Rotational acceleration

1. Apr 12, 2007

Rudipoo

I'm getting confused with different types of acceleration when dealing with rotating systems. There is centripetal acceleration, tangential acceleration, and angular acceleration as far as i know. How do you derive that linear momentum equals angular momentum multiplied by the radius?

And also, in which types of accleration are unit vectors required?

Thanks

2. Apr 12, 2007

lpfr

Almost! Angular momentum of a particle is the product of it linear momentum times the distance from de reference point to the straight line where the particle moves. NOT to the particle. In vector representation you can write:
$$\vec L = \vec r\times \vec p$$
This time $$\vec r$$ is the vector from the center to the particle and $$\vec p$$ the linear momentum. Beware: $$\times$$ stand for vectorial product.

3. Apr 12, 2007

Rudipoo

Ah I see (I think!). Is the straight line an extension either way of velocity vector line? I might be talking rubbish here...

How does the cross product differ from the dot product? And also, i've seen that
a=rA where a is the linear acceleration r is the radius and A is the angular acceleration. How does one derive this from w=v/r , because I know angular acc. is the derivative of angular velocity?

Thanks again

4. Apr 12, 2007

lpfr

Even if it is rubbish, it is clear enough for me, and yes it is "the extension of the vector".

Vector product is very different to dot product. The first gives a vector and the second a scalar. You can look in wikipedia.
You write
[tex]V_T=R\omega[/TEX]
and you derive both sides.

5. Apr 12, 2007

Rudipoo

Cheers that makes things clearer. I'm afriad my experience at differentials is sufficiently small that I don't know how to derive both sides. V_t goes to a_t by definition of acceleration I suppose, but I haven't got any t's on the RHS of the equation, and as its differentiating w.r.t t, I'm stuck... Help!! Thankyou for your time

6. Apr 12, 2007

lpfr

The time derivative of linear speed is linear acceleration, the time derivative of angular speed is angular acceleration. R does not change. You let it as it is.

7. Apr 12, 2007

Rudipoo

Oh yes of course. Thanks for your help.