Two discs having moment of inertia I_1 and I_ | vedclass

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Question

(A) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$.Since

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$I_1 t_1 = I_2 t_2$

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(A) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} M R^2$.Since both discs have the same mass $M$,we have $I_1 = \frac{1}{2} M R_1^2$ and $I_2 = \frac{1}{2} M R_2^2$.Therefore,the ratio is $\frac{I_1}{I_2} = \frac{R_1^2}{R_2^2}$.Since both discs are made of the same material,they have the same density $ho$. The mass is given by $M = 	ext{Volume} 	imes ho = (\pi R^2 t) ho$.Since $M$ and $ho$ are the same for both,we have $\pi R_1^2 t_1 ho = \pi R_2^2 t_2 ho$,which simplifies to $R_1^2 t_1 = R_2^2 t_2$.From this,we get $\frac{R_1^2}{R_2^2} = \frac{t_2}{t_1}$.Substituting this into the ratio of moments of inertia,we get $\frac{I_1}{I_2} = \frac{t_2}{t_1}$.Cross-multiplying gives $I_1 t_1 = I_2 t_2$.

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