# Change in centripetal acceleration if tangential acceleration is non-zero

I recently read about circular motion. They showed that acceleration $$\vec a$$ of a object in circular motion is given by

$$\vec a = -\omega^2r\vec e_r + \frac{dv}{dt}\vec e_t$$

where $$r$$ is the radius, $$\omega$$ is the angular velocity, $$v$$ is the magnitude of velocity, $$\vec e_r = \hat i\cos \theta + \hat j\sin \theta$$ and $$\vec e_t = -\hat i \sin \theta + \hat j\cos \theta$$ are the unit vectors along radius and tangent respectively.

The text book showed that in uniform circular motion, since speed doesn't change(i.e, $$\frac{dv}{dt} = 0$$), the acceleration reduces to $$\omega^2r\vec e_r$$. Then I wondered about non-uniform circular motion. I was like "Is it possible to have a circular path even with changing speed?". I got the answer in a question on this site. It says that for that the centripetal acceleration must change in accordance with speed to keep the path circular. So my question is by how much has the centripetal acceleration to be changed to keep the path circular path if the acceleration along tangent is non-zero?

I tried to find out. But I was not even able to find out from where to start.

• I got the answer in a question on this site. Please give a link to the question/answer. Commented Jan 13, 2022 at 11:02
• @Farcher Actually, I don't know how to insert link. Commented Jan 13, 2022 at 11:03
• Commented Jan 13, 2022 at 11:06
• @Farcher Yes! It is Commented Jan 13, 2022 at 11:09

The centripetal acceleration is $$r\omega^2$$ provided by a force $$mr\omega^2$$.

If the trajectory is to be a circle the radius $$r$$ must stay constant whilst $$\omega$$ is changing.

If it were a satellite orbiting the Earth such a change could not happen as for the same radius of orbit (equal to the separation between satellite and Earth) the gravitational force of attraction between the Earth and the satellite would need to change.

In terms of a rocket attached to a merry-go-round it can happen because in such a case the force applied to the rocket by the merry-go-round could increase in order to compensate for the increasing speed of the rocket (and merry-go-round).

As $$a=\frac{v^2} {r}$$ the rate of change of centripetal acceleration is $$\dot a = \frac{2v}{r}\frac {dv}{dt}$$.

• Thank you very much. I got my answer. But what will happen if both radius and centripetal acceleration have freedom to change? Do both will change by some amount or any one will be given preference? Commented Jan 13, 2022 at 14:37

Acceleration is derivative of the velocity:

$$\vec{a} = \frac{d\vec{v}}{dt}$$

where both acceleration $$a$$ and velocity $$v$$ are vectors. What this means is that there must be some acceleration in order for velocity to change, and this applies to both direction and magnitude of the velocity vector. Radial (centripetal) acceleration changes only velocity vector direction, while tangential acceleration changes only velocity vector magnitude.

Since velocity can be defined via angular velocity as

$$v = r \omega$$

the equation for radial acceleration from your question can be written as

$$\boxed{a_\text{rad} = \frac{v^2}{R}} \tag 1$$

This means that any value of $$a_\text{rad}$$ will result in circular motion. If you keep the radial acceleration at a fixed value and there is some tangential acceleration which changes the velocity vector magnitude, the radius will keep increasing or decreasing proportional to $$v^2$$

$$R = \frac{v^2}{a_\text{rad}} \quad \text{where} \quad a_\text{rad} = \text{const.}$$

This is still circular motion, but the path looks like growing or decaying spiral, depending on the tangential acceleration sign. In order for path to remain perfect circuit, i.e. not to spiral away, you need to change the radial acceleration proportionally to $$v^2$$, which is shown by the Eq. (1).