# The direction of the centripetal force in a vertical circular motion under uniform gravity

Consider the vertical circular motion of a point mass connected to the centre by rigid string. Here the uniform gravity $$m\vec{g}$$ acts.

I illustrated the situation in the diagram below.

Here if we do a vector addition of $$\vec{T}$$ and $$m\vec{g}$$ then we get the centripetal force of a weird direction. It is supposed to direct towards the centre, isnt' it?

I will further decompose the gravity into the radial and tangential components. See below.

So what happens to that $$mg \sin \theta$$ component? Doesn't it disturb the motion from being circular?

• Note: If I try to make the net force direct towards the centre I have to deliberately change the direction of the tension, and that seems very weird to me as we are considering an object confined by string. So if we keep it "natural" (tension towards the centre) can we really say that the object undergoes a circular motion?
• Another question: I understand that in this situation, as $$mg \cos \theta$$ changes the magnitude of the radial force has to change, and thus the velocity of the object has to change. Are we thinking of it as a local circular motion where for the velocity $$\vec{v}(t_1)$$ at a certain time $$t=t_1$$, the centripetal force $$\frac{m|\vec{v}(t_1)|^2}{r} \hat{r}$$ is only valid for the infinitesimally small time interval $$[t, t + dt]$$?
• Summing up the two questions just above - we can consider when the object is at the top or at the bottom. Then we do not have to think about the components of forces as they all lie in the same vertical line. Can we then argue that it is locally a circular motion for the short time interval $$[t, t + dt]$$?
• is the string bendable to the sides? – Umaxo Aug 17 at 5:58
• @Umaxo Nope. I edited the question. – curious Aug 17 at 6:38

In circular motion it is not always the case that $$F_\text{net}=mv^2/r$$. This is only valid for uniform circular motion. In general $$mv^2/r$$ is equal to the component of the net force that points towards the center of the circle. There is another component you should be considering: the component tangent to the circular path.

For planar motion in polar coordinates we break the net force into two components: centripetal (or radial) and tangential:

$$\mathbf F_\text{net}=m\mathbf a=m\left(\ddot r-r\dot\theta^2\right)\,\hat r+m\left(r\ddot\theta+2\dot r\dot\theta\right)\,\hat\theta$$

Where $$r$$ is the distance from the origin, $$\theta$$ is the polar angle, and a dot represents a time rate of change. For circular motion, $$r$$ is constant, so for circular motion Newton's second law reduces to

$$\mathbf F_\text{net}=m\mathbf a=-mr\dot\theta^2\,\hat r+mr\ddot\theta\,\hat\theta$$

So for your object moving in the vertical circle centered at the origin in a constant gravitational field, we can look at the two components (note that negative is towards the origin) $$F_r=-mg\cos\theta-T=-mr\dot\theta^2=-\frac{mv^2}{r}$$ $$F_\theta=mg\sin\theta=mr\ddot\theta$$

$$F_r$$ changes only the direction of the velocity, since this force component is always perpendicular to the velocity, and $$F_\theta$$ changes only the magnitude of the velocity, since this force component is always parallel/anti-parallel to the velocity.

The magnitude of the net force is then given by $$F_\text{net}=\sqrt{F_r^2+F_\theta^2}=mr\sqrt{\dot\theta^4+\ddot\theta^2}$$

Which reduces to $$mv^2/r$$ for uniform circular motion ($$\ddot\theta=0$$, and $$\dot\theta=v/r=\text{constant}$$).

The above should alleviate your worries that we are only considering local circular motion. This is just circular motion. No need to bring in unnecessary complications.

$$mg\sin\theta$$ does not contribute to the centripetal force, it is tangential acceleration that is provided to the mass m. It causes the decrease in speed of the mass during the ascent and the increase during the descent. This is not a case of uniform circular motion. Because of this complication we generally use work energy theorem to solve questions related to this subtopic. Also the centripetal force is not the vector addition of the gravitational force and the tension, it is the sum of the forces that are directed towards the center of the circle. So Centripetal force is equal to tension + $$mg\sin\theta$$ which is $$mv^2/R$$.