Why Do Angular Velocity and Acceleration Differ from Other Rotational Analogues?

[asdofindia]

Angular momentum is equal to cross product of radius and linear momentum (moment of momentum)
Torque is the cross product of radius and force (moment of force)
But angular velocity is not the moment of linear velocity, and angular acceleration is not the moment of linear acceleration..

Why's this so?

I can't get myself to think properly about where the switch is happening.

 

[Penn.6-5000]

Think of it in terms of r's having to cancel.

Work = Force * distance = Torque * theta. If Torque = Force*radius then theta must equal distance(around the circumference)/radius so that the radii cancel.

Applied power = Force * momentum = Torque * angular velocity. If Torque = Force*radius then angular velocity = regular velocity / radius.

So now we have this system where we can see that some things are times r but other things are divided by r. How do we tell which is which? In general, the expressions that have masses in them end up having r multiplied. Momentum is mass times velocity, so angular momentum is regular momentum TIMES the radius; regular velocity has no mass, so angular velocity is regular velocity DIVIDED BY the radius. Force is mass times acceleration so angular force (torque) is regular force TIMES the radius; regular acceleration has no mass so angular acceleration is regular acceleration DIVIDED BY the radius.

Why is that the case? Hrumm... I guess I would explain it in terms of the way two properties transform from linear to rotational quantities. First: You convert a displacement to an angle by dividing by the radius. The car went 300 meters along the outside of the track? What was that in radians? Well, it was a 150-meter-radius track, so the car went 300m/150m = 2 radians around the track. Angular velocity and angular acceleration work the same way for the same reason: if I'm driving 30m/s around the track, that means that in one second I go 30 meters, or 0.2 radians, so my angular velocity is 0.2 radians/second---note that I still had to divide by the radius to get there.

So what about things with mass? It turns out that the "rotational mass" of a particle about a point is mr^2, where m is the particle's regular mass and r is the distance from the particle to that point. So when you form angular momentum by multiplying this rotational mass by the angular velocity, one of the r's in mr^2 cancels out the r from the angular velocity, and the other r sticks around to be a part of the formula.

Why is the "rotational mass" (more properly called the "moment of inertia") equal to mr^2? The derivation is that if a particle is moving with in a circle of radius r with a velocity v, then its kinetic energy will be (1/2)mv^2, but since v = wr because that's the definition of angular velocity, then the kinetic energy is (1/2)mr^2w^2, so if we think of I = mr^2 as its own interesting quantity, then the kinetic energy of a particle moving in a circle can be written as (1/2)Iw^2, which has the same basic form of (1/2)mv^2.

Is that good enough? We could derive things more rigorously if you'd prefer.

 

[asdofindia]

Is that good enough?

Of course. That's the kind of insight only betterexplained.com used to provide me (in maths) earlier. Thanks a lot.

Why don't all the people here start a website or a book with such intuitive, insightful explanations? I'll sure be ready to invest.