Rotating uniform rod about an arbitrary axis

[Dustgil]

Homework Statement

A thin uniform rod of length l and mass m is constrained to rotate with constant angular velocity omega about an axis passing through the center O of the rod and making an angle alpha with the rod.
a) Show the the angular momentum L about O is perpendicular to the rod and is of magnitude
$$\frac {ml^{2} \omega} {12} sin(\alpha)$$

b) Show that the torque vector N is perpendicular to the rod and to L and is of magnitude
$$\frac {ml^{2} \omega^{2}} {12} sin(\alpha) cos(\alpha)$$

Homework Equations

I think these two:

$$L = I_{x} \omega_{x} i + I_{y} \omega {y} j + I_{z} \omega {z} k$$

$$\frac {dL} {dt} = N$$

The Attempt at a Solution

This chapter has me really confused at the moment. Where do I define the axes? I tried assigning the k axis to be along the length of the rod. then L should be perpendicular to the rod because the moment of inertia is zero in the k direction (because every point on the rod is at zero distance from that axis). The moment of inertia about the i and j axes is
$$I = \frac {ml^{2}} {12}$$

This is where i get lost. I'm not sure what the component of omega is in the x and y directions. Soooo not sure if I'm doing this right.

 

[haruspex]

Consider the plane containing both rod and axis at some instant. Which way is the instantaneous velocity of a point on the rod?

 

[Dustgil]

the instantaneous velocity should be perpendicular to the plane containing both the rod and the axis. if the plane coincides with the computer screen, it will be into the screen if the rotation is counterclockwise and out of the screen if clockwise.

 

[haruspex]

into the screen if the rotation is counterclockwise and out of the screen if clockwise

That is not quite right. Can you correct that?

 

[Dustgil]

Yeah, I see. half of it is rotating into the screen and half is rotating out. the piece at the center of mass isn't translating, just rotating.

 

[haruspex]

Yeah, I see. half of it is rotating into the screen and half is rotating out. the piece at the center of mass isn't translating, just rotating.

Good. So can you figure out the direction of the angular momentum vector?

 

[Dustgil]

Well it's perpendicular to the rod. My guess for why is that the two halves cancel and we just consider the piece that's not translating. I'm not sure if that fully answers your question.

 

[haruspex]

Well it's perpendicular to the rod. My guess for why is that the two halves cancel and we just consider the piece that's not translating. I'm not sure if that fully answers your question.

No, they don't cancel. This is angular momentum, not linear.
Try to forget the problem statement for the moment and just consider what you deduced about the instantaneous velocities. How does the linear velocity of a point on the rod vary with its distance from the centre? What axis would you think the rod was rotating about?

 

[Dustgil]

the linear velocity grows as you increase distance from the center. Going from that statement alone it implies that the axis of rotation is through the center of mass.

 

[haruspex]

the linear velocity grows as you increase distance from the center. Going from that statement alone it implies that the axis of rotation is through the center of mass.

Yes, but you can say more than that. Consider a pair of points on the rod equidistant from the centre. Taken as a pair, what axis do they appear to be rotating about, instantaneously?

 

[Dustgil]

Ah, that's an axis at an angle alpha to the whole rod.

 

[haruspex]

Ah, that's an axis at an angle alpha to the whole rod.

That's not what I see.
Consider two equal particles at (-1,-1) and (1,1). One is heading straight into the page and one straight out. Taken as a pair, what axis is the pair rotating about at that instant? I.e., which axis maximises the angular momentum?

 

[Dustgil]

I've thought about this for a while. I tried to find some visualizations to help me through it but I couldn't locate any online. I suppose it has to be perpendicular, But if the path of each particle traces a circle out in space due to rotation, the center of each circle is an axis at an angle alpha to the whole rod. instantaneously, maybe it's perpendicular...

 

[haruspex]

instantaneously, maybe it's perpendicular...

That's the point - the axis of rotation keeps moving. If you just look at the instantaneous velocities of points along the rod you cannot distinguish its movement from rotating about an axis perpendicular to the rod. Since that choice maximises the moment, that is what it is doing.
The rotation of the rod about the axis at angle alpha is a second order phenomenon involving accelerations.

 

[TSny]

One approach is to recall that the angular momentum of any system of particles about a point O is a sum over the particles of the form $\vec{L} = \sum_i \vec{r}_i \times \vec{p}_i$ where $\vec{r}_i$ is the position of the ith particle relative to O. The sum is replaced by an integral for continuous mass distributions. Using this formula you can easily see the direction of $\vec{L}$.

Alternately, the answer is easily determined from your formula $\vec{L} = I_x \omega_x \hat{i} + I_y \omega_y \hat{j}+I_z \omega_z \hat{k}$ if you interpret it correctly. Have you covered the notion of moment of inertia tensor and principal axes?

 

[Dustgil]

We have, I'm just unsure of how to apply it. I'd imagine one of the principal axes is along the length of the rod, correct?

 

[TSny]

We have, I'm just unsure of how to apply it. I'd imagine one of the principal axes is along the length of the rod, correct?

Yes. The formula assumes that you choose your coordinate axes to align with principal axes of the rod. So, you'll need to choose one of the coordinate axes to be along the length of the rod.

It can be a little confusing because the rod is changing its orientation relative to the inertial "lab" frame. But for a specific instant of time, you can choose the coordinate axes so that one of the axes is along the length of the rod.

 

[Dustgil]

so if i choose the z axis to be along the length of the rod then will omega work out like this?

$$\omega_{x} = cos(\alpha)$$
$$\omega_{y} = sin(\alpha)$$

This was the problem I was really having with the route. I couldn't understand the geometry well enough to break omega in components.

 

[TSny]

so if i choose the z axis to be along the length of the rod then will omega work out like this?

$$\omega_{x} = cos(\alpha)$$
$$\omega_{y} = sin(\alpha)$$

No. If you choose the z axis along the rod, wouldn't $\vec{\omega}$ have a nonzero z component? The x and y components of $\vec{\omega}$ would depend on your choice of orientation of the x and y axes. Choose the orientation to simplify the calculation.

 

[Dustgil]

I guess you're suggesting that omega would go to zero along either the x or y axis. the moment of inertia around the z axis is zero so that would leave one component. I think you can choose the axes so that either x or y is orthogonal to the direction of omega. Is this the correct track?

 

[TSny]

I think you can choose the axes so that either x or y is orthogonal to the direction of omega. Is this the correct track?

Yes.

the moment of inertia around the z axis is zero so that would leave one component.

One component of what?

 

[Dustgil]

of the angular momentum.

 

[TSny]

of the angular momentum.

OK, yes.

 

[Dustgil]

Okay. I have to go into work now but the second part should just involve taking the derivative of the angular momentum, right? Not sure how that shakes out to both sin and cos in the term.

 

[TSny]

the second part should just involve taking the derivative of the angular momentum, right?

Yes, that's right (the time derivative).