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

The ratio is
`(RE)/(TE) = (2)/(3)`

Step-by-step explanation:

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

`I = (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 = (1)/(2)* I * w^2`

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

`w = (v)/(r)`

=>
`w^2 = [(v)/(r)] ^2`

So

`RE = (1)/(2)* [(2)/(3) *mr^2] * [(v)/(r) ]^2`

`RE = (1)/(3) * mv^2`

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

`TE = (1)/(2) mv^2`

So

`(RE)/(TE) = ((1)/(3) * mv^2)/((1)/(2) * m*v^2)`

`(RE)/(TE) = (2)/(3)`