Physics · Motion In A Plane · NEET
No. Centripetal acceleration a_c = v^2/R is only ONE part of the total. In uniform circular motion (constant speed) a_t = 0, so the total acceleration equals the centripetal acceleration and points straight to the centre. But in non-uniform circular motion the speed changes, so there is also a tangential part a_t = dv/dt. Then the total is a = sqrt(a_c^2 + a_t^2), which is larger than a_c alone and does NOT point to the centre.
The tangential acceleration lies along the velocity (along the tangent to the circle) because it only changes the speed. The centripetal acceleration points along the radius toward the centre, which is always at 90 degrees to the tangent. Since the radius and the tangent are perpendicular, a_c and a_t are perpendicular. This is exactly why you can add them with the Pythagoras rule: a = sqrt(a_c^2 + a_t^2).
Three steps. (1) Centripetal part: a_c = v^2/R (also = omega^2 R). (2) Tangential part: a_t = dv/dt (the rate at which speed increases or decreases; also a_t = R*alpha). (3) Combine as perpendicular vectors: a = sqrt(a_c^2 + a_t^2). The angle theta the total makes with the radius (centre direction) is given by tan(theta) = a_t / a_c.
It points at an angle, between the centre and the tangent. Because a_c pulls toward the centre and a_t acts along the tangent, the resultant leans inward but tilts forward (if speeding up) or backward (if slowing down). It only points exactly to the centre when the speed is constant (a_t = 0).
Tangential acceleration changes the SPEED (the magnitude of velocity). Centripetal acceleration changes the DIRECTION of velocity. In non-uniform circular motion both happen at once, so both speed and direction are changing every instant.
The total acceleration of a particle moving clockwise in a circle of radius R = 2.5 m is a = 15 m/s^2 at a given instant, directed at 30 degrees to the radius (i.e. the total acceleration makes 30 degrees with the line to the centre). The speed of the particle is:
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 8x10^-4 J by the end of the second revolution after the start, is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
a = sqrt(a_c^2 + a_t^2), where a_c = v^2/R is the centripetal (radial) acceleration and a_t = dv/dt is the tangential acceleration. The direction is set by tan(theta) = a_t/a_c, measured from the radius.
Only in uniform circular motion, where the speed is constant so dv/dt = 0 and a_t = 0. Then a = a_c = v^2/R directed to the centre.
Centripetal acceleration (v^2/R) points toward the centre and changes the direction of velocity. Tangential acceleration (dv/dt) points along the tangent and changes the speed. They are always perpendicular.
No. Even if the speed is momentarily constant (a_t = 0), the centripetal part a_c = v^2/R is non-zero as long as the particle moves (v not 0). So total acceleration is never zero for a moving particle on a circle.
a_t = R*alpha, where alpha is the angular acceleration (rate of change of angular velocity). Similarly a_c = omega^2 * R. Both combine to give the total acceleration.
