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written 4.9 years ago by |

Dist of CG from surface = $(\frac{4r}{3\pi})$

ANSLOM

IT = $\sum$ of all moments

+$ I_o \theta = -mg sin \theta \times (\frac{4r}{3\pi})$

$sin \theta \approx \theta$

$\therefore$ $Io \theta = -mg \theta (\frac{4r}{3\pi})$

$I_o \theta + mg (\frac{4r}{3\pi}) \theta = 0$

$ \theta + mg (\frac{4r}{3\pi}) \theta = 0$

$\frac{\theta + mg (\frac{4r}{3 \pi}) \theta}{I_o}$ = 0

$w_n = \sqrt{ \frac{mg (\frac{4r}{3 \pi})}{I_o}}$

where $= I_o = \frac{mR^2}{2}$

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