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)
