Given are two solid spheres of the same size and weight. They both have their center of mass at their geometric center.
One of them (A), however, has most of its mass distributed near the center (heavy center, light shell),
while the other (B) has most of its mass at the shell (light center, heavy shell).
Now I apply a tangential force ($F$) to them (they are at rest, floating in empty space). (When the ball starts to move, let's assume the force will stay tangential to the ball, i.e., the spot on the ball's surface, that the force will be applied to, changes with rotation. So the directions of the force will continue to be the same as it was at the start.)
Assuming I put the same amount of work into both ($F \cdot s$), I guess A will spin faster than B, because of the different moments of inertia.
Also, both will not only gain rotation but translation too. Will they both have the same (translational) speed, or does the difference in mass distribution make a difference here too?
Best Answer
For equal total energy, and equal force, the total time will be different, leading to different translational speed. This is because the energy fraction in translation vs rotation will be different. And that will affect how much energy per time is needed to provide equal force.
For equal force for equal time (not equal energy), the speed must be the same. Another way to see this, is to imagine the force as being applied by some particle at the edge somehow colliding tangentially. The particle feels an opposite and equal force, and thus develops a momentum. The spheres, regardless of their content, must have the same total opposite momentum, so that the sphere+particle momentum remains 0. If the force is equal for equal time (not equal work), then both spheres have the same total linear momentum and mass, hence the same translational speed.
Edit: Notes on energy
Note $Fr = I\dot{\omega}$ and $F = m\dot{v}$.
$2E = I\omega^{2} + mv^2$
$2\dot{E} = 2I\omega\dot{\omega} + 2mv\dot{v}$
$= 2\omega Fr + 2vF$
$= 2F(\omega r + v)$
What is $\omega r + v$? The instantaneous velocity of the part of the sphere on the very edge experiencing the force.
In time $\Delta{}t$ it moves a distance of $(\omega r + v)\Delta{}t$ so the work done is $\dot{E}\Delta{}t = F(\omega r + v)\Delta{}t = F\Delta{}x$.
In the particle analogy, note that in order for a stream of particles to provide the same force by collisions, the particles must have higher velocity commensurate with the edge velocity $\omega{}r+v$. It requires higher energy particles to provide the same force to a faster moving edge, so energy conservation is not broken--it is just that the energy input is higher to provide the same $F$, in a case where $\omega r + v$ is higher. But with Force and time specified, rather than total energy, it intuitively hold that the translational velocity of both spheres should be the same. With Power (energy/time) specified, the spheres would differ, as they would have the same energy at time $t$ but a different fraction of the energy would be rotational vs translational.
Edit 2:
I will try to argue that the relevant $F\cdot{}s$ is the edge of the ball, not the center of mass of the ball, and it depends on the velocity of the edge. Let us use a pole to fix the ball to rotate on its center of mass, so that $s_{COM}=0$. The COM clearly does not move. The pole fixing the COM to not move provides a force but moves no distance, and does no work. The force $F$ is spinning the ball and increasing the energy of the ball and doing work on the ball, but $F \cdot s_{COM}=0$. So if you believed that the relevant distance is the COM distance, you would find a contradiction. If you believe instead that $F\omega r dt$ is the instantaneous energy added because the edge moves with velocity $\omega r$ over a distance $\omega r dt$, then you can integrate. $E = \int F\omega r dt = \int F \dot{\omega}trdt = \int I\dot{\omega}^{2}tdt = \frac{1}{2}t^{2}I\dot{\omega}^{2} = \frac{1}{2}I\omega^{2}$
Assume $\dot{\omega}t = \omega$ (constant acceleration) and $Fr = I\dot{\omega}$ (torque definition)