A solid sphere is in rolling motio | vedclass

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Question

Translational kinetic energy, ${K_t} = \frac{1}{2}m{v^2}$

Rotational kinetic energy, ${K_r} = \frac{1}{2}I{\om

Vedclass · Accepted Answer

$5:7$

Vedclass · Answer

Translational kinetic energy, ${K_t} = \frac{1}{2}m{v^2}$

Rotational kinetic energy, ${K_r} = \frac{1}{2}I{\omega ^2}$

$\begin{array}{ccccc}
	herefore \,{K_t} + {K_r} = \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2}\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}m{v^2} + \frac{1}{2}\left( {\frac{2}{5}m{r^2}} ight){\left( {\frac{v}{r}} ight)^2}\
	herefore \,{K_t} + {k_r} = \frac{7}{{10}}m{v^2}\,\,\,\,\,\,\left[ {I = \frac{2}{5}m{r^2}\left( {for\,sphere} ight)} ight]\
So,\,\frac{{{K_t}}}{{{K_t} + {K_r}}} = \frac{5}{7}
\end{array}$

A solid sphere is in rolling motion. In rolling motion a body possesses translational kinetic energy $(K_t)$ as well as rotational kinetic energy $(K_r)$ simultaneously. The ratio $K_t : (K_t + K_r)$ for the sphere is

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