The potential energy of a particle of mass m at a distance r from a fixed point O is given by V(r) =

Q: Why do we take the negative derivative of potential energy to find the force?

In physics, a conservative force always acts in the direction that minimizes potential energy. The negative gradient mathematically represents this 'downhill' push.

Q: How do we know the force is directed towards the center?

The potential energy $V(r) = kr^2/2$ increases as you move away from the center. Since force pushes towards lower potential energy, it must point inwards, making it an attractive central force.

Q: Why is the centripetal force equated to the magnitude of the derived force?

Centripetal force isn't a new, separate force; it's just the name for whatever net force is keeping an object in circular motion. Here, the central force we derived *is* the centripetal force.

Q: Why is the angular momentum simply $mvR$? Doesn't it involve a cross product?

Yes, angular momentum is $\vec{L} = \vec{r} \times \vec{p}$. But in a circular orbit, the position vector $\vec{r}$ and velocity vector $\vec{v}$ are always at exactly 90 degrees to each other. The sine of 90 degrees is 1, so the magnitude simplifies beautifully to $mvR$.

Q: Could we have solved this using conservation of total mechanical energy?

While total energy is conserved, it doesn't directly give us the velocity without knowing the total energy value first. The force balance equation is the most direct path to finding the orbital speed.

Question

The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $V(r) = \frac{kr^2}{2}$, where $k$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $R$ about the point $O$. If $v$ is the speed of the particle and $L$ is the magnitude of its angular momentum about $O$, which of the following statements is (are) true ?

Options:

(A) $v = \sqrt{\frac{k}{2m}}R$

(B) $v = \sqrt{\frac{k}{m}}R$

(C) $L = \sqrt{mk}R^2$

(D) $L = \sqrt{\frac{mk}{2}}R^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a particle of mass $m$ gracefully sweeping through a perfect circular orbit of radius $R$ around a central point $O$. It's not just floating aimlessly; it's bound to this path by an invisible tether. The problem gives us a crucial clue about this tether: the potential energy of the particle is $V(r) = \frac{kr^2}{2}$.

Whenever you see potential energy in a physics problem, your first instinct should be to think about the force. In the realm of conservative forces, potential energy is like a landscape, and the force is the tendency to roll downhill. Mathematically, this is expressed as the negative gradient of the potential energy.

### The Master Equation

Let's find that force. By taking the derivative of the potential energy with respect to the distance $r$, we get:

$$F = -\frac{dV}{dr} = -\frac{d}{dr}\left(\frac{kr^2}{2}ight) = -kr$$

The negative sign is incredibly important here. It tells us that the force is directed opposite to the outward radial direction. In other words, it's an **attractive central force** pulling the particle towards the center $O$.

Now, let's bridge this with what we know about circular motion. For any object to travel in a circle, it needs a centripetal force to constantly change its direction. The magnitude of this required force is $\frac{mv^2}{R}$. In our scenario, the attractive force we just calculated is the one playing the role of the centripetal force.

So, at the orbit's radius $r = R$, we can equate the magnitude of our central force to the centripetal force requirement:

$$kR = \frac{mv^2}{R}$$

### Final Calculation

This equation is the key to unlocking the particle's speed. Let's solve for $v$:

$$v^2 = \frac{kR^2}{m}$$

$$v = \sqrt{\frac{k}{m}}R$$

This perfectly matches option (B).

But the journey isn't over; we also need to find the angular momentum, $L$. For a point mass, angular momentum is the cross product of the position vector and linear momentum ($\vec{L} = \vec{r} 	imes \vec{p}$). However, circular orbits offer a beautiful simplification. Because the velocity vector is always perfectly perpendicular to the position vector, the magnitude of the angular momentum is simply the product of mass, speed, and radius:

$$L = mvR$$

Let's substitute the speed $v$ we just found into this equation:

$$L = m\left(\sqrt{\frac{k}{m}}Right)R$$

By bringing the mass $m$ inside the square root (where it becomes $m^2$) and multiplying the $R$ terms, we get:

$$L = \sqrt{m^2 \cdot \frac{k}{m}} R^2 = \sqrt{mk}R^2$$

This matches option (C). Thus, by carefully unpacking the relationship between potential energy, force, and circular motion, we've found that both the speed and angular momentum expressions in options (B) and (C) are correct.

Select Answer:

Visualized Solution

$R, m$

$F = - \frac{d V}{d r}$

$F = - k r$

$F_{c} = \frac{m v ^{2}}{R}$

$k R = \frac{m v ^{2}}{R}$

$v = \frac{k}{m} R$

$L = m v R$

$L = mk R^{2}$

$B, C$

The Sigma Insight: Kinetic Energy, Potential Energy and Power

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R,m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F=−drdV​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F=−kr

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Fc​=Rmv2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } kR=Rmv2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=mk​​R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L=mvR

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L=mk​R2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B,C