Question

The center of mass of a pitched baseball (3.80-cm radius) moves at 38.0 m/s. The ball spins about an axis through its center of mass with an angular speed of 125 rad/s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere.

Answer 1

The center of mass of a pitched baseball (3.80-cm radius) moves at $38.0\,\mathrm{m/s}$. The ball spins about an axis through its center of mass with an angular speed of 125 rad/s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere.

Translational kinetic energy:

$$W_{tr} = \frac{m v^2}{2}$$

Rotational kinetic energy:

$$W_{r} = \frac{l \omega^2}{2} = \frac{\frac{2}{5} m r^2 \omega^2}{2} = \frac{1}{5} m r^2 \omega^2$$

The ratio of the rotational energy to the translational kinetic energy:

$$\frac{W_{r}}{W_{tr}} = \frac{\frac{1}{5} m r^2 \omega^2}{\frac{m v^2}{2}} = \frac{2}{5} \frac{r^2 \omega^2}{v^2}$$ $$\frac{W_{r}}{W_{tr}} = \frac{2}{5} \frac{(0.038\,\mathrm{m})^2 (125\,\mathrm{rad/s})^2}{(38\,\mathrm{m/s})^2} = 6.25 \times 10^{-3}$$

Answer: $\frac{W_{r}}{W_{tr}} = 6.25 \times 10^{-3}$