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Question Number 70543 by mr W last updated on 05/Oct/19

Commented by mr W last updated on 05/Oct/19

$$asemi-cylinderwithradiusRand$$$$massMrestsonaroughtableasshown.$$$$whenasmallimpulseisgiventothe$$$$semi-cylinder,itbeginstooscillate.$$$$findthefrequencyoftheoscillation$$$$intermsofR.$$

Commented by ajfour last updated on 06/Oct/19

Commented by mr W last updated on 06/Oct/19

$$dA=\frac{R^{2}d\theta }{2}$$$$y_{C}A=2{\int }_0^{\frac{\pi }{2}}\frac{R^{2}d\theta }{2}\times \frac{2R}{3}\sin \theta =\frac{2R^3}{3}{\int }_0^{\frac{\pi }{2}}\sin \theta d\theta =\frac{2R^3}{3}$$$$y_C=\frac{2R^3}{3\times \frac{\pi R^2}{2}}=\frac{4R}{3\pi }$$$${dI}_0={\int }_0^R\rho rd\theta r^{2}dr=\frac{\rho R^{4}d\theta }{4}$$$$I_0=\frac{\rho R^4\pi }{4}=\frac{\rho \pi R^2}{2}\times \frac{R^2}{2}=\frac{{MR}^2}{2}$$$$I_0=I_C+{My}_C^2$$$$I_P=I_C+M{(R-y_C)}^2$$$$\Rightarrow I_P=I_0+M{(R-y_C)}^2-{My}_C^2$$$$\Rightarrow I_P=I_0+MR(R-2y_C)$$$$\Rightarrow I_P=\frac{{MR}^2}{2}+MR(R-\frac{8R}{3\pi })=\frac{{MR}^2(9\pi -16)}{6\pi }$$$$f=\frac{1}{2\pi }\sqrt{\frac{4MgR}{3\pi I_P}}$$$$f=\frac{1}{2\pi }\sqrt{\frac{4MgR}{3\pi \times \frac{{MR}^2(9\pi -16)}{6\pi }}}$$$$\Rightarrow f=\frac{1}{\pi }\sqrt{\frac{2g}{(9\pi -16)R}}$$

Commented by mr W last updated on 06/Oct/19

Commented by ajfour last updated on 06/Oct/19

$$Sir,youmissed2from{2}^{nd}to$$$$3^{rd}line.y_C=\frac{4R}{3\pi }$$$$Andthanks,irealise{I}_P{isn}^{\prime }t$$$$thatdifficulttoobtain.$$

Commented by mr W last updated on 06/Oct/19

$$thankstoosir!$$$$nowitshouldbecorrect.$$

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