Exploding Sphere Debris

The position of an infinitesimal mass on the sphere is given by

$p = ( \sin \theta \cos \phi, \sin \theta \sin \phi, 2+\cos \theta)$

The initial velocity is radial, so it is given by

$v_0 = 5 ( \sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$

Obviously, the mass is symmetric about the z-axis, so we need only consider the projectile problem along a single direction, and therefore, we can choose, $\phi = 0$

The dynamic equations for the infinitesimal mass are

$x = \sin \theta + 5 \sin \theta t$

$z = 2 + \cos \theta + 5 \cos \theta t - 1/2 g t^2$

Using $g = 10$ and setting $z = 0$ , it follows that the time of impact with the $xy$ plane is given by,

$t_1 = \dfrac{1}{10} ( 5 \cos \theta + \sqrt{ 25 \cos^2 \theta + 20 (2 + \cos \theta)} )$

From which the position of the debris is given by

$r(\theta) = x = \sin \theta (1 + 5 t_1 )$

Finally, to complete the inertia integral, we need the expression for the infinitesimal mass,

and this is given by

$d m = \rho \sin \theta d \theta d \phi$

Hence,

$I_F = \displaystyle \int_{\phi=0}^{\phi=2\pi} \int_{\theta=0}^{\pi} \rho r(\theta)^2 \sin \theta d \theta d \phi$

Since the integrand is constant with respect to $\phi$ , this becomes,

$I_F = 2 \pi \rho \displaystyle \int_{\theta=0}^{\pi} r(\theta)^2 \sin \theta d \theta$

On the other hand, the moment of inertial of the original spherical shell is given by

$I_0 = 2 \pi \rho \displaystyle \int_{\theta=0}^{\pi} \sin^3 \theta d \theta$

Evaluating the integral for $I_F$ numerically (Simpson's Rule for example), and directly integrating for $I_0$ , we find that

$\dfrac{I_F}{I_0} = 20.88$