Rotational Motion

JEE Main level, numerical answer, medium

15 of 280

Question 15 of 280

A circular disc has radius R1R_{1} and thickness T1T_{1}. Another circular disc made of the same material has radius R2R_{2} and thickness T2T_{2}. If the moment of inertia of both discs are same and R1R2=2\frac{R_{1}}{R_{2}}=2 then T1T2=1α\frac{T_{1}}{T_{2}}=\frac{1}{\alpha}. The value of α\alpha is ____\_\_\_\_.

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Answer: 16

For a circular disc made of material with density ρ\rho, moment of inertia is I=12MR2=12ρπR2T⋅R2=12ρπR4TI = \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: I1=12ρπR14T1I_1 = \frac{1}{2}\rho \pi R_1^4 T_1
For disc 2: I2=12ρπR24T2I_2 = \frac{1}{2}\rho \pi R_2^4 T_2
Given I1=I2I_1 = I_2:
R14T1=R24T2R_1^4 T_1 = R_2^4 T_2
With R1R2=2\frac{R_1}{R_2} = 2:
(2R2)4T1=R24T2(2R_2)^4 T_1 = R_2^4 T_2
16R24T1=R24T216 R_2^4 T_1 = R_2^4 T_2
T1T2=116\frac{T_1}{T_2} = \frac{1}{16}
Therefore α=16\alpha = 16.

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