Question

What is the direction of acceleration in uniform circular motion?

Answer 1

Hint: In a uniform circular motion performed by any object, it moves with a constant speed in a circle. Hence, acceleration is only due to change in direction of the object. Object moving in a circle means the object moves around a fixed point and a fixed distance from it.


Complete step by Step Explanation:
In a uniform circular motion, the body moves in a circle with a constant speed. But it’s direction is constantly changing. Hence there is a change in the velocity because we also consider direction of motion of any object when using its velocity.
Hence, there must be some acceleration and this acceleration is in radial inwards direction.
This acceleration is known as centripetal acceleration.

Let an object is moving in a circle of radius $r$ with a uniform velocity of $v$ then angular velocity $\omega$ of the object is given by $\omega ={\displaystyle \frac{v}{r}}$ .

Angular acceleration of the object $\alpha$ is given by $\alpha ={\displaystyle \frac{d\omega }{dt}}$ . It means rate of change of angular velocity.

The centripetal acceleration of that object is given by $${\omega }^{2}r$$ . This is only due to change in direction of velocity.

Due to this acceleration, it’s obvious there must be a force acting inwards, radially inwards the centre. If an object has mass $m$ then centripetal force $F=m{\omega }^{2}r$ .
Replacing $\omega$ by ${\displaystyle \frac{v}{r}}$ in formula of centripetal force, $F={\displaystyle \frac{m{v}^2}{r}}$ .

Using vector notation, $v=\omega \times r$ , where $\times$ is the cross-product.

Note: Since, object is moving in uniform circular motion, the tangential velocity must be constant, so there is no tangential acceleration. Mathematically, tangential velocity is constant and tangential acceleration $={\displaystyle \frac{d\omega }{dt}}$ . This is equivalent to ${\displaystyle \frac{1}{r}}{\displaystyle \frac{dv}{dt}}$ and ${\displaystyle \frac{dv}{dt}}=0$ because $v=0$ . Hence, tangential acceleration is $0$ .