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  • A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is -<div

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is -

Option: 1

\mathrm{\frac{2}{5}}


Option: 2

\mathrm{\frac{2}{7}}


Option: 3

\mathrm{\frac{1}{5}}


Option: 4

\mathrm{\frac{7}{10}}


Answers (1)

best_answer

For solid spherical ball,
\mathrm{I=\frac{2}{5} m r^{2}}
\mathrm{KE_{\text {Rotationd }} =\frac{1}{2} I \, \omega^{2} }
                          \mathrm{=\frac{1}{2} \times\left(\frac{2}{5} m r^{2}\right) \frac{v^{2}}{r^{2}}}  
                           =\mathrm{\frac{2}{5}\left(\frac{1}{2} m v^{2}\right)}
\mathrm{K E_{\text {Rotation }}=\frac{2}{5}\,\, K E_{\text {transalalion }}}
\mathrm{K E_{\text {total }}=K E_{R}+K E_{T}}
                  \mathrm{= \frac{7}{5}\, K E_{T} }
\mathrm{\frac{K E_{R}}{K E_{\text {total }}}=\frac{2}{7} }

The correct option is (2)
   
                  

                          
 

Posted by

Deependra Verma

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