Cannot understand dL/dt (angular momentum changes)

[TonyEsposito]

Hi guys...i'm not able to understand the rate of change of angular momentum. Suppose i have a disk rotating with an angular velocity vector w along the axis, so the angular momentum is: L=(InertiaTensor)*w, now if i apply a torque the L will change according to: M=dL/dt=d/dt[(InertiaTensor)*w]=(inertia)*(dw/dt)...right? Now the thing that is bugging me: why the disk change orientation? Should not change only the way it rotates? if w was along the z axis after w should be of axis and the disk star to rotate in a weird way because of a w component along is radius...can anybody explain it to me, I'm a little dumb so explain steo by step...please!

 

[Chandra Prayaga]

M = dL/dt is a vector equation. If M is in the same direction as L, then dL is in the same direction as M, and therefore the change in angular momentum is along the same direction as the angular momentum, so the disk will either speed up or slow down, but continue to rotate along the same axis.

Question is, what happens if the torque M is not in the same direction as L?

Let me give you a corresponding situation which may be more familiar
acceleration = rate of change of velocity: a = dv/dt. This is also a vector relation (in fact it is the definition of acceleration). If a is in the same direction as v, as happens in a ball thrown vertically up or down, then either the ball speeds up or slows down but the direction of the velocity does not change.

On the other hand, if the acceleration is in some direction different from the direction of the velocity, as happens in projectile, or circular motion, then the acceleration contributes to the change in velocity, and that change is in the same direction as the acceleration: dv = a dt, so the velocity changes direction.

Hope this helps

 

[TonyEsposito]

No, it does not help, I'm adding a picture to help explain my doubt :

https://ibb.co/bxZzBx

 

[A.T.]

why the disk change orientation? Should not change only the way it rotates? if w was along the z axis after w should be of axis and the disk star to rotate in a weird way because of a w component along is radius

Not every possible axis of rotation is stable, and the symmetry axis also has the least air resistance.
http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node71.html

 

[CWatters]

Hi guys...i'm not able to understand the rate of change of angular momentum. Suppose i have a disk rotating with an angular velocity vector w along the axis, so the angular momentum is: L=(InertiaTensor)*w, now if i apply a torque the L will change according to: M=dL/dt=d/dt[(InertiaTensor)*w]=(inertia)*(op/dt)...right?

Correct.

Now the thing that is bugging me: why the disk change orientation? Should not change only the way it rotates? if w was along the z axis after w should be of axis and the disk star to rotate in a weird way because of a w component along is radius...can anybody explain it to me, I'm a little dumb so explain steo by step...please!

If the applied torque is around the z axis the orientation of the disc stays the same. There is no "component of w along its radius".

Torque is a pseudo vector like angular velocity...

https://en.m.wikipedia.org/wiki/Torque

Typically (for example accelerating a flywheel) the direction of the torque vector points in the same direction as the angular velocity vector (eg both point in the z axis).

 

[FactChecker]

Maybe I am missing something. A rotation is, by it's definition, a change of orientation. If the rotation is around an axis of symetry, then the change may not be noticable, but it is there. Any rotation that is not around an axis of symmetry would be noticable. Therefore, any torque which causes a rotation around an axis that is not an axis of symmetry would be noticable. I guess you can call that "rotate in a weird way".

 

[CWatters]

I think the OP believes the axis of rotation tilts when it shouldn't. At least that how I interpret his drawing.

 

[Charles Link]

The drawing of the OP is somewhat unclear, but I think he is basically puzzled by the gyroscope phenomenon if I interpret his drawing correctly. See this video=perhaps this is what his puzzle is. $\\$ In the video, gravity tries to pull the axis one way, and surprisingly, the axis moves at 90 degrees to the applied force. If the wheel were not spinning, when gravity is applied, it would do just what you would expect it to do.

 

[FactChecker]

The drawing of the OP is somewhat unclear, but I think he is basically puzzled by the gyroscope phenomenon if I interpret his drawing correctly. See this video=perhaps this is what his puzzle is. $\\$ In the video, gravity tries to pull the axis one way, and surprisingly, the axis moves at 90 degrees to the applied force. If the wheel were not spinning, when gravity is applied, it would do just what you would expect it to do.

Oh, of course -- precession! I think this may make intuitive sense of it:

 

[A.T.]

The above is a good video on gyroscopic precession.

But I think the OP asks, how changing the angular momentum vector from L to L' by applying a torque M implies a certain new rotation, when the same L' could be achieved by different rotations of the same body. Here I think it also matters how M is applied and which rotation axes are stable:

A gyroscope supported at a point against gravity has the rotation constrained to axes through that point.

A gyroscope floating in space has more freedom, but still preferred stable axes of rotation.

 

[TonyEsposito]

"
But I think the OP asks, how changing the angular momentum vector from L to L' by applying a torque M implies a certain new rotation, when the same L' could be achieved by different rotations of the same body."
yeah, somehow this is the source of my doubt

 

[FactChecker]

"
But I think the OP asks, how changing the angular momentum vector from L to L' by applying a torque M implies a certain new rotation, when the same L' could be achieved by different rotations of the same body."
yeah, somehow this is the source of my doubt

Are you saying that a given angular momentum does not uniquely determine the rotation vector?

 

[Charles Link]

Are you saying that a given angular momentum does not uniquely determine the rotation vector?

The fact that the body being rotated has a significant angular momentum attached to the body makes for a completely different result of how the body responds when a torque is applied. We'll see of the OP @TonyEsposito concurs, but I believe that is what puzzled him.

 

[TonyEsposito]

The key of my doubt is in the drawing, initially the w vector is aligned with the axis of the disk then a torque is applied changing the direction of w, so why the disk does not start to spin around the new direction of w but instead it changes also orientation?

 

[Charles Link]

The answer seems to be that although a torque $\vec{ \tau}$ can cause a change in the angular velocit/rate of rotation $\omega$, so that with moment of inertia $I$, $\vec{\tau}= \frac{d \vec{L}}{dt}=I (\frac{d \omega }{dt}) \hat{z}$, it is a different case when the body has an inherent angular momentum attached to it. $\\$ For something that is has an angular momentum already associated with it, the change in angular momentum is not found simply by $\Delta \vec{L}=I \Delta {\omega} \, \hat{z}$, but rather the change $\Delta \vec{L}=\vec{L}_2-\vec{L}_1$ , also needs to be considered, where the change may simply be a change in the direction of the vector $\vec{L}$. $\\$ If the applied torque $\vec{\tau}$ goes into a changing the direction of the inherent angular momentum $\vec{L}$ of the system, there may be very little change of the rate of rotation of the form $\Delta \vec{L}=I \Delta{ \omega } \hat{z}$ that needs to occur. $\\$ And if you notice, in the case of precession, the precession occurs at a steady rate. The applied torque is steady, but there is no angular acceleration that occurs. The angular momentum vector $\vec{L}$ changes at a constant rate simply by changing its direction at a constant rate.

 

[FactChecker]

The key of my doubt is in the drawing, initially the w vector is aligned with the axis of the disk then a torque is applied changing the direction of w, so why the disk does not start to spin around the new direction of w but instead it changes also orientation?

It is spinning around a different axis $\vec{w}$. That might be hard to see because we like to think that it is still spinning around the new axis of symmetry, $\vec{w_2}$, with a separate slow rotation of $\vec{w_2}$. But those two motions are really a simple spin around $\vec{w}$.

EDIT: The Euler Rotation Theorem proves that the sum of the two rotations is a single rotation around a single axis. (see https://en.wikipedia.org/wiki/Euler's_rotation_theorem )

 

[A.T.]

Are you saying that a given angular momentum does not uniquely determine the rotation vector?

A given angular momentum vector doesn't uniquely describe the motion of a rigid body. The below T-handle has approx. constant angular momentum, while it rotates around various body-axes, especially during the quick transitions:

 

[A.T.]

why the disk does not start to spin around the new direction of w but instead it changes also orientation?

Did you see the points I made?

A gyroscope supported at a point against gravity has the rotation constrained to axes through that point.

A gyroscope floating in space has more freedom, but still preferred stable axes of rotation.
http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node71.html

 

[TonyEsposito]

Yeah, thanks A.T. i think you cleared my head from a lot of clouds! :) :)

 

[TonyEsposito]

Sorry guys, one last question...we are doing a very simple lab experiment, we have a disk rotating around is axis with a weight acting on it (the very basic experiment)...the disk have a mass of 0.1 Kg, the radius is 0.3 so the Inertia is 0.5*0.1*0.3^2=0.0045 the torque acting is 0.1 so the angular acceleration is:
acc=0.1/0.0045=22.22 so in one second the rotation done is: 0.5*0.22*1^2*(180/Pi)=636 Degrees...but this result is nosense, too large...what I'm doing wrong?

 

[Charles Link]

Sorry guys, one last question...we are doing a very simple lab experiment, we have a disk rotating around is axis with a weight acting on it (the very basic experiment)...the disk have a mass of 0.1 Kg, the radius is 0.3 so the Inertia is 0.5*0.1*0.3^2=0.0045 the torque acting is 0.1 so the angular acceleration is:
acc=0.1/0.0045=22.22 so in one second the rotation done is: 0.5*0.22*1^2*(180/Pi)=636 Degrees...but this result is nosense, too large...what I'm doing wrong?

Please show a diagram, and/or describe the force/torque on the disk in more detail. A question like this really belongs in the homework section though.

 

[TonyEsposito]

https://ibb.co/kkyfKH

This is the diagram, i did not want to open a new thread its a very silly question :p

 

[Charles Link]

https://ibb.co/kkyfKH

This is the diagram, i did not want to open a new thread its a very silly question :p

There are two forces on the mass: $F_g=-mg$ and $T$ upward. Unless the motion/acceleration of the disk is minimal, $T \neq mg$, so that the torque which is $\tau=T \cdot R$ may not be $\tau=mg \cdot R$.

 

[TonyEsposito]

Sorry i do not understand what do you mean, the torque was calculated to be Torque=m*g*r=0.1

 

[Charles Link]

Sorry i do not understand what do you mean, the torque was calculated to be Torque=m*g*r=0.1

What is the mass $m$ that is hanging from a string? What are your units for radius? $.3 \, m \approx \, 1 \, ft$?

 

[Charles Link]

What is the mass $m$ that is hanging from a string? What are your units for radius? $.3 \, m \approx \, 1 \, ft$?

This is why it really belongs in the homework section. The homework helpers would make sure you showed your work, and put some effort into it.

 

[TonyEsposito]

https://ibb.co/kn1OpH

 

[Charles Link]

https://ibb.co/kn1OpH

How was the torque computed? Unless you can show that, it is pointless to try to see what might have led to an incorrect result.

 

[TonyEsposito]

A mass of 0.034 was tied to a string as in the picture posted

 

[Charles Link]

A mass of 0.034 was tied to a string as in the picture posted

$\\$ Your calculation of $\alpha=22$ radians/sec^2 looks correct for this torque. At a radius of $R=.3 \, m$ , the outer portion of the disc as well as the mass that is supplying the downward force would then accelerate at $a=- 6.6$ m/sec^2 . $\\$ In order for the tension $T$ to be $mg$, the resulting acceleration that we just computed needed to be much much less than 9.8 m/sec^2. A more complete calculation of the torque appears necessary, where the tension $T$ in the string is first computed.