Answer:
A. Its translational kinetic energy is larger than its rotational kinetic energy.
Explanation:
Given that
Radius = R
Mass = M
We know that mass moment of inertia for the solid sphere
[tex]I=\dfrac{2}{5}MR^2[/tex]
Lets take angular speed =ω
Linear speed =V
Condition for pure rolling , V= ω R
Rotation energy ,RE
[tex]RE=\dfrac{1}{2}I\omega^2[/tex]
[tex]RE=\dfrac{1}{2}\times \dfrac{2}{5}MR^2\times \omega^2[/tex]
[tex]RE=\dfrac{1}{2}\times \dfrac{2}{5}MR^2\times \omega^2[/tex]
[tex]RE=\dfrac{1}{5}\times MR^2\times \omega^2[/tex]
[tex]RE=\dfrac{1}{5}\times MV^2[/tex]
RE= 0.2 MV²
The transnational kinetic energy TE
[tex]TE=\dfrac{1}{2}MV^2[/tex]
TE= 0.5 MV²
From above we can say that transnational energy is more than rotational energy.
Therefore the answer is A.
