A particle is moving in a circle of radius R with a constant speed v. Its average acceleration over the time when it moves over half the circle is :
- (a)v²/R
- (b)πv²/2R
- (c)2v²/πR
- (d)0
Correct — C, 2v²/πR. Average acceleration is the change in velocity divided by the time taken. Over half a circle the speed stays v but the direction reverses, so the velocity goes from v to −v; the magnitude of the change is |Δv| = 2v. The time to cover half the circumference is the arc length πR divided by the speed: t = πR/v. Dividing, average acceleration = 2v ÷ (πR/v) = 2v²/(πR). This differs from the instantaneous centripetal acceleration v²/R, because average acceleration uses the straight-line change in velocity, not the moment-to-moment turning.
- (a)v²/R — This is the instantaneous (centripetal) acceleration at any point, not the average over half the circle. The average uses the net change in velocity (2v) over the elapsed time (πR/v), giving 2v²/πR.
- (b)πv²/2R — This puts the π in the wrong place. The correct average is 2v²/πR — the factor is 2/π, not π/2.
- (d)0 — The average acceleration would be zero only if the velocity returned to its starting value, as over a full circle. Over half a circle the velocity reverses, so Δv = 2v, not zero.
Average acceleration is the vector change in velocity divided by the time interval, which is generally different from the instantaneous acceleration. In uniform circular motion the speed is constant but the velocity direction constantly changes, so there is always acceleration. Over half a revolution the velocity vector flips to the opposite direction, giving a net change of twice the speed.
The trap is to quote the centripetal value v²/R, which is the instantaneous acceleration. For an average, use Δv over Δt: the velocity reverses (|Δv| = 2v) and the half-lap takes t = πR/v, so the average is 2v²/πR. Only over a whole circle would the average be zero.
- Average acceleration = (change in velocity) ÷ (time), a vector quantity.
- Over half a circle at constant speed v, the velocity reverses, so |Δv| = 2v.
- Time for half the circle = πR/v.
- Average acceleration = 2v ÷ (πR/v) = 2v²/πR, distinct from the instantaneous v²/R.
- Quoting v²/R (the instantaneous centripetal value) when the question asks for the average.
- Saying the average is zero over half a circle — that is true only over a full circle.
A particle in uniform circular motion is asked for its average acceleration over a quarter, half, or full circle — take the vector change in velocity over the time, not the centripetal formula.
A spherical body moves with a uniform angular velocity ω around a circular path of radius r. Which one of the following statements is correct?
- (a) The body has no acceleration
- (b) The body has a radial acceleration ω²r directed towards the centre of the path
- (c) The body has a radial acceleration 2/5 ω²r directed away from the centre of the path
- (d) The body has an acceleration ω²r tangential to its path
Answer(b) The body has a radial acceleration ω²r directed towards the centre of the path
Same topic — circular motion always involves acceleration (the centre-directed centripetal acceleration), which is why a particle going round a circle at constant speed is never unaccelerated.
A uniform motion of a car along a circular path experiences
- (a) a change in speed due to a change in its direction of motion.
- (b) a change in velocity due to a change in its direction of motion.
- (c) a change in momentum due to no change in its direction of motion.
- (d) a constant momentum due to a change in its direction of motion.
Answer(b) a change in velocity due to a change in its direction of motion.
Makes the key point behind this question — in circular motion the velocity changes because its direction changes, which is exactly why the average acceleration over half a circle is non-zero.
- practice — not a real PYQ
A particle moves in a circle of radius R at constant speed v. Its average acceleration over one complete revolution is:
- (a)v²/R
- (b)2v²/πR
- (c)0
- (d)πv²/R
Answer(c) 0 — over a full circle the velocity returns to its start, so the net change in velocity, and hence the average acceleration, is zero.
- practice — not a real PYQ
In uniform circular motion, which one of the following remains constant?
- (a)velocity
- (b)speed
- (c)acceleration (as a vector)
- (d)momentum
Answer(b) speed — the speed is constant, but velocity, acceleration and momentum keep changing direction.