inertia of both discs are same and R_ 1 R_ 2 =2 then T_ 1 T_ 2

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For a circular disc made of material with density

\rho

, moment of inertia is

I = \frac{1}{2}MR^2 = \frac{1}{2}\rho \pi R^2 T \cdot R^2 = \frac{1}{2}\rho \pi R^4 T

.
For disc 1:

I_1 = \frac{1}{2}\rho \pi R_1^4 T_1

For disc 2:

I_2 = \frac{1}{2}\rho \pi R_2^4 T_2

Given

I_1 = I_2

:

R_1^4 T_1 = R_2^4 T_2

With

\frac{R_1}{R_2} = 2

:

(2R_2)^4 T_1 = R_2^4 T_2

16 R_2^4 T_1 = R_2^4 T_2

\frac{T_1}{T_2} = \frac{1}{16}

Therefore

\alpha = 16

.

A circular disc has radius R_ 1 and thickness T_ 1 . Another...

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