Answer:
[tex]\dfrac{I_2}{I_1}=\dfrac{32}{1}[/tex]
Explanation:
The momentum of inertia of the solid sphere is :
[tex]I_1=\dfrac{2}{5}mr^2[/tex]
m is the mass of solid sphere
r is the radius of sphere
Since, [tex]m=\rho V[/tex]
[tex]m=\dfrac{4\pi r^3\rho }{3}[/tex]
[tex]I_1=\dfrac{2}{5}r^2\dfrac{4\pi r^3\rho }{3}[/tex]........................(1)
Let r' is the radius of second sphere such that, r' = 2r. New moment of inertia is given by :
[tex]I_2=\dfrac{2}{5}mr'^2[/tex]
[tex]I_2=\dfrac{2}{5}m(2r)'^2[/tex]
Similarly,
[tex]I_2=\dfrac{2}{5}r'^2\dfrac{4\pi r'^3\rho }{3}[/tex].
[tex]I_2=\dfrac{2}{5}(2r)^2\dfrac{4\pi (2r)^3\rho }{3}[/tex]........................(2)
Dividing equation (1) and (2) as :
[tex]\dfrac{I_2}{I_1}=\dfrac{2^5}{1}[/tex]
[tex]\dfrac{I_2}{I_1}=\dfrac{32}{1}[/tex]
So, the ratio of their moments off inertia is 32:1
The ratio of the moments of inertia of the two solid steel spheres,
I₂ : I₁ = 32 : 1
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Given:
radius of sphere 1 = R₁ = R
radius of sphere 2 = R₂ = 2R
Asked:
ratio of the moments of inertia = I₂ : I₁ = ?
Solution:
We will use moment of inertia for sphere as follows:
[tex]I_2 : I_1 = \frac{2}{5} m_2 (R_2)^2 : \frac{2}{5} m_1 (R_1)^2[/tex]
[tex]I_2 : I_1 = \frac{2}{5} \rho V_2 (R_2)^2 : \frac{2}{5} \rho V_1 (R_1)^2[/tex]
[tex]I_2 : I_1 = V_2 (R_2)^2 : V_1 (R_1)^2[/tex]
[tex]I_2 : I_1 = [\frac{4}{3} \pi (R_2)^3] (R_2)^2 : [\frac{4}{3} \pi (R_1)^3] (R_1)^2[/tex]
[tex]I_2 : I_1 = (R_2)^5 : (R_1)^5[/tex]
[tex]I_2 : I_1 = (2R)^5 : (R)^5[/tex]
[tex]I_2 : I_1 = 32R^5 : R^5[/tex]
[tex]I_2 : I_1 = 32 : 1[/tex]
[tex]\texttt{ }[/tex]
The ratio of the moments of inertia, I₂ : I₁ = 32 : 1
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Circular Motion
