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

fractions of its total kinetic energy is in the form ofrotational

kinetic energy about the center of mass?

Fractions of total kinetic energy in the form of rotational

kinetic energy about the center of mass = 2/5

Explanation:

Moment of Inertia for a thin walled spherical Shell, I = (2/3) * m*r^2

Translational Kinetic Energy = (1/2) * m*v^2

Rotational Kinetic Energy = (1/2) * (I) * (w^2)

Angular Velocity, w = v/r, where v is the linear velocity

Rotational Kinetic Energy = (1/2) * ((2/3) * m*r^2) * ((v/r) ^2)

Rotational Kinetic Energy = (1/3) * m*v^2

Total Kinetic Energy = (1/3 + 1/2) * m*v^2

Total Kinetic Energy = (5/6) * m*v^2

Fraction = Rotational Kinetic Energy / Total Kinetic Energy

Fraction = ((1/3) * m*v^2) / ((5/6) * m*v^2)

Fraction = 2/5