My tentative understanding from two previous stackexchange posts listed at the end of this post (one my own) is that an object being swung in a circle with a rope possesses an internal rotational energy about its center of mass. However I don't feel totally confident in my conclusion because this perspective is new to me, and I want to ask a follow-up question about a pendulum (which was not addressed in the previous stackexchange posts) to check my understanding.

Is the simple basic conservation of energy equation relating gravitational potential energy and translational kinetic energy

$$ mgh_{i} + 1/2mv^{2}\, _{i} = mgh_{f} + 1/2mv^{2}\, _{2}$$

not totally exact for any real pendulum? It seems to me that some of the initial gravitational potential energy contributes to the internal rotational energy component at the expense of the translational energy component. Only at the top of the pendulum's swing, when the angular rotation direction changes direction does the angular rotational energy equal zero. Thus at the bottom of the swing, when the pendulum is moving the fastest, its translational speed will be less than that of a free-falling body dropped from the same height.

It is surprising to me that I haven't seen this point mentioned in texts that I've read, especially since the texts mention the small-angle approximation for the derivation of the equation of motion. I think that I see that this internal rotational energy term can be neglected under conditions of very long pendulum arm length. My purpose in asking this question is not to do a gotcha of the textbooks, as I realize that physics frequently makes use of limiting cases. I am just trying to better understand rotational motion, which my previous question had also addressed.

Along the same line of reasoning, would the full energy analysis of other common systems like a frictionless rollercoaster or brachistochrone also require a rotational energy term?

See Does angular momentum exist around the moving center of mass of a released swinging rigid body?

and this stackexchange question

Rotational Kinetic Energy of a Pendulum

  • 1
    $\begingroup$ Look at compound physical pendulum in link, where they mention the effects of rotation of extended masses in the energy balance. The reason why this is not mentioned often in physics textbooks is because in most cases the mass at the end of the pendulum is considered punctual, the extended case is maybe mentioned when covering rotational dynamics. $\endgroup$
    – secavara
    Commented Apr 4, 2021 at 20:03
  • $\begingroup$ thanks I was wondering about that exact case, that makes perfect sense $\endgroup$
    – lamplamp
    Commented Apr 4, 2021 at 20:58
  • 1
    $\begingroup$ @lamplamp to capture the rotational aspect you can use energy = $(1/2)I\omega^2$ where $I$ would be the moment of inertia about the pivot and $\omega$ is the angular velocity at the bottom of the swing. Here the entire pendulum including the rope is assumed to be a rigid body with a defined mass and center-of-mass (for calculating potential energy). $\endgroup$
    – Roger Wood
    Commented Apr 4, 2021 at 22:30
  • $\begingroup$ thanks that makes sense $\endgroup$
    – lamplamp
    Commented Apr 5, 2021 at 4:40
  • $\begingroup$ Are we talking about a solid body, or a point particle in the end of the pendulum? $\endgroup$
    – JAlex
    Commented Apr 6, 2021 at 14:07

1 Answer 1


The most general case is that of a solid rigid body attached to the end of the pendulum. The initial potential energy is converted into kinetic energy at the bottom of the swing.


Kinematically the speed of the center of mass is linked to the rotation of the pendulum $$ v_f = \ell \omega_f$$

The solid object has not only mass $m$ but mass moment of inertia $I$ about the center of mass.

$$\begin{array}{r|l} \text{quantity} & \text{expression} \\ \hline \text{initial potential energy} & PE = m g h_i \\ \text{final kinetic energy} & KE = \tfrac{1}{2}m v_f^2 + \tfrac{1}{2} I \omega_f^2 \\ \end{array}$$

Equating the two gives

$$ m g h_i = \tfrac{1}{2}m v_f^2 + \tfrac{1}{2} I (v_f/\ell)^2 $$


$$ m g h_i = \tfrac{1}{2} (m + I/\ell^2 ) v_f^2 $$

$$ \Rightarrow \;\;\; v_f = \sqrt{ \frac{2 m g\, h_i}{m + I/\ell^2} } $$

So only if the object is a point mass with $I=0$ the above equals the falling speed of

$$ v_f = \sqrt{ 2 g\,h_i } $$

The way to think about it is in terms of energy. If there is a rotational energy component $\tfrac{1}{2} I \omega^2$ included then the translational kinetic energy is going to be less than the potential energy.


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