Uniform Circular MotionJEE Main

Centripetal acceleration & angular velocity — interactive Physics simulation for IIT-JEE.

Concept

In uniform circular motion the speed is constant but the velocity direction changes continuously, so the particle is accelerating. That acceleration points toward the centre (centripetal) with magnitude $v^2/r$ , and a real force must supply it — tension, gravity, friction, or a normal force.

Key formula

\omega = \frac{v}{r}, \quad a_c = \frac{v^2}{r} = \omega^2 r, \quad F_c = \frac{mv^2}{r}

Derivation

Over a small time $\Delta t$ the velocity vector rotates by $\Delta\theta = \omega\Delta t$ but keeps magnitude $v$ . The change in velocity $|\Delta \vec v| = v\,\Delta\theta$ points toward the centre.

Dividing by $\Delta t$ : $a_c = v\,\dfrac{\Delta\theta}{\Delta t} = v\omega = \dfrac{v^2}{r}$ since $\omega = v/r$ .

The net inward force is therefore $F_c = ma_c = mv^2/r$ . There is no outward 'centrifugal' force in the inertial frame — that is a pseudo-force seen only in the rotating frame.

Scenarios to explore

Uniform Circular Motion — Centripetal acceleration, angular velocity and period.

Real-world applications

Satellites and planetary orbits (gravity is the centripetal force).
Cars rounding a flat bend (friction supplies $F_c$ ).
Centrifuges separating particles by density.

JEE exam tips

Identify which physical force plays the role of $F_c$ before writing equations.
For a conical pendulum or banked road, resolve forces — the horizontal component is $F_c$ .

Common mistakes

Treating the centrifugal force as a real outward force in the ground frame.
Forgetting that acceleration is non-zero even though speed is constant.
Mixing up $a_c = v^2/r$ with $a_c = \omega^2 r$ — they are equal, use whichever data you have.

Exam traps to avoid

Non-uniform circular motion also has a tangential acceleration $a_t = r\alpha$ .
$F_c$ is a requirement, not an extra force you add to the free-body diagram.