Consider two uniform discs of the same thickness and different radii R 1= R and R 2=α 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 α is:

Question

Consider two uniform discs of the same thickness and different radii R 1= R and R 2=α 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 α is: (A) √2 (B) 2 (C) 4 (D) 2 √2

Tardigrade Admin · Accepted Answer

I 1=( MR 2/2)=(ρ(π R 2) t ⋅ R 2/2) I propto R 4 ( I 1/ I 2)=( R 14/ R 24)=(1/16) ∴ ( R 1/ R 2)=(1/2)

Q. Consider two uniform discs of the same thickness and different radii $R_{1} = R$ and $R_{2} = α R$ made of the same material. If the ratio of their moments of inertia $I_{1}$ and $I_{2}$ , respectively, about their axes is $I_{1} : I_{2} = 1 : 16$ then the value of $α$ is:

$2$

2

4

$22$

$I_{1} = \frac{M R ^{2}}{2} = \frac{ρ ( π R ^{2} ) t \cdot R ^{2}}{2}$
$I \propto R^{4}$
$\frac{I _{1}}{I _{2}} = \frac{R _{1}^{4}}{R _{2}^{4}} = \frac{1}{16}$
$∴ \frac{R _{1}}{R _{2}} = \frac{1}{2}$

Q. Consider two uniform discs of the same thickness and different radii R1​=R and R2​=αR 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 α is:

2​

2

4

22​

I1​=2MR2​=2ρ(πR2)t⋅R2​ I∝R4 I2​I1​​=R24​R14​​=161​ ∴R2​R1​​=21​

Q. Consider two uniform discs of the same thickness and different radii $R_{1} = R$ and $R_{2} = α R$ made of the same material. If the ratio of their moments of inertia $I_{1}$ and $I_{2}$ , respectively, about their axes is $I_{1} : I_{2} = 1 : 16$ then the value of $α$ is:

$2$

$22$

$I_{1} = \frac{M R ^{2}}{2} = \frac{ρ ( π R ^{2} ) t \cdot R ^{2}}{2}$
$I \propto R^{4}$
$\frac{I _{1}}{I _{2}} = \frac{R _{1}^{4}}{R _{2}^{4}} = \frac{1}{16}$
$∴ \frac{R _{1}}{R _{2}} = \frac{1}{2}$