# When calculating centripetal force, do we ignore non-radial or tangential forces

Suppose an object moving in circular motion in the vertical plane (ie such that gravity points directly downwards) around a central point attached by a string; the object is constantly accelerating as its direction is constantly changing, therefore a resultant force is acting on it . This centripetal force acts radially inwards and is the resultant force of forces acting on the object.

However when I attempt to calculate such a scenario I end up with a force unaccounted for. Take the following diagram My question is what happens to this the component of weight that is parallel to the objects tangential velocity (ie mgsinθ), surely this should be included when calculating resultant force and therefore centripetal force. Unless my understanding of centripetal force is wrong and resultant force is not always equal to it.

This can be extended to when the object's velocity is perpendicular to the ground and tension in the string/rope is parallel to the ground. We calculate that centripetal force = tension in string, why is it we completely ignore the weight of the object in our calculations. I know the tension is radial and the weight is tangential and so perpendicular, but why do we not carry out a vector sum to work out the centripetal force? For planar motion in polar coordinates, Newton's second law becomes:

$$\mathbf F=m(\ddot r-r\dot\theta^2)\hat r+m(r\ddot\theta+2\dot r\dot\theta)\hat\theta$$ where the each dot represents a time derivative (or time rate of change if you aren't familiar with calculus). We have two forces acting on the object: the tension force and the weight. Now, we know the tension force is always radially inward, i.e. $$\mathbf T=-T\hat r$$. And we can break the weight into two components: $$\mathbf w=w\cos\theta\ \hat r-w\sin\theta\ \hat\theta$$. Therefore we have $$(w\cos\theta-T)\ \hat r+(-w\sin\theta)\ \hat\theta=m(\ddot r-r\dot\theta^2)\ \hat r+m(r\ddot\theta+2\dot r\dot\theta)\ \hat\theta$$

The radial component is essentially what you have if you replace $$\dot\theta$$ with $$v/r$$ and recognize that $$\ddot r$$ must be $$0$$ since the object stays at a constant radius $$r$$: $$w\cos\theta-T=-\frac{mv^2}{r}$$

What you are asking about that is left over is the tangential component: $$-w\sin\theta=m(r\ddot\theta+2\dot r\dot\theta)$$ or since $$\dot r=0$$ $$\ddot\theta=-\frac w{mr}\sin\theta=-\frac gr\sin\theta$$

which you might recognize is the differential equation for the angular motion of a simple pendulum.

• Thanks for the reply, so essentially the centripetal force is just the resultant of forces acting radially whereas tangential forces effect the tangential acceleration of the object? Similar to how an object in projectile motion has independent horizontal and vertical motion, where gravity only effects vertical motion. In the case of circular motion are we merely rotating our axis from vertical and horizontal to radial and tangential? – Patrick Apr 15 '19 at 18:25
• @Patrick Kind of. The centripetal force is just the radial component, and the tangential component is, well, tangent. But I hesitate to say they are independent. You can see how these equations are fairly coupled. Another analogy that fails is that there is a net radial force, but the radial coordinate $r$ is not changing. Really we are look at orthogonal components of the force, but these components change direction depending on your position. – BioPhysicist Apr 15 '19 at 18:29
• Thanks for the clarification, I will look into the subject further. – Patrick Apr 15 '19 at 18:33

The component of force tangent to the circle ends up being the net tangential force on the object, the force that drives the tangential acceleration. The radial component of the gravitational force adds to the tension force to form the centripetal force, driving the centripetal acceleration.

It's important to keep in mind that in this case the object is not executing uniform circular motion. The analysis is done at some arbitrary point at which $$T-mg\cos\theta = mv^2/r$$, but unlike the case of uniform circular motion there is a tangential acceleration. So as time goes on, $$T$$, $$\theta$$, and $$v$$ all change, but the relationship between them stays fixed.

I'm not sure what you mean in your last paragraph. If the pendulum is horizontal, then $$T=v=0$$ and $$\theta = \pi/2$$. Indeed the centripetal force is equal to the tension force, but they are both zero.

• Thanks for the quick reply, I was previously unaware of tangential acceleration but this would appear to explain this, thanks. In the last paragraph about the point when the object/ pendulum is horizontal I have added a diagram to my post to illustrate it. I was wondering about your last point where T=v= 0, if v=0 and the only force acting on it is the objects weight pointing downwards won't the object accelerate downwards and not complete the circle? (note I am not concerned with Simple harmonic motion in this question). – Patrick Apr 15 '19 at 18:04
• In the horizontal position, the net force is straight down. But a moment later, the constraint of the tension device will add a small horizontal force. In analyzing circular pendula problems (and similar), one is wise to focus attention at one moment in time, considering the instantaneous values of $T$, $\theta$ and $v$. They will change in the next instant, but the relationship among them will remain the same. – garyp Apr 15 '19 at 19:24