દળ m અને ત્રિજ્યા r નો ઘન ગોળો ઢોળાવવાળા સમ | vedclass

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Question

When a solid sphere having

$	ext { mass }=m$

Radius $=r$

Now the Kinetic energy wllbe

The total Kinetic energy (KE) of an object is given

Vedclass · Accepted Answer

$\frac{2}{5}$ ચાકગતિ  $\,\frac{3}{5}$ સ્થાનાંતરિત

Vedclass · Answer

When a solid sphere having

$	ext { 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

$KE =\frac{1}{2} mv ^2+\frac{1}{2} 	imes \frac{2}{5} mr ^2\left(\frac{ v ^2}{ r }ight)$

$=\frac{1}{2} mv ^2+\frac{2}{10} mv ^2$

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.

દળ $m $ અને ત્રિજ્યા $ r$ નો ઘન ગોળો ઢોળાવવાળા સમતલ પરથી રોલિંગ કરીને નીચે આવે છે ત્યારે ગતિઊર્જા....

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