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

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(A) This is a case of a conical pendulum.Let $	heta$ be the angle the string makes with the vertical.From the geometry of the setup,we have $\sin 	h

Vedclass · Accepted Answer

$\sqrt{rg}$

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(A) This is a case of a conical pendulum.Let $	heta$ be the angle the string makes with the vertical.From the geometry of the setup,we have $\sin 	heta = \frac{r}{L}$.Given $r = \frac{L}{\sqrt{2}}$,we get $\sin 	heta = \frac{L/\sqrt{2}}{L} = \frac{1}{\sqrt{2}}$.This implies $	heta = 45^{\circ}$.The forces acting on the particle are the tension $T$ in the string and the gravitational force $mg$.Resolving the tension into horizontal and vertical components:Horizontal component: $T \sin 	heta = \frac{mv^2}{r}$ (providing the necessary centripetal force).Vertical component: $T \cos 	heta = mg$ (balancing the weight).Dividing the two equations: $\frac{T \sin 	heta}{T \cos 	heta} = \frac{mv^2/r}{mg} \Rightarrow 	an 	heta = \frac{v^2}{rg}$.Since $	heta = 45^{\circ}$,$	an 45^{\circ} = 1$.Therefore,$1 = \frac{v^2}{rg} \Rightarrow v^2 = rg \Rightarrow v = \sqrt{rg}$.

$A$ particle of mass $m$ is suspended from a ceiling through a string of length $L$. The particle moves in a horizontal circle of radius $r$ such that $r = \frac{L}{\sqrt{2}}$. The speed of the particle will be:

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