Rotational vs Translational Kinetic Energy of a Baseball

[UrbanXrisis]

The center of mass of a pitched base ball (radius=3.8cm) moves at 38m/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.

$$0.5Iw^2:0.5mv^2$$
$$I=mr^2=m(.038m)^2=m0.001444$$
$$0.5m0.001444(125rad/s)^2:0.5m(39m/s)^2$$
$$22.5625:1444$$
$$1:64$$

is this correct?

 

[mooshasta]

The moment of inertia for a uniform sphere is as follows:

$$I = \frac{2}{5}m\,r^{2}$$

 

[UrbanXrisis]

thanks:

$$0.5Iw^2:0.5mv^2$$
$$I=(2/5)mr^2=m(2/5)(.038m)^2=m5.776E-4$$
$$0.5m(5.776E-4)(125rad/s)^2:0.5m(39m/s)^2$$
$$9.025:1444$$
$$1:160$$

is this correct?

 

[mooshasta]

Looks good to me..

 

[UrbanXrisis]

why is the kinetic energy so much larger than the rotational energy?