Answer:
[tex]=\frac{1/3}{5/6} = 0.4[/tex]
Explanation:
Moment of inertia of given shell[tex] = \frac{2}{3} MR^2[/tex]
where
M represent sphere mass
R -sphere radius
we know linear speed is given as [tex]v = r\omega[/tex]
translational [tex]K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2[/tex]
rotational [tex]K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2[/tex]
total kinetic energy will be
[tex]K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2[/tex]
[tex]K.E =\frac{5}{6} MR^2 \omega^2[/tex]
fraction of rotaional to total K.E
[tex]=\frac{1/3}{5/6} = 0.4[/tex]
