Trying (desperately) to understand angular momentum

[zkhandwala]

Hi - I'm undertaking a self-study of calculus-based physics 101 using Sears/Zemansky's 'University Physics' text (12th ed). So far so good, but I've hit a mental stumbling block on the topic of angular momentum and am hoping this forum can help me get past it...

The text begins by referring to an arbitrary 3-D rigid body rotating around the z-axis. This same body was used to discuss rotational energy, inertial moment calculation, etc. The text then considers one arbitrary particle of the body, and defines the angular momentum of the particle as \vec{r} x m\vec{v}. What's immediately odd to me about this is that \vec{r} is considered relative to the origin of the coordinate system, when in previous sections it had been given as relative to the axis of rotation. Annoyed , I accept this and move on, but...

...Then the notion is introduced that the rate of change of this particle's angular momentum is equal to the torque applied to the particle. Fair enough, but here again the torque is given as relative to the origin, which really makes no sense to me, since the particle in question is not revolving about the origin, but rather around the z-axis (when torque was introduced in 2-D space, it was calculated relative to the point around which the body was rotating, which made sense). I try to accept this 'new' notion of torque as applied to a 3-D body, but can't, since it seems to me to imply the following: consider a solid cylinder rotating around the z-axis, with the base of the cylinder lying on the x-y plane. Now consider that I apply a tangential force somewhere on the surface of cylinder, in order to create rotational acceleration around the z-axis. According to the definition of torque as relative to the origin, the higher up on the cylinder that I apply the force, the greater the magnitude of \vec{r}, and thus the greater the magnitude of torque, which doesn't make sense to me.

Anyway, at this point the text goes on to confuse me further, but before we go there I'd like to resolve my confusion about the above. Any advice that might help me think through this a little more clearly? Thanks in advance!

Cheers,
Zain

 

[tiny-tim]

Welcome to PF!

Hi Zain! Welcome to PF!

… consider a solid cylinder rotating around the z-axis, with the base of the cylinder lying on the x-y plane. Now consider that I apply a tangential force somewhere on the surface of cylinder, in order to create rotational acceleration around the z-axis. According to the definition of torque as relative to the origin, the higher up on the cylinder that I apply the force, the greater the magnitude of \vec{r}, and thus the greater the magnitude of torque, which doesn't make sense to me.

The higher up on the cylinder, the greater the torque for tipping the cylinder over!

You're assuming, however, that the cylinder can only rotate about the z-axis …

in that case, only the z-component of the angular momentum and of the torque matter …

and the z-component of the torque does not depend on how far up the force is applied (because that only adds a multiple of k to r, and that won't affect k.(r x F))

 

[zkhandwala]

Thanks, TT! I guess what still confuses me is that the origin is not a point of rotation for this system. Then again, is there even such a thing as a "point of rotation" when we're dealing with three dimensions, or can we only talk about an axis of rotation in such a space?

In any case, is this a meaningful way to think about torque in 3-D space?: we calculate the torque vector for any particle with respect to the origin. This vector represents the rotational force that would apply to that particle if we were to have it revolve around the origin (in a 2-D sort of way). Then, if we want to know what the effective torque on that particle would be if we were to have it instead revolve around the z-axis, we would simply take the z-component of the torque vector we had calculated.

Does that make sense? If so, then I presume I can make the same argument for momentum vectors as well as torque vectors?

Thanks again. This discussion is very helpful!

 

[tiny-tim]

Hi Zain!

(just got up :zzz: …)

… is there even such a thing as a "point of rotation" when we're dealing with three dimensions, or can we only talk about an axis of rotation in such a space?

there's always an (instantaneous) axis of rotation …

that's why we're able to define an angular velocity vector ω, defined by v = ω x r

the torque vector … represents the rotational force that would apply to that particle if we were to have it revolve around the origin (in a 2-D sort of way).

Sorry, i don't understand this.

Then, if we want to know what the effective torque on that particle would be if we were to have it instead revolve around the z-axis, we would simply take the z-component of the torque vector we had calculated.

Torque vector = rate of change of angular momentum vector.

So z component of torque vector = rate of change of z component of angular momentum vector.

 

[zkhandwala]

Thanks again. It's taking me a little time to digest this, and I'm not really sure why, but I'm at least comforted by the fact that Walter Lewin, in his MIT 8.01 lectures on the topic, does constantly stress that the concept of angular momentum is the most challenging one (for instructors and students) in all of introductory physics :-).