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

To solve the problem of finding the speed of a particle moving in a horizontal circle while being suspended by a string, we need to analyze the forces and the geometry involved.

1. Consider a particle of mass m suspended from the ceiling with a string of length L. The particle moves in a horizontal circle of radius r = \frac{L}{\sqrt{2}}.
2. The particle is moving in a circle, meaning it experiences a centripetal force that keeps it in this path. The tension in the string provides the necessary centripetal force.
3. Let's consider the vertical and horizontal components of the forces acting on the particle. The tension, T, has two components: one balancing the weight of the particle (mg) and the other providing the centripetal force required for circular motion.
4. The balance of forces in the vertical direction is given by:
   T \cos \theta = mg
   where \theta is the angle that the string makes with the vertical.
5. The centripetal force required in the horizontal direction is:
   T \sin \theta = \frac{mv^2}{r}
   where v is the speed of the particle.
6. From the geometry of the problem:
   \cos \theta = \frac{\sqrt{L^2 - r^2}}{L} = \frac{\sqrt{L^2 - \left(\frac{L}{\sqrt{2}}\right)^2}}{L} = \frac{\sqrt{\frac{L^2}{2}}}{L} = \frac{1}{\sqrt{2}}
7. Plug this back into the vertical force equation:
   T \cdot \frac{1}{\sqrt{2}} = mg \Rightarrow T = mg\sqrt{2}
8. Substitute T = mg\sqrt{2} in the centripetal force equation:
   mg\sqrt{2} \cdot \frac{r}{L} = \frac{mv^2}{r} \Rightarrow g\sqrt{2} \cdot \frac{1}{\sqrt{2}} = \frac{v^2}{r}
   \Rightarrow g = \frac{v^2}{r}
9. Therefore, the speed of the particle is:
   v = \sqrt{rg}
10. Thus, the correct answer is \sqrt{rg}.

Given these calculations, the correct option is \sqrt{rg}.