angular momentum - Confusion regarding rotational motion!

Let us assume I have a rod of some mass m, moment of inertia I, length l and center C.

If I apply a force F on C for a duration of time t, it will accelerate forward. If I apply it elsewhere, the body will rotate. Now, my confusion arises here:

• Will the body still accelerate as much as it did when F was applied on C?


• If so, why?


• How can it rotate and yet accelerate with the same velocity as that of without rotation?


• Where is it getting this extra energy from?


• How can I calculate the respective velocities of linear and angular motion given F and t?

Please cite any reliable sources if possible and state all the formulas.

Answer

I think this should help clear things up. Suppose you take a rod at rest and apply a force $F$ perpendicular to the rod at a distance $r$ away from its center of mass for a short time $\delta t$ - short enough that the orientation of the rod does not change much during the time the force is applied. The rod's linear momentum will become

$$F\delta t$$ (from $F = \mathrm{d}p/\mathrm{d}t$) and its angular momentum will become approximately $$rF\delta t$$ (from $\tau = rF\sin\theta = \mathrm{d}L/\mathrm{d}t$).

You can then calculate the rod's kinetic energy after this process:

$$K = K_\text{lin} + K_\text{rot} = \frac{p^2}{2m} + \frac{L^2}{2I} = \frac{F^2 \delta t^2}{2m} + \frac{r^2 F^2 \delta t^2}{2mL^2/12} = \frac{F^2 \delta t^2}{2m} + \frac{6r^2F^2}{mL^2}\delta t^2\tag{1}$$

where $L$ is the length of the rod. If you hit the rod away from the center of mass, it acquires more energy.

Now, why is that? Well, let's think about what happens if we do consider the fact that the rod's orientation changes as the force is being applied to it, if the force is off center. We know the rod acquires an angular momentum $rF\delta t$ in time $\delta t$, which corresponds to angular velocity

$$\omega = \frac{L}{I} = \frac{rF\delta t}{mL^2/12} = \frac{12rF}{mL^2}\delta t$$

Assuming the torque is constant, the angular acceleration will also be constant,

$$\alpha = \frac{\Delta\omega}{\Delta t} = \frac{12rF}{mL^2}$$

and for constant angular acceleration we can figure out the total angular displacement as

$$\Delta \phi = \frac{1}{2}\alpha\Delta t^2 = \frac{6rF}{mL^2}\delta t^2$$

The linear displacement corresponding to this angular displacement is just

$$r\Delta\phi = \frac{6r^2 F}{mL^2}\delta t^2$$

which means that when you apply the force off center, the point at which you are applying the force moves $\frac{6r^2 F}{mL^2}\delta t^2$ further than when you apply the force at the center. (Of note: this is zero when $r=0$, as it should be.)

Work is force times distance, when the force is constant, so this means the off-center force does a small amount of extra work relative to the on-center force because of the increased distance:

$$W = \frac{6r^2 F^2}{mL^2}\delta t^2$$

which is exactly the same as the extra amount of energy that the rod acquires when you apply the force off center, from equation (1). This is the origin of that extra energy: the extra distance that the point of application of the force moves.