Respuesta :
					                   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
					                
					                
					             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 ) = [tex]\frac{2}{5} mr^{2}[/tex]
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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