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Colt1911 [192]
3 years ago
8

A thin walled spherical shell is rolling on a surface. What

Physics
1 answer:
ExtremeBDS [4]3 years ago
8 0

Answer:

=\frac{1/3}{5/6} = 0.4

Explanation:

Moment of inertia of given shell= \frac{2}{3} MR^2

where

M represent sphere mass

R -sphere radius

we know linear speed is given as v = r\omega

translational K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2

rotational K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2

total kinetic energy will be

K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2

K.E =\frac{5}{6} MR^2 \omega^2

fraction of rotaional to total K.E

=\frac{1/3}{5/6} = 0.4

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