Question

One end of 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 circle with speed ‘v’, the net force on the particle (directed towards center) will be (T represents the tension in the string)

• A. \(T\)
• B. \(T+\frac {mv^2}{l}\)
• C. \(T-\frac {mv^2}{l}\)
• D. Zero

Answer 1

The Correct Option is A

To solve this problem, we need to understand the forces acting on the particle as it moves in a circle on a smooth horizontal table.

When a particle moves in a circular path with constant speed, it experiences centripetal force, which always acts towards the center of the circle. This force is given by the formula:

\( F_c = \frac{mv^2}{l} \),

where:

• \( m \) is the mass of the particle,
• \( v \) is the speed of the particle, and
• \( l \) is the length of the string.

In this scenario, the string provides the necessary tension to keep the particle in its circular path. Therefore, the tension in the string is the net force that acts towards the center of the circle, serving as the centripetal force.

Thus, the tension \( T \) in the string is equal to the centripetal force. Hence, the net force on the particle which is directed towards the center is simply the tension \( T \) itself.

Therefore, the correct answer is:

\( T \)

Let's evaluate the given options to ensure our answer is accurate:

• \(T\): This is correct as explained above.
• \(T+\frac{mv^2}{l}\): This is incorrect as it suggests an additional force beyond the required centripetal force.
• \(T-\frac{mv^2}{l}\): This would imply the centripetal force is less than the tension, which is not possible in this context.
• Zero: A net force of zero would not allow for circular motion as described.