A thin rod of length L is bent to form a semi | vedclass

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(D) The length of the rod is $L$,which forms a semicircle of radius $R$. Thus,$L = \pi R$,which implies $R = \frac{L}{\pi}$.The linear mass density of

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$ - \frac{\pi GM}{L} $

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(D) The length of the rod is $L$,which forms a semicircle of radius $R$. Thus,$L = \pi R$,which implies $R = \frac{L}{\pi}$.The linear mass density of the rod is $\lambda = \frac{M}{L}$.Consider a small element of the rod of length $dl = R \, d	heta$ at an angle $	heta$. The mass of this element is $dm = \lambda \, dl = \left( \frac{M}{L} ight) R \, d	heta$.The gravitational potential $dV$ at the centre $O$ due to this element is $dV = -\frac{G \, dm}{R}$.Substituting $dm$ and $R$,we get $dV = -\frac{G}{R} \left( \frac{M}{L} ight) R \, d	heta = -\frac{GM}{L} \, d	heta$.To find the total potential $V$,we integrate $dV$ from $	heta = 0$ to $	heta = \pi$:$V = \int_{0}^{\pi} -\frac{GM}{L} \, d	heta = -\frac{GM}{L} [	heta]_{0}^{\pi} = -\frac{\pi GM}{L}$.

$A$ thin rod of length $L$ is bent to form a semicircle. The mass of the rod is $M$. What will be the gravitational potential at the centre of the circle?

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