In this picture the acceleration vector $\vec{a}$ points upward when the pendulum is halfway

Click To see animated GIF

But according to this picture, the force acts tangentially:

enter image description here

Which means the acceleration should be tangential too, and never pointing upward?

So whats right?

  • $\begingroup$ Looks like the diagram didn't label the tension force by the string very well. $\endgroup$
    – BMS
    Aug 29, 2014 at 16:38

3 Answers 3


The diagram is misleading. Look at this:

enter image description here

At any moment in time, you have the following forces on the particle:

  1. Gravity
  2. Tension in the string

When you are at the bottom of the path, the tension in the string is equal to the tension needed to counter gravity, PLUS the tension needed to keep the mass in its path (in other words, to keep the string constant length). The net effect of these two forces is a force that is pointing exactly upwards - and since $a=F/m$, this means that at that precise moment the mass is accelerating upward.


Please note that in the picture, there are two forces acting: 1) the weight, mg, which acts vertically downward, and does not change, and 2) the tension in the string, Z, which points from the mass to the point the string connects to the ceiling, provided the string remains taut. Z varies with time periodically.

These two forces combine to give the resultant force, and it is the resultant force which occurs in the same direction as the acceleration, as seen in the gif.

The green arrows in the picture are actually just the tangential and normal components of gravity.

Edit: also, I believe the source of confusion might have lied with assuming the normal component of gravity cancels with the tension. This is not the case: you cannot use the equations of equilibrium if the system is not in equilibrium, i.e. accelerating.

  • $\begingroup$ Okay, guess I understand now. But regarding your last paragraph, I think the normal component is always equals the tension because this is the force which pulls on the string? For example, in the picture, the green normal components equals Z, or not? $\endgroup$
    – StefanH
    Aug 29, 2014 at 12:09
  • $\begingroup$ @Stefan - no, Z (which I think is meant to be the tension) can be larger than the normal component of gravity. The difference provides the net acceleration to keep the mass in the circular orbit. $\endgroup$
    – Floris
    Aug 29, 2014 at 12:10
  • $\begingroup$ The two actually don't equal. By saying they're equal, you are using an equation of equilibrium. However, you cannot assume that as the system is not in equilibrium. $\endgroup$
    – Involute
    Aug 29, 2014 at 12:12
  • $\begingroup$ when it passes its equilibrium position, then the tension Z equals the normal component of the force (because this is where the equilibrium equation applies), or when it has reached its maximum displacement (because then $a$ indeed points tangentially)? $\endgroup$
    – StefanH
    Aug 29, 2014 at 12:23
  • 1
    $\begingroup$ At maximum displacement the velocity is zero, so the only tension needed is the one that opposes the normal force of gravity. At that point $a$ is indeed tangential. I assume that when you say "equilibrium position" you mean the lowest point of the arc. At that point the tension is equal to the force of gravity plus the centripetal force - net force straight up. $\endgroup$
    – Floris
    Aug 29, 2014 at 13:19

True: But the complete solution to this problem is

1) Use ODE

yielding Theta(t)= Theta0*cos(w*t) w=Sqrt(g/L):

From ODE: d2(Theta)/dt2+(g/L) *(Theat)=0: Homogeneous Response, Initial Conditions yield Theta(t) above

2) Apply Dynamic Equations in 2 D Polar: these ACCELERATIONS[relative to the Rotating COORDINATE SYSTEM] are:

Remember that "w" is from solution of the ODE and is a function of time : w(t)=d(Theta(t))/dt

Radially= (dL2/d2t)-(wsquared)*L :Radial Acceleration/Centripital Acceleration

Tangentially=[d2(Theta)/dt2]L+2(Theta/dt)*L: Tangential Acceleration/Coriolis Acceleration.

In the case of a Pendulum, Radial and Coriolis are zero.

Now just "plug and chug" w(t) into the Dynamic Equations for Tangential and Centripital Acceleration.

NOTE: This can easily be done from Velocity , as well ,since:


Radial: dL/dt

Tangential: [(Theta(t))/dt]*L



d(Tangentialunitvector)/dt=-w*Radialunitvector Again: (d(Theta(t))/dt=w(t)

Hope this helps:

PPM Ph.D Applied Physics


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