Question

A particle moves on a circular path of radius R with uniform speed v. The magnitude of change in acceleration of the particle in half revolution is 1)0 2)v²/R 3)2v²/R 4)v²/2R

Answer 1

4. v²/2R is the correct answer of this question.

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Answer 2

Answer:

The magnitude of change in acceleration of the particle in half revolution is 2v²/R i.e.option (3).

Explanation:

The centripetal acceleration is given as,

$a=\frac{v^2}{r}$               (1)

Where,

a=acceleration of the particle

v=velocity of the particle

r=radius of the particle

In a half-circle, the direction of acceleration is reversed as the velocity is acting in opposite direction.

Let us suppose the particle starts its motion from point A and ends its motion on the half-circle at B. So, the acceleration at respective points will be given as,

$a_A=\frac{v^2}{r}$        (2)

$a_B=\frac{-v^2}{r}$     (3)

aA=acceleration at point A

aB=acceleration at point B

The change in acceleration,

$\Delta a=a_A-a_B=\frac{v^2}{r}-\frac{-v^2}{r}=\frac{2v^2}{r}$

Hence, the magnitude of change in acceleration of the particle in half revolution is 2v²/R i.e.option (3).