Can a Running Dog Have Angular Momentum Relative to a Nearby Merry-Go-Round?

[kehler]

If two things were not connected or touching each other and one of those things were moving with linear rater than rotational velocity - let's say a dog running down the road and a merry go-round on the side of the road - can we still calculate the angular momentum of the dog relative to the centre of the merry go round? Does the dog even have angular momentum since it's running in a straight line?
I'm having trouble grasping the concepts of angular momentum and torque. I've read the text I can't apply what I've read given a question that doesn't just involve plucking in numbers into a formula. Any suggestions on what I can do?

Any help would be appreciated :)

 

[Doc Al]

If two things were not connected or touching each other and one of those things were moving with linear rater than rotational velocity - let's say a dog running down the road and a merry go-round on the side of the road - can we still calculate the angular momentum of the dog relative to the centre of the merry go round?

Sure.

Does the dog even have angular momentum since it's running in a straight line?

About that point it does (but not about its center of mass, which is what you might be thinking of). After all, it is kind of "rotating" about that point as it passes by.

Reading these might help a bit: http://hyperphysics.phy-astr.gsu.edu/Hbase/amom.html"

 

[kehler]

Thanks. So if the dog has angular momentum, its angular momentum with respect to that point must be changing every second since it's distance from that point keeps increasing. I guess this would mean that torque is acting on the dog. Is that right?
Does this mean that the natural state of objects is to rotate unless an external torque acts upon it?

 

[Doc Al]

So if the dog has angular momentum, its angular momentum with respect to that point must be changing every second since it's distance from that point keeps increasing.

No. Its angular momentum is given by:
\vec{L} = \vec{r}\times\vec{p}

Where p is the dog's linear momentum and r is the position vector vector of the dog as measured from the reference point in question. The magnitude of that vector product equals rp\sin\theta. Even though the distance r changes, the product r\sin\theta does not change. (It's the distance from the reference point to the line of the dog's motion.)

So the dog's angular momentum about that point doesn't change and thus no torque is required.

I guess this would mean that torque is acting on the dog. Is that right?

Not at all. It would be pretty weird if the dog felt a torque acting just by running in a straight line.

Does this mean that the natural state of objects is to rotate unless an external torque acts upon it?

Not at all.

 

[kehler]

Thanks for that. I think I might have to read the chapter again. It's rather difficult to get it in my head after years of just dealing with linear momentum and energy.