Answer:
Explanation:
The kinetic energy of the rotating sphere consist of two parts:
Let [tex]r[/tex] denote the radius of the sphere, and let [tex]v[/tex] denote the speed at which the center of this sphere is moving down the incline. Since the sphere is not slipping, the angular speed of the sphere would be [tex]\omega = v / r[/tex].
At a linear speed of [tex]v[/tex], the kinetic energy of an object of mass [tex]m[/tex] would be [tex](1/2)\, m\, v^{2}[/tex].
The rotational kinetic energy of an object depends on its moment of inertia [tex]I[/tex] and angular velocity [tex]\omega[/tex]. For a sphere of uniform density, the moment of inertia of the sphere would be [tex]I = (2/5)\, m\, r^{2}[/tex], where [tex]m[/tex] is the mass of the sphere and [tex]r[/tex] is the radius of the sphere. At an angular velocity of [tex]\omega = v / r[/tex], the rotational kinetic energy of the sphere in this question would be:
[tex]\begin{aligned} \frac{1}{2}\, I\, \omega^{2} &= \frac{1}{2}\, \left(\frac{2}{5}\, m\, r^{2}\right)\, \left(\frac{v}{r}\right)^{2} = \frac{1}{5}\, m\, v^{2}\end{aligned}[/tex].
Hence, the total kinetic energy of this sphere (linear and rotational) would be:
[tex]\displaystyle \frac{1}{2}\, m\, v^{2} + \frac{1}{5}\, m\, v^{2} = \frac{7}{10}\, m\, v^{2}[/tex].
The ratio between rotational kinetic energy and total kinetic energy would be:
[tex]\begin{aligned}\frac{\displaystyle \frac{1}{5}\, m\, v^{2}}{\displaystyle \frac{7}{10}\, m\, v^{2}} = \frac{2}{7}\, m\, v^{2}\end{aligned}[/tex].