I have a question bothering me for a long while. It is kind of a combination of the questions and answers of two other posts:

The setting: we have a rigid body, for example a thin rod, in space. We give it a nudge at the center of mass or at one of its tips.

A explains, that in both cases the Force will change the linear velocity of the body in the same way and in the second case it will additionally increase the angular velocity.

B explains, that this does not violate conservation of energy because the force is applied for a shorter path in case 1 than in case 2 due to the rotation. Thus, more work is done which explains the higher energy.

Did I get this right so far?

But what if the example is like the following:


Here we use a compressed Spring to apply the force. So in the end, the same amount of work $W = \frac{1}{2}k\Delta x$ will be done, right? Can we now answer the question of A: how much of the work will end up in rotational and how much in linear kinetic energy?

EDIT 1 Okay, I tried to do, what R.W.Bird suggested:

  • Linear: $$ F = k\Delta x = ma = m\dot{v} $$

  • Angular (force is acting on tip of rod with length $L$): $$ \tau = \frac{L}{2}k\Delta x = I\dot{\omega}\\ I = \frac{L^2m}{12} $$

  • Movement of the tip point $x$, assuming that vertical movement is very small and can be neglected: $$ \dot{x} = v + \frac{L}{2}\omega\\ \ddot{x} = \dot{v} + \frac{L}{2}\dot{\omega}\\ \Delta x = max(s - x,0) $$ For the last one I set the x-axis of the coordinate system aligned with the spring's axis and its origin at the contact point with the rod. $s$ is the spring's rest length.

  • Now putting it all together:

\begin{align} \ddot{x} &= \dot{v} + \frac{L}{2}\dot{\omega} = \frac{k}{m}\Delta x + \frac{L^2k}{4I}\Delta x = \left(\frac{k}{m} + \frac{3k}{m}\right)\Delta x\\ \lambda &= \frac{4k}{m}\\ \\ \mathbf{\ddot{x}} &\mathbf{= \lambda s - \lambda x} \quad (for \ x < s) \end{align}

  • Solving this ODE with $x(0) = 0$ and $\dot{x}(0) = 0$ results in:

$$ x(t) = s(1 - cos(\sqrt{\lambda}t)) $$

  • Solving $x(t) = s $ for $t$ results in $t_s = \frac{\pi}{2\sqrt{\lambda}}$
  • Plugging in $x(t)$ in the first two equations for $\dot{\omega}$ and $\dot{v}$ and integrating once results in: $$ v(t) = \frac{ks}{m\sqrt{\lambda}}sin(\sqrt{\lambda}t) \quad v(t_s) = \frac{ks}{m\sqrt{\lambda}}\\ \omega(t) = \frac{Lks}{2I\sqrt{\lambda}}sin(\sqrt{\lambda}t) \quad \omega(t_s) = \frac{Lks}{2I\sqrt{\lambda}} $$

Now those should be the final values for linear and angular velocity when the spring is fully relaxed.

  • Finally computing the rotational and translational KE: $$ TKE = \frac{1}{2}mv^2 = \frac{k^2s^2}{2m\lambda} = \frac{ks^2}{8}\\ RKE = \frac{1}{2}I\omega^2 = \frac{L^2k^2s^2}{8I\lambda} = \frac{12L^2k^2s^2m}{8mL^2 4k} = \frac{3ks^2}{8} $$

Wohoo that was fun :D And the Energies sum up to $\frac{ks^2}{2}$ like a charm.

And now, the answer of my own question is: 75% of the Energy ends up in rotational and 25% for translational kinetic Energy. (I hope, I didn't mess something up...)

Maybe I will also compute the result for a general Rigid Body and a general distance $r$ from the center, where the force is applied and share the results here.


After I wrote all this, I saw, that the result was maybe already clear at: $$ \ddot{x} = \dot{v} + \frac{L}{2}\dot{\omega} = \left(\frac{k}{m} + \frac{3k}{m}\right)\Delta x $$

The more general solution should be:

$$ \ddot{x} = \dot{v} + r\dot{\omega} = \frac{F}{m} + \frac{r^2F}{I}\\ \\ tKE : rKE = 1 : \frac{mr^2}{I} $$

Which means for something like a thin ring or cylinder it should be 50% for rotational KE and 50% for translational KE. And for any body it could never be more than 50% translational KE if the force acts at the out most point. (Given that, I did not mess up something which is not uncommon)


1 Answer 1


The question implies that in both cases the spring exerts the same force over the same distance. The resulting total kinetic energy should be the same in both cases. We can note that in case B, the spring encounters less inertial resistance. It expands faster, and reaches full extension in less time. With less time, the linear impulse is smaller, and the change in linear velocity is also smaller. The gain in rotational K.E. Is matched by a loss of translational K.E.

  • $\begingroup$ That was clear to me, but can we compute the ratio of rotational KE/translational KE? Assuming that inertia, length, mass of the rod and initial potential energy of the spring and the point where the force is applied are given of course... $\endgroup$
    – Catrexis
    Commented Nov 10, 2020 at 21:42
  • 1
    $\begingroup$ From the linear impulse (the integral of F dt): J = mv. From the angular impulse J(L/2) = (1/12)m$L^2$ω (relative to the C.M.). Divide these two and you get, ω, in terms of, v. $\endgroup$
    – R.W. Bird
    Commented Nov 11, 2020 at 14:04

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