Question

The center of mass of a pitched baseball or radius 2.42 cm moves at 23.3 m/s. The ball spins about an axis through its center of mass with an angular speed of 158 rad/s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere.

Answer 1

To solve the problem, it will be necessary to define the rotational and translational kinetic energy in order to determine the relationship between the two. Rotational energy is defined as,

$KE_{Rotational} = \frac{1}{2} I\omega^2$

Here,

I = Moment of Inertia

$\omega$ = Angular velocity

Now the translational energy will be,

$KE_{Translational} = \frac{1}{2} mv^2$

Here,

m = Mass

v = Velocity

Therefore the relation between them will be,

$\frac{KE_{Rotational} }{KE_{Translational}} = \frac{\frac{1}{2} I\omega^2 }{\frac{1}{2} mv^2 }$

Applying the moment of inertia of a sphere we have,

$\frac{KE_{Rotational} }{KE_{Translational}} = \frac{\frac{1}{2} (\frac{2}{5}mr^2)\omega^2 }{\frac{1}{2} mv^2 }$

$\frac{KE_{Rotational} }{KE_{Translational}} = \frac{2}{5} \frac{r^2\omega^2}{v^2}$

$\frac{KE_{Rotational} }{KE_{Translational}} = \frac{2}{5} \frac{(2.42*10^{-2})^2(158)^2}{23.3^2}$

$\frac{KE_{Rotational} }{KE_{Translational}} = 0.01077$

Therefore the ratio will be 0.01077