# How can we say that a body is doing circular motion while doing a non-uniform circular motion if the centripetal force is changing?

I get it that for body to do non-uniform motion there should be some component of a force which acts tangentially along the direction of linear velocity of the particle so in the case of non uniform circular motion there are two forces which act on the particle. One is acting tangentially along the linear velocity and the other is forming a right angle with the linear velocity and is directed towards the centre. Now my question is that the radial or centripetal acceleration which is acting towards the centre is given by $$a_R=v^2/r$$ which means that if the linear velocity changes then the centripetal acceleration will change and I think the change in centripetal acceleration will cause somewhat spiral motion.

Please explain this to me I searched and studied it on the internet but could not understand it.

The basic idea is that for a body to keep moving in a circle with radius r and uniform speed v , the centripetal force should be (v^2)/r as you mentioned. Now consider a situation where a ball tied to a string is whirled around In a circle in the horizontal plane (Neglect gravity). In this case the tension in the string is providing the centripetal force to the ball . If the speed of the ball increases by means of some other tangential force , notice that to keep the ball moving in the same circle , the string has to apply a greater force on the ball (the string seems to tighten up) . So in fact the change in centripetal force in this case is what keeps the uniform circular motion going . Just to clear things , consider the motion of satellites around the earth. In this case it's the gravitational force of the earth which provides the necessary centripetal force . Say a satellite gets hit by another satellite causing it to slow down ; but the gravitational force still says the same . So the centripetal force is more than what is required to keep it moving in the same circle with the reduced speed . Hence , the satellite gets pulled inward towards the earth and undergoes "spiral motion"

• So from this we can also say that suppose if we apply force F1 on one face of the cube(not top or bottem face) and apply another force F3 on the other face( not opposite to the face on which F1 has been applied ) and then on a similar cube we apply force F2 on the same face where F1 has been applied and then force F4 on the same face where force F3 has been applied then F4>F3 if F2>F1. Commented May 4, 2020 at 13:12

It is exactly the change in centripetal force what makes the body to keep on doing circular motion. If at each instant in time the centripetal acceleration is $$a_R = v^2/r$$, then the motion is circular. Therefore if the tangential velocity $$v$$ changes, the centripetal force has to change, in order for the motion to stay circular.

The acceleration vector in polar coordinates for a general non-uniform 2D planar motion can be described as (see eq. (6.5.14) on page 6 in reference for derivation):

$$\mathbf{a} = \left[\frac{d^2r}{dt^2} - r \left(\frac{d\theta}{dt}\right)^2\right] \hat{r} +\left[ 2 \frac{dr}{dt} \frac{d\theta}{dt} + r \frac{d^2\theta}{dt^2}\right]\hat{\theta}$$

Now, for non-uniform circular motion, the distance $$r$$ from axis of motion is fixed:

$$\frac{dr}{dt} = 0 \Rightarrow \frac{d^2r}{dt^2}=0$$

Which gives the acceleration vector as:

$$\mathbf{a} = - r \left(\frac{d\theta}{dt}\right)^2 \hat{r} + r \frac{d^2\theta}{dt^2}\hat{\theta}$$

The $$\hat{r}$$ component of acceleration is the centripetal acceleration:

$$a_r = - r \left(\frac{d\theta}{dt}\right)^2 = - \frac{v^2}{r}$$

where $$v = r\frac{d\theta}{dt}$$, is time-dependent. Hence, we can have a time-varying centripetal acceleration while keeping $$r$$ fixed, which is the essential constraint for circular motion of any kind.

Now, for a spiral motion, $$r$$ is not fixed and is usually expressed as a function of $$\theta$$, such that $$r=r(\theta)$$. You can get different spirals depending on the exact form of $$r(\theta)$$, the equations of which can be found here.