Physics · Motion In A Plane · NEET
Acceleration means a change in the velocity VECTOR, not just its size. In uniform circular motion the speed (magnitude) stays the same, but the direction of the velocity keeps changing every instant. A changing direction is still a changing velocity, so there is a real acceleration. This acceleration is centripetal acceleration and it points toward the centre. For NEET, remember: constant speed does NOT mean zero acceleration in circular motion.
Take a particle moving with constant speed v around a circle of radius r. In a small time Δt it turns through a small angle Δθ. The velocity vectors at the start and end have the same length v but different directions, so the change in velocity Δv forms a small triangle similar to the position triangle. From the two similar triangles: Δv/v = Δr/r, so Δv = (v/r)·Δr. Dividing by Δt: Δv/Δt = (v/r)·(Δr/Δt). As Δt tends to zero, Δr/Δt becomes the speed v, giving a = v²/r. Since v = ωr, you can also write a = ω²r.
It always points from the particle toward the centre of the circle (radially inward). As Δt becomes very small, the change-in-velocity vector Δv turns to point straight at the centre. This is why it is called 'centripetal', which means 'centre-seeking'. It is always perpendicular to the velocity, which is along the tangent.
They are the same acceleration written two ways, linked by v = ωr. Use a = v²/r when you are given the linear speed v and radius r. Use a = ω²r when you are given the angular speed ω (rad/s) and radius r. You can also combine them as a = vω. Pick whichever matches the data in the question.
There is no new force called 'centripetal force'. Centripetal is just the NAME for the net inward force (or acceleration) needed to bend the path into a circle. A real force provides it: tension in a string, gravity for a satellite, or friction for a car on a curve. By Newton's second law, F = ma = mv²/r, so the required inward force is mv²/r.
In the figure, a = 15 m/s² represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 2.5 m at a given instant of time. The total acceleration makes an angle of 30° with the radius. The speed of the particle is:
A particle moves so that its position vector is given by r = cos(ωt) x-hat + sin(ωt) y-hat, where ω is a constant. Which of the following is true?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
a = v²/r, where v is the linear speed and r is the radius. Using v = ωr, it can also be written as a = ω²r or a = vω. Its SI unit is m/s².
No. Centripetal acceleration is always perpendicular to the velocity, so it changes only the DIRECTION of motion, not the speed. In uniform circular motion the speed stays constant while the direction keeps turning.
It always points radially inward, from the particle toward the centre of the circle. That is why it is also called radial or centre-seeking acceleration.
It links motion in a plane with laws of motion (F = mv²/r) and appears in banking of roads, satellite motion, and vertical circular motion. NEET regularly tests the formula, its direction, and the fact that acceleration is non-zero even at constant speed.
Centripetal acceleration (v²/r) points toward the centre and changes direction of velocity. Tangential acceleration points along the tangent and changes the SPEED. In uniform circular motion tangential acceleration is zero; in non-uniform circular motion both exist.
