# How is there kinetic energy at the bottom of the pendulum swing if velocity = 0 at the lowest point?

Say a mass $$m$$ is attached to a massless string that is attached to the ceiling. The mass is pulled 30 degrees to the left of the vertical and then let go. At the lowest point of its swing it has kinetic energy but no potential, but at the lowest point velocity = 0 like at the highest point of a projectile trajectory velocity = 0. So how is there kinetic energy?

• Think about a real pendulum that you have seen while it is swinging. If you have never seen one, here is an example, Is the velocity of the pendulum really zero when the bob is at its lowest point? Mar 27, 2022 at 2:56
• v = dr/dt. since the slope of the path = 0, v = 0
– Anna
Mar 27, 2022 at 15:55
• If you write the height of the pendulum bob $y$ as a function of the $horizontal displacement$ $x$, then it it is true that $dy/dx=0$ at the minimum. But this does not imply that $dr/dt=0$. Mar 27, 2022 at 15:59
• Anyway, I think you are getting confused with the mathematics, and it would help to think physically about the problem. Don't use any math. Just picture a real pendulum that is swinging. Does the pendulum stop when it reaches the bottom of its swing? Mar 27, 2022 at 15:59
• ok, i see it now. dy/dt = 0 but dx/dt =/= 0
– Anna
Mar 27, 2022 at 16:31

At the bottom of the swing the vertical component of the velocity is (ideally) zero, but there is definitely a horizontal non-zero component to the velocity. In other words, at that point it is swinging from right to left or from left to right.

In polar coordinates,

$$\mathbf v=\frac{\text dr}{\text dt}\hat r+r\frac{\text d\theta}{\text dt}\hat\theta$$

Throughout the entire motion of the pendulum, $$\text dr/\text dt=0$$, so we have in this case

$$\mathbf v=r\frac{\text d\theta}{\text dt}\hat\theta$$

At the lowest point of the pendulum, $$r\neq0$$ and $$\text d\theta/\text dt\neq0$$, therefore, $$\mathbf v\neq0$$.

In regards to some of your comments, note that $$\text dr/\text dt$$ is not the slope of the trajectory. That is $$\text dy/\text dx$$, which in this case is just equal to $$\tan\theta$$.