Answer:
The ratio of the rotational energy to the translational kinetic energy is 0.00618.
Explanation:
The rotational kinetic energy of a solid sphere is given by
[tex]\text{KE}_R = \frac{1}{5}mr^2\omega^2[/tex]
where m is its mass, r is its radius and ω is its angular speed.
The translational kinetic energy is given by
[tex]\text{KE}_T = \frac{1}{2}mv^2[/tex]
where v is its linear speed.
The ratio of the rotational energy to the translational kinetic energy is
[tex]\dfrac{\text{KE}_R}{\text{KE}_T} = \dfrac{\frac{1}{5}mr^2\omega^2}{\frac{1}{2}mv^2} = \dfrac{2r^2\omega^2}{5v^2}[/tex]
From the question,
r = 2.11 cm = 0.0211 m
ω = 139 rad/s
v = 23.6 m/s
[tex]\dfrac{\text{KE}_R}{\text{KE}_T} = \dfrac{2(0.0211\ \text{m})^2(139\ \text{rad/s})^2}{5(23.6\ \text{m/s})^2} = 0.00618[/tex]
Answer:
The ratio of the rotational energy to the translational kinetic energy is 0.00618
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
[tex]\frac{KE_{rotational}}{KE_{translational}} = \frac{\frac{1}{2} Iw^2}{\frac{1}{2} MV^2}[/tex]
But the rotational inertia of a uniform sphere is I = 2mr²/5
[tex]\frac{1}{2}Iw^2 = \frac{1}{2} \times \frac{2}{5} MR^2w^2[/tex]
[tex]\frac{KE_R}{KE_T} =\frac{2}{5} \frac{R^2W^2}{v^2} \\\\[/tex]
[tex]\dfrac{\text{KE}_R}{\text{KE}_T} = \dfrac{2(0.0211\ \text{m})^2(139\ \text{rad/s})^2}{5(23.6\ \text{m/s})^2} = 0.00618[/tex]
Therefore, the ratio of the rotational energy to the translational kinetic energy is 0.00618.
