# How does tension work for a simple pendulum? What force is at play to keep a rigid body from stretching?

Here recently I have been working on programming a physics framework for simulations, but I've ran into a problem...

I was testing forces, so I created a simple pendulum like in the photo, and I created the two forces acting on it, the force of tension and the force of gravity...

When the angle of the pendulum starts at 0 degrees, everything seems fine, the tension force is equal (and opposite) to the gravitational force and the bob of the pendulum is in equilibrium (not moving), but as soon as I raise the bob, I notice a "problem".

Lets say I raise it up 90 degrees, and let it go, as soon as I let it go, the gravitational force starts accelerating the bob downwards, and the tension force is 0, therefore, the bob gains a velocity downwards.... as the angle starts to decrease though, instead of the bob slowing down, it just keeps traveling downwards, it reaches a sort of terminal velocity.

The force of tension and the force of gravity are equal and opposite, therefore there is no acceleration, but the bob of the pendulum also already had a velocity downwards, so now it just keeps traveling down with no net force to slow it down...

How does the tension of the string stop the bob from going down any further if the bob already had a downwards velocity, because the tension force can't slow it down because the tension force can only have a magnitude of the weight of the bob?

During the motion of the pendulum the tension force is not equal to the weight of the bob. This is because the bob is accelerating. The tension and the perpendicular component of the weight serve to cause the centripetal acceleration. Using Newton's second law:

$$T-F_{perp}=F_{cent}=ma_{cent}=m\omega^2L$$ where $$\omega$$ is the (time dependent) angular velocity of the bob about the pivot of the pendulum, and $$L$$ is the length of the string.

The tangent component of the weight serves to change the angular velocity by exerting a torque on the bob about the pivot of the pendulum. Once again, using Newton's second law and using the fact that the moment of inertia of the bob about the pivot is $$mL^2$$ $$-F_{tan}L=\tau=mL^2\dot\omega$$

The issue with this analysis is that $$T$$, $$F_{perp}$$, and $$F_{tan}$$ are not constant in time. They change during the oscillations, and hence the accelerations change. This is seen by expressing the components of the weight in terms of the angle the string makes with the vertical: $$F_{perp}=F\cos\theta$$ $$F_{tan}=F\sin\theta$$ Recognizing that $$\dot\theta=\omega$$ and $$F=mg$$

$$T-mg\cos\theta=m\dot\theta^2L$$ $$-mgL\sin\theta=mL^2\ddot\theta$$

The second of these differential equations can be numerically solved (or analytically solved in the limit of small angles) to determine $$\theta(t)$$, which you can then use to determine what $$T$$ should be with the first equation. You can also use $$\theta(t)$$ to determine the relevant components of the weight as the pendulum swings.

The key misconception I wanted to for sure hit on though is that the tension and weight forces do not cancel out while the pendulum swings. There is acceleration, so the net force cannot be $$0$$.

• The tension force is the dot product of the gravitational force and the radius unit vector, so when the angle of the pendulum is 0 degrees, the force of tension and the force of gravity are equal and opposite, so the two cancel out... So therefore, what force is the force that accelerates the bob upwards after reaching the equilibrium point? If you model out all of the forces, with their vectors and all, the bob will start to accelerate downwards, but once it reaches an angle of 0 degrees, it will have a sort of terminal -velocity because the net force on the bob is 0 (tension - gravity) – Ayden Cook Mar 28 '19 at 0:03
• @AydenCook I must stress it again: When the pendulum is swinging and $\theta=0$ the tension force and the weight are not equal and opposite. The tension force is larger than the weight, which is what is causing the upward acceleration to change the direction of the velocity. Also, the tension force is not equal to $\mathbf F_g\cdot\hat r$. The dot product is a scalar, and the tension force is a vector. – Aaron Stevens Mar 28 '19 at 0:05
• @AydenCook What is being stretched here? I have been assuming a string of constant length. Were you assuming a string with elasticity? – Aaron Stevens Mar 28 '19 at 0:10
• Yeah, I'm trying to model only forces at play, with no inference to a rigid body or constant length.... What you said helps though, so if the tension is larger than the weight when the angle is 0, is there an equation to determine the magnitude of the tension force during swinging? – Ayden Cook Mar 28 '19 at 0:12
• Ohhhh, would I be able to use rotational acceleration as the other acceleration that would account for the increase in tension?? I think you actually said that in your first response... Sorry about that – Ayden Cook Mar 28 '19 at 0:17