A particle is moving with uniform speed along the circumference of a circle of radius R under the ac

Q: Why is the force called 'fictitious' here?

In physics, 'fictitious' usually refers to pseudo-forces in non-inertial frames. However, in the context of this specific problem, it simply implies a hypothetical or mathematically defined central force rather than a fundamental force like gravity or electromagnetism.

Q: Why did we use $m\omega^2 R$ instead of $\frac{mv^2}{R}$?

Both expressions are perfectly valid since $v = \omega R$. However, because we ultimately need to find the time period $T$, and $T$ is directly related to angular velocity by $T = \frac{2\pi}{\omega}$, using the $\omega$ form saves us an extra substitution step.

Q: What happens to the mass $m$ and constants like $2\pi$ when we use proportionalities?

Proportionality ($\propto$) allows us to ignore constant factors. Since the mass $m$ of the particle and the number $2\pi$ do not change as the radius $R$ changes, they are absorbed into the proportionality constant and do not affect the final relationship between $T$ and $R$.

Q: How did $\omega^2 \propto \frac{1}{R^4}$ become $\omega \propto \frac{1}{R^2}$?

We simply took the square root of both sides. Mathematically, if $A \propto B^2$, then $\sqrt{A} \propto \sqrt{B^2}$, which simplifies to $\sqrt{A} \propto B$. Here, the square root of $\frac{1}{R^4}$ is exactly $\frac{1}{R^2}$.

Q: What if the central force was proportional to $R$ instead of inversely proportional to $R^3$?

If $F \propto R$, then $m\omega^2 R \propto R$. The $R$ cancels out, leaving $\omega^2 \propto 1$, meaning $\omega$ is a constant. Consequently, the time period $T$ would be independent of the radius $R$. This is exactly what happens in simple harmonic motion!

Question

A particle is moving with uniform speed along the circumference of a circle of radius $R$ under the action of a central fictitious force $F$ which is inversely proportional to $R^3$. Its time period of revolution will be given by

Options:

(A) $T \propto R^2$

(B) $T \propto R^{3/2}$

(C) $T \propto R^{5/2}$

(D) $T \propto R^{4/3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a particle gracefully sweeping along the circumference of a circle of radius $R$. For any object to maintain a circular path, there must be a net force constantly pulling it towards the center. This is what we call the **centripetal force**.

In this problem, we are introduced to a "central fictitious force" $F$. While the term "fictitious" might sound like a pseudo-force from a non-inertial frame, here it simply represents a mathematically defined hypothetical force acting towards the center.

The problem gives us a crucial piece of information: this force is inversely proportional to the cube of the radius.

$$F \propto \frac{1}{R^3}$$

### The Master Equation

From the dynamics of circular motion, we know that the required centripetal force can be expressed in terms of the particle's mass $m$, its angular velocity $\omega$, and the radius $R$:

$$F_c = m \omega^2 R$$

Since our central force $F$ is the one providing this necessary centripetal pull, we can equate the two concepts. Because we are dealing with proportionalities, we can drop the equality and write:

$$m \omega^2 R \propto \frac{1}{R^3}$$

### Solving for Angular Velocity

Our goal is to find the time period $T$, which is intimately connected to the angular velocity $\omega$. Let's isolate $\omega$ in our proportionality.

First, the mass $m$ is a constant for the particle, so it doesn't affect how things scale with $R$. We can safely absorb it into the proportionality. Next, we divide both sides by $R$ (or move $R$ to the denominator on the right side):

$$\omega^2 \propto \frac{1}{R^3 \cdot R}$$

$$\omega^2 \propto \frac{1}{R^4}$$

To find $\omega$, we simply take the square root of both sides. The square root of $\frac{1}{R^4}$ is $\frac{1}{R^2}$:

$$\omega \propto \frac{1}{R^2}$$

### Final Calculation: The Time Period

We are in the home stretch! The time period $T$ of a particle in circular motion is the time it takes to complete one full revolution ($2\pi$ radians). The relationship is:

$$T = \frac{2\pi}{\omega}$$

Since $2\pi$ is just a constant number, the time period is inversely proportional to the angular velocity:

$$T \propto \frac{1}{\omega}$$

Now, we substitute our finding for $\omega$ into this relation:

$$T \propto \frac{1}{\frac{1}{R^2}}$$

When we divide by a fraction, we multiply by its reciprocal. The $R^2$ flips up to the numerator, revealing our final, elegant result:

$$T \propto R^2$$

The time period of revolution is directly proportional to the square of the radius.

Animated Solution for Physics - Laws of Motion: A particle is moving with uniform speed along the circumference of a circle of radius $R$ under the action of a central fictitious force $F$ which is inversely proportional to $R^{3}$ . Its time period of revolution will be given by

Select Answer:

Visualized Solution

$Circular Motion Setup$

$Centripetal Force$

$Given Force Relation$

$Equating the Forces$

$Solving for ω$

$Time Period Relation$

$Final Proportionality$

The Sigma Insight: Dynamics of Circular Motion

Analyzing the Setup

The Master Equation

Solving for Angular Velocity

Final Calculation

Similar Questions

A particle is moving with a uniform speed in a circular orbit of radius $R$ in a central force inversely proportional to the nth power of $R$ . If the period of rotation of the particle is $T$ , then :

A particle of mass $m$ is suspended from a ceiling through a string of length $L$ . The particle moves in a horizontal circle of radius $r$ such that $r = \frac{L}{2}$ . The speed of particle will be

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that

A tube of length $L$ is filled completely with an in-compressible liquid of mass $M$ and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $ω$ . The force exerted by the liquid at the other end is

Consider a binary star system of star A and star B with masses $m_{A}$ and $m_{B}$ revolving in a circular orbit of radii $r_{A}$ and $r_{B}$ , respectively. If $T_{A}$ and $T_{B}$ are the time period of star A and star B respectively, then

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Circular Motion Setup

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Force Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating the Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for ω

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Period Relation

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