Physics · Motion In A Plane · NEET
Centripetal acceleration a_c = v^2/r points toward the centre of the circle. Its only job is to change the DIRECTION of the velocity, not its size. Tangential acceleration a_t = r*alpha points along the tangent (the direction of motion). Its only job is to change the SPEED (the size of velocity). They are always perpendicular to each other, because one points to the centre and the other points along the tangent.
In circular motion the velocity vector can change in two independent ways: its direction and its magnitude. A curved path always forces the direction to change, and that needs a centripetal (centre-pointing) acceleration. If the speed is ALSO changing (speeding up or slowing down), a second, tangential acceleration is needed. Non-uniform means the speed changes, so both are present. Uniform circular motion has constant speed, so a_t = 0 and only a_c survives.
Yes. As long as the object moves on a circle, its direction is changing, so there must be a centre-pointing acceleration a_c = v^2/r at every instant (except momentarily when v = 0). Tangential acceleration, however, is present only when the speed is changing. So a body can have a_c without a_t (uniform circular motion), but on a real circular path it can never have a_t without a_c.
Tangential acceleration changes only the SPEED. Because it points along the line of motion (parallel or anti-parallel to velocity), it can only make the particle faster or slower, never turn it. Turning (change of direction) is done entirely by the centripetal acceleration. This clean split is why NEET questions love it: a_t handles v^2 = u^2 + 2*a_t*s type speed problems, a_c handles the v^2/r direction part.
Yes, always. The tangent to a circle is always at 90 degrees to the radius at that point. Since a_t lies along the tangent and a_c lies along the radius (toward the centre), the two are perpendicular. That is why the total acceleration is found by the Pythagoras rule a = sqrt(a_t^2 + a_c^2) and why they never add up simply as numbers.
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. The magnitude of this acceleration, if the kinetic energy of the particle becomes equal to 8 x 10^-4 J by the end of the second revolution after the beginning of the motion, is:
a = 15 m/s^2 represents the total acceleration of a particle moving clockwise in a circle of radius R = 2.5 m at a given instant. The total acceleration makes 30 degrees with the radius. The speed of the particle is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Centripetal (radial) acceleration a_c = v^2/r = omega^2 * r, directed toward the centre. Tangential acceleration a_t = dv/dt = r*alpha, directed along the tangent, where alpha is the angular acceleration. Total acceleration a = sqrt(a_t^2 + a_c^2).
Tangential acceleration is zero when the speed is constant, that is, in uniform circular motion (alpha = 0). Then the only acceleration is centripetal, a_c = v^2/r toward the centre.
Yes. If the particle is slowing down, a_t points opposite to the velocity, so it is taken as negative. A negative a_t simply means deceleration along the path; the centripetal part a_c is always positive and toward the centre.
The net acceleration points somewhere between the tangent and the radius, tilted toward the centre. Its magnitude is sqrt(a_t^2 + a_c^2) and it makes an angle theta with the radius given by tan(theta) = a_t / a_c.
Centripetal acceleration is the reason objects move on curved paths and links directly to centripetal force F = m*v^2/r used in banking of roads, motion in a vertical circle and charged particles in magnetic fields. Almost every year NEET asks a circular-motion question that uses a_c = v^2/r.
