One Each Day Till They Go Away (Part 1)

This recent news comes as a disappointment to me. But I am sure to enjoy this last streak of problems.

Consider an elementary mass being projected from the point $(r \cos{\theta},r \sin{\theta},0)$ , where $0 \le r \le 1$ and $0 \le \theta \le 2\pi$ .

The velocity of projection is:

$\vec{v}_o = v_{xo} \ \hat{i} + v_{yo} \ \hat{j} + v_{zo} \ \hat{k}$ $\vec{v}_o = v_o \left(\frac{\left(r \cos{\theta} \ \hat{i} + r \sin{\theta} \ \hat{j}\right) - \left( -1 \ \hat{i} + 0 \ \hat{j} -1 \ \hat{k}\right)}{\biggr\lvert \left(r \cos{\theta} \ \hat{i} + r \sin{\theta} \ \hat{j}\right) - \left( -1 \ \hat{i} + 0 \ \hat{j} -1 \ \hat{k}\right) \biggr\rvert} \right)$

This leads to:

$v_{xo} = \frac{10(r \cos{\theta}+1)}{r^2 + 2r\cos{\theta} + 2}$ $v_{yo} = \frac{10r\sin{\theta}}{r^2 + 2r\cos{\theta} + 2}$ $v_{zo} = \frac{10}{r^2 + 2r\cos{\theta} + 2}$

Now, as the mass element is projected from the specified point, its motion is governed by the following set of ODEs:

$\ddot{x} = 0 \ ; \ \ddot{y} = 0 \ ; \ \ddot{z} = -10$ $\dot{x}(0) = v_{xo} \ ; \ \dot{y}(0) = v_{yo} \ ; \ \dot{z}(0) = v_{zo}$ $x(0) = r \cos{\theta} \ ; \ y(0) = r \sin{\theta} \ ; \ z(0) = 0$

Solving this system of equations leads to:

$x = v_{xo}t + r \cos{\theta} \ ; \ y = v_{yo}t + r \sin{\theta} \ ; \ z = v_{zo}t - 5t^2$

Now, the mass element's final position is when its z coordinate becomes zero. This happens at time:

$t_f = \frac{v_{zo}}{5}$

The final x and y coordinates of the mass element are:

$x_f = v_{xo}t_f + r \cos{\theta}$ $y_f = v_{yo}t_f + r \sin{\theta}$

After simplification:

$x_f = r \cos{\theta} + \frac{20(r \cos{\theta}+1)}{r^2 + 2r\cos{\theta} + 2}$ $y_f = r \sin{\theta} + \frac{20r\sin{\theta}}{r^2 + 2r\cos{\theta} + 2}$

Now, the mass of the mass element is: $dm = \sigma \ r \ dr \ d\theta \ ; \ \sigma = 1$ . Therefore, the moment of inertia of the mass element about the Z axis when it reaches its final position is:

$dI = \left(x_f^2 + y_f^2\right) \ dm$

Simplifying:

$dI = \left( r^2 + \frac{40r^2 + 40r\cos{\theta}}{r^2 + 2r\cos{\theta} + 2} + \frac{400\left(r^2 + 2r\cos{\theta} + 1 \right)}{\left(r^2 + 2r\cos{\theta} + 2\right)^2} \right)r \ dr \ d\theta$

Finally, the total moment of inertia about the Z axis is:

$I = \int_{0}^{2\pi} \int_{0}^{1} \left( r^2 + \frac{40r^2 + 40r\cos{\theta}}{r^2 + 2r\cos{\theta} + 2} + \frac{400\left(r^2 + 2r\cos{\theta} + 1 \right)}{\left(r^2 + 2r\cos{\theta} + 2\right)^2} \right)r \ dr \ d\theta \approx 276.1474$

The mass distribution is as follows:

$r$ is varied in steps of 0.04 and $\theta$ is varied is steps of $\pi/50$ . The points in red indicate the initial mass distribution which is obviously a unit disk. The points in blue indicate the final mass distribution.