Question

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 = \( \frac{L}{\sqrt{2}} \). The speed of particle will be :

• A. \( \sqrt{rg} \)
• B. \( \sqrt{\frac{rg}{2}} \)
• C. \( \sqrt{2rg} \)
• D. \( 2\sqrt{rg} \)

Hint:

For any conical pendulum problem, the key relation is \( \tan\theta = \frac{v^2}{rg} \). First, find the angle \(\theta\) from the geometry (\(L, r, h\)), and then use this relation to find the unknown quantity (like `v` or the time period).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem describes a conical pendulum. A mass `m` attached to a string of length `L` revolves in a horizontal circle of radius `r`. We are given a specific relation between `r` and `L` and are asked to find the speed `v` of the particle.
Step 2: Key Formula or Approach:
We will analyze the forces acting on the particle. The tension `T` in the string provides both the vertical force to counteract gravity and the horizontal centripetal force required for circular motion. Let \(\theta\) be the angle the string makes with the vertical.
The forces are:
1. Vertically: \( T \cos\theta = mg \)
2. Horizontally: \( T \sin\theta = \frac{mv^2}{r} \)
Step 3: Detailed Explanation:
First, let's determine the angle \(\theta\) from the given geometry. In the right-angled triangle formed by the string, the vertical axis, and the radius, we have:
\[ \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{r}{L} \] We are given that \( r = \frac{L}{\sqrt{2}} \). Substituting this in the equation for \(\sin\theta\):
\[ \sin\theta = \frac{L/\sqrt{2}}{L} = \frac{1}{\sqrt{2}} \] This means the angle is \( \theta = 45^\circ \).
Now, we use the force balance equations. Divide the horizontal force equation by the vertical force equation to eliminate the tension `T`:
\[ \frac{T \sin\theta}{T \cos\theta} = \frac{mv^2/r}{mg} \] \[ \tan\theta = \frac{v^2}{rg} \] Since we found that \( \theta = 45^\circ \), we know that \( \tan(45^\circ) = 1 \).
Substituting this value into the equation:
\[ 1 = \frac{v^2}{rg} \] Solving for the speed `v`:
\[ v^2 = rg \] \[ v = \sqrt{rg} \] Step 4: Final Answer:
The speed of the particle is \( \sqrt{rg} \). This corresponds to option (A).