"Consider two uniform discs of the same thickness and different radii R1 = R and R2 = $\alpha$R made of the same material. If the ratio of their moments of inertia I1 and I2 , respectively, about their axes is I1 : I2 = 1 : 16 then the value of $\alpha$ is :"

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Accepted Answer

"Moment of inertia of disc, $$I = {{M{R^2}} \over 2} = {{\left[ {p\left( {\pi {R^2}} \right)t} \right]{R^2}} \over 2}$$

$I = K{R^4}$

${{{I_1}} \over {{I_2}}} = {\left( {{{{R_1}} \over {{R_2}}}} \right)^4}$

$${1 \over {16}} = {\left( {{R \over {\alpha R}}} \right)^4} \Rightarrow \alpha = {\left( {16} \right)^{{1 \over 4}}} = 2$$"

Consider two uniform discs of the same thickness and different radii R₁ = R and
R₂ = $\alpha$R made of the same material. If the ratio of their moments of inertia I₁ and I₂ , respectively, about their axes is I₁ : I₂ = 1 : 16 then the value of $\alpha$ is :

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Consider two uniform discs of the same thickness and different radii R1 = R and R2 = $\alpha$R made of the same material. If the ratio of their moments of inertia I1 and I2 , respectively, about their axes is I1 : I2 = 1 : 16 then the value of $\alpha$ is :

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More Rotational Motion questions

Drill 25 more like these. Every day. Free.

Consider two uniform discs of the same thickness and different radii R₁ = R and
R₂ = $\alpha$R made of the same material. If the ratio of their moments of inertia I₁ and I₂ , respectively, about their axes is I₁ : I₂ = 1 : 16 then the value of $\alpha$ is :