- #1
Benny
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I am having trouble with the following question.
Q. Derive the expression for the moment of inertia of a spherical ball of mass m and radius r about the axis z through its centre of gravity.
Ans: [itex]I_G = \frac{2}{5}mr^2 [/itex].
I did the following.
[tex]I_G = \int\limits_m^{} {r^2 } dm[/tex]
[tex]dm = \rho dV[/tex] where rho is the constant density.
[tex]
V = \frac{4}{3}\pi r^3 \Rightarrow dV = 4\pi r^2 dr
[/tex]
[tex]
I_G = 4\pi \rho \int\limits_0^r {r^4 dr} = \frac{{4\pi r^5 \rho }}{5} = \frac{3}{5}\left( {\frac{4}{3}\pi r^3 \rho } \right)r^2 = \frac{3}{5}mr^2 \ne \frac{2}{5}mr^2
[/tex]
I can get the answer using a triple integral but the question seems to only require a single integral. At the moment I can't see where I've gone wrong. Can someone help me out? Thanks.
Q. Derive the expression for the moment of inertia of a spherical ball of mass m and radius r about the axis z through its centre of gravity.
Ans: [itex]I_G = \frac{2}{5}mr^2 [/itex].
I did the following.
[tex]I_G = \int\limits_m^{} {r^2 } dm[/tex]
[tex]dm = \rho dV[/tex] where rho is the constant density.
[tex]
V = \frac{4}{3}\pi r^3 \Rightarrow dV = 4\pi r^2 dr
[/tex]
[tex]
I_G = 4\pi \rho \int\limits_0^r {r^4 dr} = \frac{{4\pi r^5 \rho }}{5} = \frac{3}{5}\left( {\frac{4}{3}\pi r^3 \rho } \right)r^2 = \frac{3}{5}mr^2 \ne \frac{2}{5}mr^2
[/tex]
I can get the answer using a triple integral but the question seems to only require a single integral. At the moment I can't see where I've gone wrong. Can someone help me out? Thanks.