newtonian mechanics - Conical vs Simple Pendulum

I don't understand why the Tension $T$ on a conical pendulum and a simple pendulum are different.

In a simple pendulum, one would say that the tension of the rope is $T=mg \cos(\theta)$.

simple pendulum http://n8ppq.net/site_n8ppq/Physics/pendulum_files/image001.gif

However, in a conical pendulum (describing a circular motion), $mg=T \cos(\theta)$.

The only difference I see in the set up of the two cases is that in the second one there is a velocity component that makes the bob go around in a circle.

I know that in the conical pendulum, the component $T \sin(\theta)$ would give the centripetal acceleration of the circular motion.

I've seen this everywhere. The two cases look pretty much the same to me, so I would be tempted to say one of them (rather the second one) is wrong.

Answer

Both pendulums are correct in their respective situation. We must remember that Newton's second law dictates that the vector sum of forces on an object must be equal to the mass of the object times the acceleration of the object.

\begin{equation} \sum_n F_n = ma \end{equation}

In the first pendulum the object is swinging side to side so we know that the acceleration of the object is orthogonal to the arm of the pendulum pointing at an angle $\theta$ below the horizontal towards the center of oscillation of the pendulum. This means that the forces in line with the arm of the pendulum must be equal and opposite since there is no motion in this direction and we see that $T=mg\cos{\theta}$ is true for this pendulum.

For the second (conical) pendulum the object is moving at the same vertical height in a circular path of radius $r$. This tells us that the acceleration of the object points horizontally inward at an angle of $\pi/2-\theta$ with respect to the arm of the pendulum. We also know that for circular motion Newton's second law can be rewritten as

\begin{equation} \sum_n F_n = \frac{mv^2}{r} \end{equation}

Since there is no downward acceleration in the conical pendulum the vertical forces must be in equilibrium such that $T\cos{\theta}=mg$.

Moral of the story? Your choice of coordinate axes is important and net acceleration must be accounted for when making free-body diagrams.