Question

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

fraction of its total kinetic energy is in the form ofrotational
kinetic energy about the center of mass?

Answer 1

The fraction of rotational kinetic energy for a rolling thin-walled spherical shell can be found by comparing the expressions for rotational and translational kinetic energy.

Step-by-step explanation:

The fraction of the total kinetic energy in the form of rotational kinetic energy for a thin-walled spherical shell rolling on a surface can be determined by comparing the corresponding expressions for both forms of kinetic energy. For the rolling motion, the total kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy.

For a thin-walled spherical shell, the moment of inertia is given by I = 2/3 * M * R^2, where M is the mass of the shell and R is the radius.

Hence, the fraction of the total kinetic energy in the form of rotational kinetic energy is given by the expression:

Fraction of rotational kinetic energy = (2/5) * (I * w^2) / (0.5 * M * v^2)

where w is the angular velocity and v is the linear velocity of the shell.

Answer 2

`=(1/3)/(5/6) = 0.4`

Step-by-step explanation:

Moment of inertia of given shell
`= (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 = (1)/(2) mv^2 = (1)/(2) m(r\omega)^2`

rotational
`K.E = (1)/(2) I \omega^2 = (1)/(2) (2)/(3) MR^2 \omega^2`

total kinetic energy will be

`K.E = (1)/(2) m(r\omega)^2 + (1)/(2) (2)/(3) MR^2 \omega^2`

`K.E =(5)/(6) MR^2 \omega^2`

fraction of rotaional to total K.E

`=(1/3)/(5/6) = 0.4`