Question

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

• A. $\frac{2GM}{7R}(4\sqrt{2}-5)$
• B. $-\frac{2GM}{7R}(4\sqrt{2}-5)$
• C. $\frac{GM}{4R}$
• D. $\frac{2GM}{5R}(\sqrt{2}-1)$

Answer 1

Answer: (B)

$-\frac{2GM}{7R}(4\sqrt{2}-5)$

We know that the work required to take a unit mass from P to infinity =$-V_p$, where $-V_p$ is the gravitational potential at P due to the disc. To find $V_p$, we divide the disc into small elements, each of thickness dr. Consider one such element at a distance r from the center of the disc as shown.

Mass of the element $dm=\frac{M\left(2\pi rdr\right)}{\pi (4R{)}^2-\pi (3R{)}^2}$$=\frac{2Mrdr}{7R^2}$

Thus,

$V_p=-{\int }_{3R}^{4R}\frac{Gdm}{\sqrt{r^2+16R^2}}$$=-\frac{2MG}{7R^2}{\int }_{3R}^{4R}\frac{rdr}{(r^2+16R^2{)}^{1/2}}$

Putting $r^2+16R^2=x^2$,we get $2rdr=2xdx$ or $rdr=xdx$

When $r=3R,x=\sqrt{9R^2+16R^2}=5R$

When $r=4R,x=\sqrt{16R^2+16R^2}=4\sqrt{2}R$

$V_p=-\frac{2MG}{7R^2}{\int }_{5R}^{4\sqrt{2}R}dx=-\frac{2MG}{7R^2}(4\sqrt{2}-5)R$

or

$-V_p=\frac{2GM}{7R}(4\sqrt{2}-5)$