Question

A semicircular ring of mass m and radius $R$ is fixed as shown. A point mass $m_0$ is moved by an external agent from $O$ to $O'$. The work done by gravitational field in moving $m_0$ from $O$ to $O'$ is:

• A. $\frac{Gmm_0}{R}\left[\left\{\frac{2}{\pi}ln(\sqrt{2}+1)\right\}-1\right]$
• B. $\frac{Gmm_0}{R}\left[\left\{\frac{1}{\pi}ln(\sqrt{2}+1)\right\}-1\right]$
• C. $\frac{Gmm_0}{R}\left[\left\{-\frac{2}{\pi}ln(\sqrt{2}+1)\right\}-1\right]$
• D. $\frac{Gmm_0}{R}\left[\left\{-\frac{1}{\pi}ln(\sqrt{2}+1)\right\}-1\right]$

Answer 1

Answer: (A)

$\frac{G m m_0}{R}\left[\frac{2}{\pi}\ln(\sqrt2+1)-1\right]$

Answer 2

To find the work done by the gravitational field when a point mass $m_0$ is moved from point $O$ to $O'$, we need to calculate the change in gravitational potential energy.

1. Potential at O: Every element of the ring is at a distance $R$, so $V(O) = -\frac{Gm}{R}$.

2. Potential at O': The distance from a mass element at $(R \cos\theta, R \sin\theta)$ to $O'$ (which is $(0, -R)$) is $r(\theta) = R \sqrt{2(1 + \sin\theta)}$.

The potential at $O'$ is $V(O') = -\frac{Gm}{\pi R \sqrt{2}} \int_0^\pi \frac{d\theta}{\sqrt{1 + \sin\theta}}$.

3. Evaluating the integral: Using the identity $1 + \sin\theta = (\sin(\frac{\theta}{2}) + \cos(\frac{\theta}{2}))^2$, the integral becomes:

$I = \int_0^\pi \frac{d\theta}{\sin(\frac{\theta}{2}) + \cos(\frac{\theta}{2})} = -2\sqrt{2} \ln(\sqrt{2} - 1)$.

4. Potential at O': $V(O') = -\frac{Gm}{\pi R \sqrt{2}} \cdot I = -\frac{2Gm}{\pi R} \ln(\sqrt{2} + 1)$.

5. Work done by gravity: $W_{grav} = V(O) - V(O') = -\frac{Gm}{R} + \frac{2Gm}{\pi R} \ln(\sqrt{2} + 1) = \frac{Gmm_0}{R} \left[ \frac{2}{\pi} \ln(\sqrt{2} + 1) - 1 \right]$.

Thus, the work done by the gravitational field is $\frac{G m m_0}{R} \left[ \frac{2}{\pi} \ln(\sqrt{2} + 1) - 1 \right]$.