Question

The center of mass of a pitched baseball of radius 4.37 cm moves at 51 m/s. The ball spins about an axis through its center of mass with an angular speed of 199 rad/s. Treat the baseball as if it is a solid sphere rotating about its center (I = 2/5 MR^2). Calculate the ratio of the rotational energy to the translational kinetic energy.

Answer 1

To calculate the ratio of the rotational energy to the translational kinetic energy of the baseball, we first need to compute both types of energy.

### Step 1: Calculate the Translational Kinetic Energy (TKE)

The translational kinetic energy \( TKE \) is given by the formula:

\[
TKE = \frac{1}{2} M v^2
\]

where:
- \( M \) is the mass of the baseball,
- \( v \) is the translational speed of the baseball (51 m/s).

### Step 2: Determine the Mass of the Baseball

Assuming a standard baseball mass, we can use \( M = 0.145 \, \text{kg} \) (the mass of a typical baseball).

### Step 3: Calculate \( TKE \)

Now substituting the values into the TKE formula:

\[
TKE = \frac{1}{2} (0.145) (51^2)
\]
\[
= \frac{1}{2} (0.145) (2601)
\]
\[
= 0.0725 \times 2601 \approx 188.174 \, \text{J}
\]

### Step 4: Calculate the Rotational Kinetic Energy (RKE)

The rotational kinetic energy \( RKE \) is given by the formula:

\[
RKE = \frac{1}{2} I \omega^2
\]

where:
- \( I = \frac{2}{5} M R^2 \) is the moment of inertia of a solid sphere,
- \( \omega \) is the angular speed (199 rad/s),
- \( R \) is the radius of the sphere (4.37 cm = 0.0437 m).

### Step 5: Calculate Moment of Inertia \( I \)

Let's calculate \( I \):

\[
I = \frac{2}{5} M R^2 = \frac{2}{5} (0.145) (0.0437^2)
\]
\[
= \frac{2}{5} (0.145) (0.00191069) \approx \frac{2}{5} (0.0002777) \approx 0.00011108 \, \text{kg}\cdot\text{m}^2
\]

### Step 6: Calculate \( RKE \)

Now substituting \( I \) and \( \omega \) into the \( RKE \) formula:

\[
RKE = \frac{1}{2} (0.00011108) (199^2)
\]
\[
= \frac{1}{2} (0.00011108) (39601)
\]
\[
\approx 0.00005554 \times 39601 \approx 2.200 \, \text{J}
\]

### Step 7: Calculate the Ratio

Now, we can find the ratio of rotational kinetic energy to translational kinetic energy:

\[
\text{Ratio} = \frac{RKE}{TKE} = \frac{2.200}{188.174} \approx 0.0117
\]

### Conclusion

Thus, the ratio of the rotational energy to the translational kinetic energy of the baseball is approximately:

\[
\boxed{0.0117}
\]