Question

Consider a thin uniform annular disc of mass $m=4\text{kg}$ having inner and outer radius 3 m and 4 m respectively. A point mass $m=2\text{kg}$ is taken from point P to infinity so that at infinity it has kinetic energy of 20 J. If work done by external agent in this process is $W=\{20+\frac{16G}{7}(N\sqrt{2}-5\left)\right\}J$ (G is universal gravitational constant in SI unit). Find the value of N

• A. 4
• B. 4.00
• C. 4.0

Answer 1

Answer: (A), (B), (C)

We know the work requird 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 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,

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

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