Answer
The order in which the objects reach the bottom is: solid sphere, solid cylinder, spherical shell, hoop.
Work Step by Step
The total mechanical energy $E_0$ at the top of the incline ($\omega_0=0$ and $v_0=0$) is the same as the total mechanical energy $E_f$ at the bottom of the incline ($h_f=0$). Therefore, $$mgh_0=\frac{1}{2}mv_f^2+\frac{1}{2}I\omega_f^2$$
The forumla for an object's moment of inertia has the form of $I=\frac{a}{b}mr^2$. Also, $\omega_f=\frac{v_f}{r}$. Therefore, $$mgh_0=\frac{1}{2}mv_f^2+\frac{1}{2}\Big(\frac{a}{b}mr^2\frac{v_f^2}{r^2}\Big)$$ $$mgh_0=\frac{1}{2}mv_f^2+\frac{a}{2b}mv_f^2=\frac{a+b}{2b}mv_f^2$$ $$gh_0=\frac{a+b}{2b}v_f^2$$ $$v_f=\sqrt{\frac{2b}{a+b}gh_0}$$
- A hoop: $I=MR^2$, so $a=b=1$. Therefore, $$v_f=\sqrt{\frac{2}{2}gh_0}=\sqrt{gh_0}$$
- A solid cylinder: $I=1/2MR^2$, so $a=1$ and $b=2$. Therefore, $$v_f=\sqrt{\frac{4}{3}gh_0}=\sqrt{1.33gh_0}$$
- A spherical shell: $I=2/3MR^2$, so $a=2$ and $b=3$. Therefore, $$v_f=\sqrt{\frac{6}{5}gh_0}=\sqrt{1.2gh_0}$$
- A solid sphere: $I=2/5MR^2$, so $a=2$ and $b=5$. Therefore, $$v_f=\sqrt{\frac{10}{7}gh_0}=\sqrt{1.43gh_0}$$
The faster the object goes, the less time it takes for the object to reach the bottom. So the order in which they reach the bottom is: solid sphere, solid cylinder, spherical shell, hoop.
