A particle of mass m is suspended from a ceil | vedclass

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(A) Let the tension in the string be $T$ and the angle the string makes with the vertical be $	heta$.For the particle moving in a horizontal circle,t

Vedclass · Accepted Answer

$r \sqrt{\frac{g}{\sqrt{L^2-r^2}}}$

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(A) Let the tension in the string be $T$ and the angle the string makes with the vertical be $	heta$.For the particle moving in a horizontal circle,the forces acting on it are the tension $T$ and the weight $mg$.The vertical component of tension balances the weight: $T \cos 	heta = mg$.The horizontal component of tension provides the necessary centripetal force: $T \sin 	heta = \frac{mv^2}{r}$.Dividing the two equations,we get: $	an 	heta = \frac{v^2}{rg} \Rightarrow v = \sqrt{rg 	an 	heta}$.From the geometry of the figure,$\sin 	heta = \frac{r}{L}$,so $\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - \frac{r^2}{L^2}} = \frac{\sqrt{L^2 - r^2}}{L}$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{r/L}{\sqrt{L^2 - r^2}/L} = \frac{r}{\sqrt{L^2 - r^2}}$.Substituting this into the expression for $v$:$v = \sqrt{rg \cdot \frac{r}{\sqrt{L^2 - r^2}}} = r \sqrt{\frac{g}{\sqrt{L^2 - r^2}}}$.

$A$ particle of mass $m$ is suspended from a ceiling through a string of length $L$. If the particle moves in a horizontal circle of radius $r$ as shown in the figure,then the speed of the particle is

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