So imagine a rod that receives an initial impulse acting on its tip. This impulse would cause the rod to translate and rotate about CM. Since the throw is at the tip, the rotational kinetic energy is at its max, correct? The minimum would be an impulse directly at the CM, which would cause only translation.
I just want to know the ratio between the two velocities $\frac{v}{\omega}$ when the rod is thrown by the tip. A general equation would also be appreciated.
Got it. I was refusing to use the impulse theorem because i thought it wouldn't cancel the time, but it does.
So, to start, a force $F$ perpendicular to the CM is applied at any point of the rod and this force applies an impulse into the center of mass in the value of:
$$ \int Fdt = p = mv$$
This contribution goes to the translational kinetic energy: $$ E_{kT} = \frac{mv^2}{2}= \frac{p^2}{2m}$$
Now, for the rotation contribution, the same force is applied at any given point $x$ away from the center of mass, providing an impulse in the form of torque:
$$ \int \tau dt =\int Fx dt = px = I\omega $$
Bar moment of inertia about CM: $\frac{mL^2}{12}$
$$ px = \frac{mL^2 \omega}{12}$$ $$\omega =\frac{12px}{mL^2} $$
This contribution goes into the rotational kinetic energy:
$$ E_{kR} = \frac{I\omega^2}{2}=\frac{6p^2x^2}{mL^2} $$
The total kinetic energy is:
$$ E_k = \frac{p^2}{2m} + \frac{6p^2x^2}{mL^2}$$
Now, assuming that the impulse is at the tip $x=\frac{L}{2}$
$$ E_k = \frac{p^2}{2m} + \frac{6p^2L^2}{4mL^2} = \frac{p^2}{2m} + \frac{3p^2}{2m} $$
Therefore, the rotational kinetic energy is 3 times the translational kinetic energy when the rod is thrown at the tip. Thus the translational energy is 25% the total energy and the rotational energy is 75%. Proceeding to calculate the ratio of velocities
$$\omega =\frac{12px}{mL^2} = \frac{6p}{mL} = \frac{6mv}{mL} =\frac{6v}{L} $$
$$ \frac{v}{\omega} = \frac{L}{6}$$
A little quicker way
Impact = momentum.
$p=m v$
Moment of inertia
$I=\frac{L^2 m}{12}$
Angular impact = angular momentum.
$\frac{p L}{2}=I \omega=\frac{1}{12} L^2 m \omega$
Eliminate $p$ from the two momentum equations, rearrange and you get the same answer you had before.
This method is also a good way to compute forward or reverse english when hitting cue ball at a certain height above the table.
This is really a purely geometrical question.
Consider an impulse $J$ acting through point A, at a distance $d$ from the center of mass C.
Instantaneously the body is going to rotate about some point B on the other side of the center of mass at a distance $c$.
The formula for where B is is
$$ \boxed{ c = \frac{I}{m d} = \frac{\kappa^2}{d} } $$
where $m$ is the mass, $I$ is the mass moment of inertia about the center of mass, and $\kappa$ is the radius of gyration such that $I=m \kappa^2$.
As you can see, the close you hit a body to the center of mass, the smaller $d$ is, and thus the larger $c$ is. At the limit when $d=0$, $c = \infty$ and the body purely translates (rotated about a point at infinity).
The reason I say this is a geometrical question is due to the fact the above equation only contains geometrical terms, $d$, $\kappa$, and $c$.
Specifically for the thin rod $I = \frac{m}{12} L^2$ and $d = \frac{L}{2}$ which makes $\kappa = \frac{L}{2 \sqrt{3}}$.
So the center of rotation is at $$ c = \frac{\kappa^2}{d} = \frac{L}{6} $$
Finally, the velocity of the center of mass is $v_C = c \omega$ and the velocity of the end point is $v_A = (c+d) \omega$.
Finally, the motion of the body is
$$ \omega = \frac{d\,J}{I} = \frac{6 J}{m L} $$
