Question

One end of the string of length $ l $ is connected to a particle of mass $ m $ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in a circle with speed $ v $, the net force on the particle (directed towards the center) will be (T represents the tension in the string)

• A. \( T \)
• B. \( T + \frac{m v^2}{l} \)
• C. \( T - \frac{m v^2}{l} \)
• D. zero

Hint:

In circular motion, the net force acting on the particle is the centripetal force, which is provided by the tension in the string. For an object of mass \( m \) moving with speed \( v \) in a circle of radius \( l \), the centripetal force is \( \frac{m v^2}{l} \).

Answer 1

The Correct Option is A

The problem involves a particle of mass \( m \) connected to a string of length \( l \), which moves in a circular path on a smooth horizontal table. The force of tension \( T \) in the string acts as the centripetal force, which keeps the particle moving in a circle at speed \( v \). The centripetal force \( F_c \) required to keep an object moving in a circle of radius \( r \) at speed \( v \) is given by:

\[ F_c = \frac{m v^2}{r} \]

Here, the radius of the circle is the length of the string \( l \), so the expression becomes:

\[ F_c = \frac{m v^2}{l} \]

Since the only force acting towards the center is the tension in the string, we equate it to the centripetal force:

\[ T = \frac{m v^2}{l} \]

Thus, the tension \( T \) itself serves as the required centripetal force. 
Therefore, the net force on the particle directed towards the center is: \( T \)

Answer 2

To determine the net force on the particle, we start by considering the forces acting on it. The particle moves in a circle due to the tension \( T \) in the string, which acts as the centripetal force. The centripetal force \( F_c \) required for circular motion is given by the formula:

\[ F_c = \frac{m v^2}{l} \]

where:

• \( m \) is the mass of the particle.
• \( v \) is the speed of the particle.
• \( l \) is the length of the string (also equal to the radius of the circle).

Since the tension in the string provides the necessary centripetal force to keep the particle moving in a circle, we equate the tension \( T \) to the centripetal force:

\[ T = \frac{m v^2}{l} \]

Thus, the net force on the particle, directed towards the center of the circle, is simply the tension in the string:

\[ \text{Net Force} = T \]

Therefore, the correct answer is:

\( T \)