Q.One end of a string of length $l$ is connected to a particle of mass $m$ and the other 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 centre) is:

The key idea is that the tension in the string provides the centripetal force required for circular motion.

1. For a particle moving in a horizontal circle on a smooth table, the only horizontal force acting on it is the tension $T$ in the string. …

The question asks for the net force directed toward the centre of the circular path. That net force is what we call the centripetal force — the force that keeps the particle moving in a circle. The key is to identify all forces acting on the particle in the horizontal plane.

Let’s think about what’s happening physically. The particle is on a smooth horizontal table, so there is no friction. The string is taut, passing through a small peg at the centre. The only horizontal force the particle feels is the tension $T$ in the string, pulling it toward the peg. That’s it — no other horizontal forces exist.

Now, for any particle moving in a circle of radius $l$ with constant speed $v$, the required centripetal force is $\frac{mv^2}{l}$. This is not a separate force; it is the net inward force that must be supplied by real forces. Here, the tension $T$ is the only candidate. So the net inward force is simply $T$, and it must equal $\frac{mv^2}{l}$ for the motion to be circular.

Let’s go step by step.

1. Identify the forces on the particle.

   Vertically: weight $mg$ downward and normal reaction $N$ upward from the table — these cancel, so no net vertical force.

   Horizontally: only the tension $T$ in the string, directed radially inward toward the peg. No friction, no other horizontal force.

2. What is the net force toward the centre?

   Since only $T$ acts horizontally, the net radial force is exactly $T$. There is no other force to add or subtract.

3. Relate to centripetal force requirement.

   For circular motion of radius $l$ at speed $v$, the required centripetal force is $\frac{mv^2}{l}$. This must equal the net inward force. So:

   $T = \frac{mv^2}{l}$ …

Concept: Identifying the Net (Centripetal) Force from the Forces Actually Present

Step 1: List every force on the particle

Vertically: weight $mg$ down and the table's normal reaction $N$ up — these cancel (no vertical acceleration).

Horizontally: only the string's tension $T$, directed toward the peg (the centre).

Step 2: Recognise that $mv^2/l$ is a requirement, not a separate force

The quantity $\dfrac{mv^2}{l}$ is the centripetal force needed to keep the particle on its circular path — it is not an extra force that adds to or subtracts from $T$. It is satisfied by whichever real force(s) act toward the centre.

Step 3: Identify the net inward force …