# Chapter 9 - Rotational Dynamics - Check Your Understanding - Page 239: 14

In both a) and b), both balls will have the same translational speed at the bottom.

#### 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, $$mgh_0=\frac{1}{2}mv_f^2+\frac{1}{2}I\omega_f^2$$ A solid sphere has $I=2/5mr^2$. Also, $\omega_f=\frac{v_f}{r}$. Therefore, $$mgh_0=\frac{1}{2}mv_f^2+\frac{1}{2}\Big(\frac{2}{5}mr^2\frac{v_f^2}{r^2}\Big)$$ $$mgh_0=\frac{1}{2}mv_f^2+\frac{1}{5}mv_f^2=\frac{7}{10}mv_f^2$$ $$gh_0=\frac{7}{10}v_f^2$$ $$v_f=\sqrt{\frac{10}{7}gh_0}$$ As we can see from this equation, the final translational velocity $v_f$ does not depend on either the mass of the ball $m$ or its radius $r$. Therefore, in both a) and b), both balls will have the same translational speed at the bottom.

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