Q. A uniform solid sphere rolls without slipping. The ratio of translational to total kinetic energy is:

एक समान ठोस गोला बिना फिसले लुढ़कता है। अनुवादात्मक और कुल गतिज ऊर्जा का अनुपात है:

2:5

5:2

1:2

2:3

Total Kinetic Energy = Translational + Rotational

K.E_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

For a solid sphere, $I = \frac{2}{5}MR^2$ , and for rolling without slipping, $v = R\omega$ :

K.E_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{2} \cdot \frac{2}{5}MR^2 \cdot \left(\frac{v}{R}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2

Translational part is $\frac{1}{2}mv^2$ , so:

Ratio = \frac{\frac{1}{2} m v^{2}}{\frac{7}{10} m v^{2}} = \frac{5}{7}

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