A thin uniform annular disc (see figure) of mass $M$ has outer radius $4R$ and inner radius $3R$. The work required to take a unit mass from point $P$ on its axis to infinity is [image:8_image41.emf]

Question

Dr. Abubacker · Accepted Answer

(a) $W = \mathrm{\Delta}U = U_{f} - U_{i} = U_{\infty} - U_{P}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = - U_{P} = - mV_{P}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = - V_{P}(as\ m = 1)$
Potential at point $P$ will be obtained by in integration as given below. Let $dM$ be the mass of small rings as shown
[image:8_image76.emf]
$\ \ \ \ \ \ \ \ \ \ \ \ \ dM = \frac{M}{\pi(4R)^{2} - \pi(3R)^{2}}(2\pi r)dr$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = \frac{2Mr\ dr}{{7R}^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ {dV}_{P} = - \frac{G.dM}{\sqrt{16R^{2} + r^{2}}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = - \frac{2GM}{7R^{2}}\int_{3R}^{4R}{\frac{r}{\sqrt{16R^{2} + r^{2}}} \bullet dr}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = - \frac{2GM}{7R}\left( 4\sqrt{2} - 5 ight)$
$	herefore\ \ \ \ \ \ \ \ \ \ W = + \frac{2GM}{7R}(4\sqrt{2} - 5)$

Q. A thin uniform annular disc (see figure) of mass $M$ has outer radius $4R$ and inner radius $3R$. The work required to take a unit mass from point $P$ on its axis to infinity is

Explanation

Reviewed by Dr. Abubacker

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