Question

The center of mass of a pitched baseball or radius 2.11 cm moves at 23.6 m/s. The ball spins about an axis through its center of mass with an angular speed of 139 rad/s. Calculate the ratio of the rotational energy to the translational kinetic energy. Treat the ball as a uniform sphere.

Answer 1

The ratio of the rotational energy to the translational kinetic energy is 0.00618

Step-by-step explanation:

If m is the mass of the baseball and v its speed, then the translational kinetic energy is

Kt = 1/2 mv²

If I is its rotational inertia and ω is its angular speed, then the rotational kinetic energy is

Kr = 1/2 Iω²

`(KE_(rotational))/(KE_(translational)) = ((1)/(2) Iw^2)/((1)/(2) MV^2)`

But the rotational inertia of a uniform sphere is I = 2mr²/5

`(1)/(2)Iw^2 = (1)/(2) * (2)/(5) MR^2w^2`

`(KE_R)/(KE_T) =(2)/(5) (R^2W^2)/(v^2) \\\\`

`\frac{\text{KE}_R}{\text{KE}_T} = \frac{2(0.0211\ \text{m})^2(139\ \text{rad/s})^2}{5(23.6\ \text{m/s})^2} = 0.00618`

Therefore, the ratio of the rotational energy to the translational kinetic energy is 0.00618.