Answer:
\(\dfrac{I_2}{I_1}=\dfrac{32}{1}\)
Explanation:
The momentum of inertia of the solid sphere is :
\(I_1=\dfrac{2}{5}mr^2\)
m is the mass of solid sphere
r is the radius of sphere
Since, \(m=\rho V\)
\(m=\dfrac{4\pi r^3\rho }{3}\)
\(I_1=\dfrac{2}{5}r^2\dfrac{4\pi r^3\rho }{3}\)........................(1)
Let r' is the radius of second sphere such that, r' = 2r. New moment of inertia is given by :
\(I_2=\dfrac{2}{5}mr'^2\)
\(I_2=\dfrac{2}{5}m(2r)'^2\)
Similarly,
\(I_2=\dfrac{2}{5}r'^2\dfrac{4\pi r'^3\rho }{3}\).
\(I_2=\dfrac{2}{5}(2r)^2\dfrac{4\pi (2r)^3\rho }{3}\)........................(2)
Dividing equation (1) and (2) as :
\(\dfrac{I_2}{I_1}=\dfrac{2^5}{1}\)
\(\dfrac{I_2}{I_1}=\dfrac{32}{1}\)
So, the ratio of their moments off inertia is 32:1
