Answer :
Final answer:
In this case, the ratio of the rotational energy to the translational kinetic energy for the baseball is approximately [tex]\( 2.892 \times 10^{-6} \)[/tex].
Explanation:
Let's solve this step-by-step.
Step 1: Convert radius to meters. Given that the radius of the baseball is 4.96 cm, we need to convert it into meters for consistency in our calculations (since the translational speed is given in meters per second).
[tex]\[ \text{radius in meters} = \text{radius in centimeters} \times \frac{1 \text{ meter}}{100 \text{ centimeters}} \][/tex]
[tex]\[ \text{radius}_m = 4.96 \times \frac{1}{100} \][/tex]
[tex]\[ \text{radius}_m = 0.0496 \text{ meters} \][/tex]
Step 2: Calculate the moment of inertia for a solid sphere. The moment of inertia [tex]\( I \)[/tex] for a solid sphere is given by:
[tex]\[ I = \frac{2}{5}mr^2 \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the baseball (0.145 kg),
- [tex]\( r \)[/tex] is the radius of the baseball (0.0496 m).
Plugging in the values, we have:
[tex]\[ I = \frac{2}{5}(0.145)(0.0496^2) \][/tex]
[tex]\[ I = \frac{2}{5}(0.145)(0.0496 \times 0.0496) \][/tex]
[tex]\[ I = \frac{2}{5}(0.145)(0.0024576) \][/tex]
[tex]\[ I = 0.0001454 \text{ kg}.m^2 \][/tex]
Step 3: Calculate the rotational kinetic energy. Rotational kinetic energy [tex](\( E_{rot} \))[/tex] is given by: [tex]\[ E_{rot} = \frac{1}{2}I\omega^2 \][/tex]
where:
- [tex]\( \omega \)[/tex] is the angular velocity (97.7 rad/s).
Plugging in the values, we have:
[tex]\[ E_{rot} = \frac{1}{2}(0.0001454)(97.7^2) \][/tex]
[tex]\[ E_{rot} = \frac{1}{2}(0.0001454)(9556.29) \][/tex]
[tex]\[ E_{rot} = 0.0006947 \text{ J} \][/tex]
Step 4: Calculate the translational kinetic energy.
Translational kinetic energy [tex](\( E_{trans} \))[/tex] is given by:
[tex]\[ E_{trans} = \frac{1}{2}mv^2 \][/tex]
where:
-[tex]\( m \)[/tex] is the mass (0.145 kg),
- [tex]\( v \)[/tex] is the linear (translational) speed (57.6 m/s).
Plugging in the values, we have:
[tex]\[ E_{trans} = \frac{1}{2}(0.145)(57.6^2) \][/tex]
[tex]\[ E_{trans} = \frac{1}{2}(0.145)(3317.76) \][/tex]
[tex]\[ E_{trans} = 240.1908 \text{ J} \][/tex]
Step 5: Calculate the ratio of the rotational energy to the translational kinetic energy.
Now, we simply divide the rotational kinetic energy by the translational kinetic energy to find the ratio:
[tex]\[ \text{ratio} = \frac{E_{rot}}{E_{trans}} \] \\[/tex]
[tex]\[ \text{ratio} = \frac{0.0006947}{240.1908} \]\\[/tex]
[tex]\[ \text{ratio} \approx 2.892 \times 10^{-6} \][/tex]
So, the ratio of the rotational energy to the translational kinetic energy for the baseball is approximately [tex]\( 2.892 \times 10^{-6} \)[/tex].