Define centripetal acceleration. Derive expression for the centripetal acceleration of a particle for a uniform circular motion. What will be the direction of the velocity acceleration at any instant?

**OR**

What is a projectile? Find mathematically the nature of the path of pr

Question

Define centripetal acceleration. Derive expression for the centripetal acceleration of a particle for a uniform circular motion. What will be the direction of the velocity acceleration at any instant?

**OR**

What is a projectile? Find mathematically the nature of the path of projectile when projected at an angle $	heta$ with the horizontal. Under what condition is the horizontal range maximum?

Accepted Answer

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> [!TLDR]
> Geometrically analysing the small change in the velocity vector for uniform circular motion gives $a=v^2/r$, directed radially inward at every instant.
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> **Centripetal acceleration** is defined as the acceleration of a body moving along a circular path, directed always towards the centre of the circle, and it is this acceleration that continuously changes the direction of the velocity to keep the body moving in a circle.
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> **Derivation:** Consider a particle moving with constant speed $v$ on a circle of radius $r$. At time $t$, its position is $P$ and velocity $\vec v_1$ (tangential, magnitude $v$). At time $t+\Delta t$, its position is $Q$, having swept a small angle $\Delta	heta$ at the centre, with velocity $\vec v_2$ (tangential at $Q$, same magnitude $v$, but rotated by $\Delta	heta$ relative to $\vec v_1$).
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> Since $|\vec v_1|=|\vec v_2|=v$, but their directions differ by the small angle $\Delta	heta$, the magnitude of the change $|\Delta\vec v| = |\vec v_2-\vec v_1| \approx v\,\Delta	heta$ for small $\Delta	heta$ (chord length of an arc of radius $v$ subtending angle $\Delta	heta$).
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> The average acceleration over $\Delta t$ is $\dfrac{|\Delta\vec v|}{\Delta t} \approx \dfrac{v\Delta	heta}{\Delta t}$. As $\Delta t	o0$, $\dfrac{\Delta	heta}{\Delta t}	o\dfrac{d	heta}{dt}=\omega$ (angular speed), and since the particle covers arc length $v\,\Delta t = r\,\Delta	heta$, we have $\omega = v/r$. So the instantaneous acceleration magnitude is
>
> $$a = v\omega = v\cdot\dfrac{v}{r} = \dfrac{v^2}{r}$$
> …

Concept understanding — Uniform Circular Motion

Uniform Circular Motion

1. Angular Quantities

2. Centripetal Acceleration

3. Directions

4. Points on a Rotating Body

5. Connected Systems

6. Non-Uniform Circular Motion

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