Is there no translational energy?

[TheCanadian]

Homework Statement

A uniform rod of length b and mass m stands vertically upright on a floor and then tips over.
a) Assuming that the floor is rough (i.e., the end of the rod that is initially touching the floor cannot slip), what is the rod’s angular velocity when it hits the floor?

2. Homework Equations

$$ T = \frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 $$

The Attempt at a Solution

I have found the solution, but only by setting $$ U_i = T_f $$ (initial potential energy = final kinetic energy). But what I found odd is that I set $$ T_f = \frac{1}{2}I\omega_f^2 $$ and stated the translational energy term goes to 0. I got the right answer by doing this, but it doesn't seem right. I understand that the CoM of the rod remains at the same length radially, but since there is an x- and y-component to its velocity, shouldn't it have a non-zero final translational velocity? Although this is reminiscent almost of a ball kept stationary but rotated, which has only rotational energy, and zero net translational energy (at its CoM)...but are these two cases really analogous? In one case it seems very clear that the CoM is moving, while in the other, the CoM is stationary from a fixed axis.

 

[TSny]

Homework Equations

$$ T = \frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 $$

To help answer your question, think about the meaning of the above equation. In particular, the speed $v$ in the equation refers to the speed of what point? The symbol $I$ refers to the moment of inertia about what axis?

 

[TheCanadian]

To help answer your question, think about the meaning of the above equation. In particular, the speed $v$ in the equation refers to the speed of what point? The symbol $I$ refers to the moment of inertia about what axis?

$v$ would refer to the CoM, or $\frac {b}{2}sin\theta$ if $\theta$ is the angle the rod makes with the horizontal, right? And $I$ represent the inertia about the rotation axis (i.e. parallel to the rod itself), point in the radial direction. Is it wrong to think of this as non-uniform circular motion, so that there's only a tangential velocity equivalent $b\omega$? Is it because there is only a tangential velocity that we consider the velocity of the CoM to be 0?

 

[TSny]

Yes, $v$ is the speed of the center of mass when using the equation $T = \frac{1}{2} mv^2 + \frac{1}{2}I\omega^2$. However, $I$ in this equation is the moment of inertia about an axis through the center of mass, not the moment of inertia about the end of the rod. So, it might be clearer to write the equation as $$ T = \frac{1}{2} mv_c^2 + \frac{1}{2}I_c\omega^2 $$

If you are familiar with the parallel axis theorem, you can relate $I_{end}$ about one end of the rod to $I_c$ about the center of mass. Then you can show that $$ T = \frac{1}{2} mv_c^2 + \frac{1}{2}I_c\omega^2 = \frac{1}{2}I_{end}\omega^2$$

 

[TheCanadian]

Yes, $v$ is the speed of the center of mass when using the equation $T = \frac{1}{2} mv^2 + \frac{1}{2}I\omega^2$. However, $I$ in this equation is the moment of inertia about an axis through the center of mass, not the moment of inertia about the end of the rod. So, it might be clearer to write the equation as $$ T = \frac{1}{2} mv_c^2 + \frac{1}{2}I_c\omega^2 $$

If you are familiar with the parallel axis theorem, you can relate $I_{end}$ about one end of the rod to $I_c$ about the center of mass. Then you can show that $$ T = \frac{1}{2} mv_c^2 + \frac{1}{2}I_c\omega^2 = \frac{1}{2}I_{end}\omega^2$$

Wow, I was looking at it very wrong. Thank you!

 

[TheCanadian]

Yes, $v$ is the speed of the center of mass when using the equation $T = \frac{1}{2} mv^2 + \frac{1}{2}I\omega^2$. However, $I$ in this equation is the moment of inertia about an axis through the center of mass, not the moment of inertia about the end of the rod. So, it might be clearer to write the equation as $$ T = \frac{1}{2} mv_c^2 + \frac{1}{2}I_c\omega^2 $$

If you are familiar with the parallel axis theorem, you can relate $I_{end}$ about one end of the rod to $I_c$ about the center of mass. Then you can show that $$ T = \frac{1}{2} mv_c^2 + \frac{1}{2}I_c\omega^2 = \frac{1}{2}I_{end}\omega^2$$

I also just had a small related question. I was just wondering why in this image, the derivation for I begins with stating $dI = \frac{1}{2} y^2dm$. Where exactly does this 1/2 coefficient come from? Shouldn't it just be $dI = y^2dm$?

 

[TSny]

I also just had a small related question. I was just wondering why in this image, the derivation for I begins with stating $dI = \frac{1}{2} y^2dm$. Where exactly does this 1/2 coefficient come from? Shouldn't it just be $dI = y^2dm$?

Here, $dI = \frac{1}{2} y^2dm$ is the rotational inertia of a thin disk of radius $y$ and mass $dm$, as shown in the figure. So, the 1/2 factor is correct here. If $dm$ were the mass of a particle (rather than a disk), then $dI$ would equal $r^2dm$, where $r$ is the distance from the point mass to the axis of rotation.