# Centripetal force pendulum

So I was trying to figure out the free-body diagram for a pendulum, and I stumbled upon a video that explains it pretty well (see screenshot).

What I don't understand is how we can have a centripetal acceleration. Shouldn't the tension of the rope be equal to the radial component (or in this case, $x$ component) of the gravitational force? I assumed we have $$T=-G_{g_x}.$$ However, that seems odd too, because sure we have a centripetal force if our motion is circular... So could someone explain what's going on?

Just because the motion of the bob is along a circle doesn't mean that its acceleration has to be tangent to it - which is, I think, the root of your confusion. The bob has both a tangential acceleration to change the magnitude of its velocity vector (because it is speeding up or slowing down according to whether it is going down the circle or up the circle currently) and also a radial (centripetal acceleration) that changes the direction of the velocity vector.

So, no, the tension should not equal the radial component of gravity because the bob does have net acceleration in this direction.

• Right, I think I was treating this as a statics problem, which could maybe explain why I made the error. Thanks for your answer! – Sha Vuklia May 13 '17 at 12:05
• Not correct - see my answer below. – Jens Jun 27 '17 at 14:25

Sorry about my previous answer, which was wrong. I was too quick but have now done the math which I submit for the sake of completeness and hopefully restore a bit of my reputation.

• This looks like an image cut from a book. What do you actually mean by "have now done the math"? :) It's good though if you understand what is done here. – nasu Jul 1 '17 at 21:12
• Not cut from a book but from a MS Word document I decided to write to get my own mind straight. I used Word's built-in equation editor and screen grabbed the above because I don't have/use MathJax (seems difficult to use), so for me it is easier to use Word (as I always do) and just snip the content - if not too long to fit on 1 screen. – Jens Jul 2 '17 at 0:15

The tensile force Fs in the string pulls on the bob in the direction of the pivot and if the gravity component of mg is Fg in the opposite direction then the acceleration of the bob towards the pivot is Fs-Fg but since there is no motion in this direction the net acceleration is zero and thus Fs=Fg as you correctly assume. In fact Fg= Fs=mv^2/r where r is the length of the string. Compare with a stone lying on a table. Is the acceleration g. Yes if if was not for the reactionary force from the table. Any circular motion gives a centripetal acceleration of v^2/r but this is a virtual acceleration only related to the circular path.

• "No motion" does not mean zero acceleration. There is definitely acceleration along the radial direction and is not Fs but the difference Fs-Fg which equals $mv^2/2$. – nasu Jun 27 '17 at 14:42
• @nasu you made a typo. 'R' not '2' in your last word. – Abhijeet Melkani Jun 27 '17 at 15:16
• Yes, thank you for pointing this out. Unfortunately it seems that I cannot edit it anymore. It should be $Fs-Fg= mv^2/r$, sure. – nasu Jun 27 '17 at 15:22