# Rigid Body in Space - Energy Conservation

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.

EDIT 2

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)

• 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. – R.W. Bird Nov 11 '20 at 14:04