Question

A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:

Answer 1

Answer:

The ratio  is  $\frac{RE}{TE} = \frac{2}{3}$

Explanation:

Generally  the Moment of inertia of a spherical object (shell) is mathematically represented as

              $I = \frac{2}{3} * m r^2$

Where m is  the mass of the spherical object

       and   r is the radius  

Now the the rotational kinetic energy can be mathematically represented as

       $RE = \frac{1}{2}* I * w^2$

Where  $w$ is the angular velocity which is mathematically represented as

             $w = \frac{v}{r}$

=>           $w^2 = [\frac{v}{r}] ^2$

So

             $RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2$

            $RE = \frac{1}{3} * mv^2$

Generally the transnational  kinetic energy of this motion is  mathematically represented as

                $TE = \frac{1}{2} mv^2$

So  

      $\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}$

       $\frac{RE}{TE} = \frac{2}{3}$