please help.....

Asked by  | 26th Jul, 2009, 01:43: PM

Expert Answer:

When a ball rolls down without slipping , it possesses both rotational K.E. and translational K.E. Total K.E. of solid ball

K = K r + Kt

or, K = (Iω 2 / 2)  +  ( mv2 /2)

         = (1/2) ( 2MR2 / 5 ) ω 2  + ( mv2 /2)

         =  ( 1/5) MR2 ω 2   + ( mv2 /2)

         = ( 1/5)mv2  + ( mv2 /2)

or, K = ( 7/10)  mv2

Rotational K.E. =  (1/5)mv2 

Total K.E.    =   ( 7/10)  mv2

Fractional rotatinal K.E.  = (1/5)mv2  /  ( 7/10)  mv2

                                              =  2/7

Answered by  | 26th Jul, 2009, 04:11: PM

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