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દળ $m $ અને ત્રિજ્યા $ r$ નો ઘન ગોળો ઢોળાવવાળા સમતલ પરથી રોલિંગ કરીને નીચે આવે છે ત્યારે ગતિઊર્જા....
$\frac{1}{2}$ ચાકગતિ $\,frac{1}{2}$ સ્થાનાંતરિત
$\frac{2}{7} $ ચાકગતિ $\frac{5}{7}$ સ્થાનાંતરિત
$\frac{2}{5}$ ચાકગતિ $\,\frac{3}{5}$ સ્થાનાંતરિત
$\frac{5}{7}$ ચાકગતિ $\frac{2}{7}$ સ્થાનાંતરિત
Solution
When a solid sphere having
$\text { mass }=m$
Radius $=r$
Now the Kinetic energy wllbe
The total Kinetic energy (KE) of an object is given by
$KE =\frac{1}{2} mv ^2+\frac{1}{2} I \omega^2$
Where I is the moment of inertia of the object and $\omega$ is the angular momentum. For a solid sphere, the moment of inertia is given by
$I=\frac{2}{5} m r^2$
Angular velocity would be derived from its radius and linear velocity
$\omega=\frac{V}{r}$
So the total equation would
$KE =\frac{1}{2} mv ^2+\frac{1}{2} \times \frac{2}{5} mr ^2\left(\frac{ v ^2}{ r }\right)$
$=\frac{1}{2} mv ^2+\frac{2}{10} mv ^2$
To get the percentage attributed to rotational energy, we would divide the rotational part of the energy by the total energy
$\frac{\frac{2}{10} m v^2}{\frac{1}{2} m v^2+\frac{2}{10} m v^2}=\frac{\frac{2}{10} m v^2}{\frac{7}{10} m v^2}$
$=\frac{2}{7}$
Hence $\frac{5}{7}$ is rotational and $\frac{2}{7}$ translational.
