The ratio between rotational energy and translational kinetic energy is [tex]1.47\cdot 10^{-3}[/tex]
Explanation:
The translational kinetic energy of the ball is given by:
[tex]KE_t = \frac{1}{2}mv^2[/tex]
where
m is the mass of the ball
v is the speed of the ball
The rotational kinetic energy of the ball is given by
[tex]KE_r = \frac{1}{2}I\omega^2[/tex]
where
I is the moment of inertia
[tex]\omega[/tex] is the angular speed
The moment of inertia of a solid sphere through its axis is given by
[tex]I=\frac{2}{5}mR^2[/tex]
where
m is the mass of the ball
R is its radius
Substituting into the previous equation,
[tex]KE_r = \frac{1}{2}(\frac{2}{5}mR^2)\omega^2 = \frac{1}{5}mR^2 \omega^2[/tex]
The ratio between the two energies is
[tex]\frac{KE_r}{KE_t}=\frac{\frac{1}{5}mR^2 \omega^2}{\frac{1}{2}mv^2}=\frac{2R^2 \omega^2}{5v^2}[/tex]
And substituting:
R = 3.91 cm = 0.0391 m
v = 33.6 m/s
[tex]\omega=52.1 rad/s[/tex]
we find:
[tex]ratio = \frac{2(0.0391)^2(52.1)^2}{5(33.6)^2}=1.47\cdot 10^{-3}[/tex]
Learn more about kinetic energy:
brainly.com/question/6536722
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