Centripetal Acceleration Physics Notes

Centripetal Acceleration Physics Notes

Centripetal Acceleration:
We will consider the circular motion of a particle as shown in the figure. At any instant t, the particle is at P, where its angular displacement is θ.

Now, we will draw a unit vector PA = â_r on point P, in the outward direction from the radius r of the circle, and another unit vector PB = â_t, tangent to the point and in the direction of the angle θ as shown in the figure.

We call â_r as radial unit vector and â_t as a tangential unit vector.

Now, draw a line PX’ parallel to X-axis and PY’ parallel to Y-axis, as depicted in the figure.

Calculation of Acceleration in Circular Motion

According to the figure,
\(\frac{\overrightarrow{P A}}{|\overrightarrow{P A}|}\) = î cos θ + ĵ sin θ
â_r = î cos θ + ĵ sin θ ….(1)

Here, | \(\vec{PA}\) | = 1 and î and ĵ are the unit vectors along X-axis and Y-axis.
Similarly,
\(\frac{\overrightarrow{P B}}{|\overrightarrow{P B}|}\) = -î sin θ + ĵ cos θ
or
â_t =-î sinθ +ĵ cosθ …(2)
Now, the position vector of the particle at time 2
will be
\(\vec{r}=\overrightarrow{O P}\) = OP âr
or, \(\vec{r}\) = r (î cos θ + ĵ sin θ) …(3)
Now, taking the derivative of equation (3) w.r.t. t, we get the velocity of the particle.
So, the velocity of the particle

The above equation shows that the term rω is the speed of the particle at time t [: v = rω] and its direction is toward â_t or tangential.

We can find out the acceleration of the particle at any instant of time t by taking the derivative of equation (4).
∴ The acceleration of the particle,

and equations (1) and (2).

In equation (6) we observe that the acceleration of particle a has two components.
(i) \(\vec{a}\) r = -ω râ_r directed along -â_r or towards the centre of the circle.
That is why it is called centripetal acceleration.

It is directed along the tangent of the point. Its value is rex. For a non-uniform circular motion of a particle, there will be these two accelerations, but for the uniform circular motion of a particle, v is definite and so is co
as (v = rω). Therefore, \(\frac{d v}{d t}\) = r\(\frac{d \omega}{d t}\) = 0. Hence, the value of tangential acceleration at will be zero. So, the centripetal acceleration (from equation (6)).
\(\overrightarrow{a_{r}}=\vec{a}\) = -ω2 râ_r ………..(7)

This acceleration will be directed towards the centre of the circle and the magnitude of the centripetal acceleration will be
â_r = ω²r = \(\frac{v^{2}}{r^{2}} r=\frac{v^{2}}{r}\) …….(8)

Here, we should note that u is constant but the direction of the particle keeps on changing. So, there is a change in velocity of the particle and hence, acceleration is produced.

The net acceleration in a circular motion :

Centripetal force: A force required to make a body move along a circular path with uniform speed is called centripetal force. This force is always acting along the radius and towards the center of the circular path.

Expression for centripetal force: We drive the centripetal acceleration in article number 4. 10.

Physics Notes