A thin uniform annular disc of mass M has outer radius 4R and inner radius 3R. The work required to

Q: Why is the work done equal to $-V_P$ instead of $+V_P$?

The work done by an external agent in moving a mass slowly is equal to the change in potential energy: $W = U_{\infty} - U_P$. Since the potential energy at infinity is zero ($U_{\infty} = 0$), we get $W = -U_P$. For a unit mass ($m=1$), $U_P = m V_P = V_P$, which gives $W = -V_P$.

Q: How did we get the area of the annular disc as $7\pi R^2$?

The annular disc is a ring-like structure with an outer radius of $4R$ and an inner radius of $3R$. Its area is the difference between the area of the outer circle and the inner circle: $A = \pi (4R)^2 - \pi (3R)^2 = 16\pi R^2 - 9\pi R^2 = 7\pi R^2$.

Q: Why is the distance of every point on the elemental ring to point $P$ the same?

Because of the axial symmetry of the setup. Point $P$ lies on the central axis perpendicular to the plane of the disc. Every point on a concentric ring of radius $r$ is at the same horizontal distance $r$ from the axis and the same vertical height $4R$ from the center, making the slant distance $\sqrt{(4R)^2 + r^2}$ identical for all points on that ring.

Q: Can we solve this by subtracting the potential of a smaller disc from a larger disc?

Yes, absolutely! You can find the potential of a solid disc of radius $4R$ and subtract the potential of a solid disc of radius $3R$. However, you must be careful to use the correct mass of each hypothetical disc based on the uniform surface mass density $\sigma$.

Q: What would change if the particle was projected with some initial velocity instead of being moved slowly?

If the particle was projected with an initial velocity, it would have kinetic energy. In that case, you would use the conservation of mechanical energy ($K_i + U_i = K_f + U_f$) to find its final speed at infinity, rather than just calculating the work done by an external agent.

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

Options:

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

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

(C) $\frac{GM}{4R}$

(D) $\frac{2GM}{5R} (\sqrt{2} - 1)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a flat, ring-like disc—an **annular disc**—lying in the horizontal plane.

It has an inner boundary of radius $3R$ and an outer boundary of radius $4R$.

Directly above its center, at a height of $4R$ along the vertical axis, sits our point of interest, point $P$.

We want to find out how much work is needed to take a tiny unit mass from this point $P$ all the way to infinity, completely escaping the disc's gravitational pull.

### The Work-Energy Connection

To find the work done, we use the **work-energy theorem**.

When we move a mass slowly, the work done by an external agent is simply the change in potential energy:

$$W = U_{\infty} - U_P$$

Since gravity fades to zero at infinite distance, the potential energy at infinity is zero ($U_{\infty} = 0$).

This means the work done is simply the negative of the potential energy at point $P$:

$$W = -U_P$$

And since we are moving a **unit mass** ($m = 1$), the work is exactly equal to the negative of the gravitational potential at point $P$:

$$W = -V_P$$

So, our main task is to find this potential, $V_P$.

### Setting up the Calculus

Before we can integrate to find the potential, we need to know how the mass is distributed across the disc.

The disc is uniform, so we can define a constant **surface mass density**, $\sigma$.

The area of this annular shape is the area of the outer circle minus the area of the inner circle:

$$A = \pi (4R)^2 - \pi (3R)^2 = 16\pi R^2 - 9\pi R^2 = 7\pi R^2$$

Dividing the total mass $M$ by this area, we get our surface mass density:

$$\sigma = \frac{M}{7\pi R^2}$$

Now, let's use the power of calculus! We divide the entire annular disc into infinitely many thin concentric rings.

Let's focus on one such representative ring of radius $r$ and extremely small width $dr$.

The area of this thin strip is its circumference, $2\pi r$, multiplied by its width, $dr$:

$$dA = 2\pi r \, dr$$

To find its tiny mass, $dM$, we multiply this area by our surface mass density, $\sigma$:

$$dM = \sigma \, dA = \left(\frac{M}{7\pi R^2}\right) (2\pi r \, dr) = \frac{2Mr \, dr}{7R^2}$$

Next, let's find the gravitational potential at point $P$ due to just this single elemental ring.

Because of the symmetry, every single point on this ring is at the exact same distance from point $P$.

Using the Pythagorean theorem on our right-angled triangle, this distance is:

$$s = \sqrt{(4R)^2 + r^2} = \sqrt{16R^2 + r^2}$$

Since potential is a scalar quantity, we can write the elemental potential, $dV_P$, simply as:

$$dV_P = -\frac{G \, dM}{s} = -\frac{G \, dM}{\sqrt{16R^2 + r^2}}$$

Substituting our expression for $dM$:

$$dV_P = -\frac{2GM}{7R^2} \frac{r \, dr}{\sqrt{16R^2 + r^2}}$$

To find the total potential, $V_P$, produced by the entire disc, we must sum up the contributions of all such rings by integrating from the inner radius $3R$ to the outer radius $4R$:

$$V_P = -\frac{2GM}{7R^2} \int_{3R}^{4R} \frac{r}{\sqrt{16R^2 + r^2}} \, dr$$

### The Integration Journey

To solve this integral, we can use a simple substitution.

Let's define a new variable, $t^2$, equal to $16R^2 + r^2$.

Differentiating both sides gives:

$$2t \, dt = 2r \, dr \implies r \, dr = t \, dt$$

We also need to update our integration limits for this new variable:

- When $r = 3R$, $t = \sqrt{16R^2 + 9R^2} = 5R$
- When $r = 4R$, $t = \sqrt{16R^2 + 16R^2} = 4\sqrt{2}R$

Let's substitute these new terms and limits back into our integral:

$$V_P = -\frac{2GM}{7R^2} \int_{5R}^{4\sqrt{2}R} \frac{t \, dt}{t}$$

The $t$ in the numerator and denominator cancel out beautifully, leaving us with the simplest possible integral:

$$V_P = -\frac{2GM}{7R^2} \int_{5R}^{4\sqrt{2}R} dt$$

Integrating $dt$ gives us simply $t$:

$$V_P = -\frac{2GM}{7R^2} [t]_{5R}^{4\sqrt{2}R}$$

Evaluating this at the upper and lower limits, we get:

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

Factoring out $R$ from the bracket and dividing it by $R^2$ in the denominator leaves us with a single $R$ in the denominator:

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

### The Grand Finale

Finally, let's calculate the work done.

Using our relation $W = -V_P$:

$$W = -\left[ -\frac{2GM}{7R} (4\sqrt{2} - 5) \right]$$

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

This perfectly matches **option (a)**!

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

Select Answer:

Visualized Solution

Visualizing the Annular Disc and Point $P$

Connecting Work and Gravitational Potential

Calculating the Surface Mass Density

Selecting an Elemental Ring

Potential of the Elemental Ring at Point $P$

Setting up the Integral for Total Potential

Solving the Integral: Substitution

Evaluating the Integral

Simplifying the Potential Expression

Calculating the Work Done

Exploring Further Variations

The Sigma Insight: Gravitational Potential and Potential Energy

Analyzing the Setup

The Work-Energy Connection

Setting up the Calculus

The Integration Journey

The Grand Finale

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