# Centripetal force in rotating mass attached with a string in gravitational field

Say we have a mass, tied to a string, rotating around a centre point. At the bottom of the rotation, $$F_g$$ is pointing down and $$F_t$$ is pointing up. In this scenario, is $$F_c = F_t - F_g$$ (sum of all forces in the y axis), or is $$F_c = F_t$$?

My teacher said that $$F_c = F_t - F_g$$. $$F_g$$ is NOT pointing towards the centre of the circle, so how can it be a centripetal force?

Any help would be appreciated :)

• What is Ft? If its the force of the string then yes it is the centrifugal force, which is equal to -Fg sin$\alpha$ for $\alpha \in (0,\pi)$. Commented Dec 16, 2020 at 23:23
• Ah, thank you for all your answers. I see that centripetal force is actually the component of the net force pointing towards the centre of the circle. Bit of a misunderstanding Commented Dec 17, 2020 at 0:07

Its a question of names, but technically centripetal force is not the same as the sum of all forces. Centripetal force is centripetal force . There's an equation to figure out exactly what it is (specifically $$F=\frac{mv^2}{r}$$). Whether that equals the sum of other forces depends on your problem.

What you describe is a statics problem, in a rotating frame. If you consider the problem from the perspective of a rotating frame centered on the point where you are holding the string, the object is stationary. If you were to put a camera on the string near your finger (attaching the camera to the string and letting the camera move with the string), the object would appear stationary, because the camera is always pointing right at it.

In a stationary frame, we say the sum of all forces is 0, $$\Sigma F = 0$$. In this system, we have three forces involved: tension, gravity, and the pseudo-force "centripetal force". Thus $$F_t + F_g + F_c = 0$$. We have a centripetal force because we have a rotating frame. Thus, by straight mathematics, $$F_c = -(F_t + F_g)$$. You can stop there. Centripetal force has a magnitude equal to the sum of all other forces, but the opposite direction. This is no more profound than pointing out that the force of gravity has a magnitude equal to the sum of all forces ($$F_g=-(F_t + F_c)$$) or that the force of tension has a magnitude equal to the sum of all forces ($$F_t=-(F_g+F_c)$$). They are all natural consequences of $$\Sigma F = 0$$.

So something has to give, right? The key is that $$F_t\ne F_c$$ -- the force of tension is not the same as the centripetal force. We can point out that its rather obvious, the difference between the force of tension and the centripetal force must be exactly the same as the force of gravity!

How? When you swing an object like this, the string is never perfectly horizontal. If you swing it fast, you might get close, but the mass is always slightly below the level of the point you are rotating around. As the mass moves downward, that changes the direction of the force of tension. Now it points slightly upward. The object will move further and further down towards the ground until the vertical component of this tension is exactly equal to the force of gravity, but in the opposite direction. It will stay at that height.

In fact, because the centripetal force is at right angles to gravity at all times (because you're swinging it horizontally), we can calculate exactly what the relationship is. The force of tension must be diagonal upwards and inwards. This makes a triangle with the force of gravity and the centripetal force as the legs, and the tension force as a hypotenuse. $$||F_t||=\sqrt{||F_g||^2 + ||F_c||^2}$$

The centripetal force is the component of the net force that is directed towards the center of your circle and equals $$mv^2/R$$, all provided that the motion you are dealing with is circular.

In the picture above, the thick red, purple and yellow forces all act on the green particle. Their projections along the center are shown. The sum of those projections is the centripetal force.

(note that the forces represent random arbitrary forces)

The centripetal force is not the force that is pointing toward center, but the net force that points toward the center if an object is in circular motion. In your case when the object reaches the bottom of the circle the net force is $$F_T-F_g$$ and we must have $$F_T>F_g$$ so that the net force is pointing toward center. But we don't require $$F_g$$ pointing towards the center, because the centripetal force is the net force $$F_T-F_g$$. So no matter how complicated the forces are acting on the object, as long as we know it is undergoing circular motion, then we can conclude there must be a net force pointing toward center and we call this net force the centripetal force. If the object has a linear acceleration (i.e. non-uniform circular motion) then it also has a net force in the tangential direction.

The force of tension would have a horizontal and vertical component. The force of gravity would equal the vertical component of the tension force.