Question

A solid sphere is rolling on a surface. what fraction of its total kinetic energy is in the form of rotational kinetic energy about the center of mass

Answer 1

Total KE = KE (rotational) + KE (translational) Moment of inertia of sphere is I = (2/5)mr^2 So KE (rotational) = (1/2) x I x w^2 = (1/2) x (2/5)mr^2 x w^2 = (1/5) x m x r^2 x w^2 KE (translational) = (1/2) x m x v^2 = (1/2) x m x (rw)^2 = (1/2) x m x r^2 x w^2 Hence KE = (1/5) x m x r^2 x w^2 + (1/2) x m x r^2 x w^2 = m x r^2 x w^2 ((1/5) + (1/2)) KE = (7/10) m x r^2 x w^2 Calculating the fraction of rotational kinetic energy to total kinetic energy, = rotational kinetic energy / total kinetic energy = (1/5) x m x r^2 x w^2 / (7/10) m x r^2 x w^2 = (1/5) / (7/10) = 2 / 7 The answer is 2 / 7

Answer 2

The fraction of its total kinetic that is in the form of rotational kinetic energy about the center of mass :  2 / 7

Given that :

Moment of inertia of solid sphere ( I ) = $\frac{2}{5} mr^{2}$

Determine the fraction Kinetic energy that is rotational

Total kinetic energy(  K.E )  = Rotational K.E + Translational K.E

                            K.E (total) =  ( 1/2 )Iw² + ( 1/2 )mv² ---- ( 1 )

where : I (moment of inertia )

equation ( 1 ) becomes

K.E (total) = ( 1/5 )mr²w²  +  ( 1/2 )mr²w²

                = mr²w²( 1/5 + 1/2 )

Final step : determine the fraction of Total K.E that is rotational K.E

K.E ( rotational ) / K.E ( total )

= ( 1/5 )mr²w² / mr²w²( 1/5 + 1/2 )

= 2 / 7

Hence we can conclude that The fraction of its total kinetic that is in the form of rotational kinetic energy about the center of mass :  2 / 7.

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