Homework Help: Tangential and angular velocity and acceleration

1. Dec 10, 2009

Panda040

Angular veloity, tangential velocity and acceleration ect.. blahh

okay so weve gon over it is class but it still confuses me...
some of the forumal's ive remembered are
s=(theta)r (theta)=s/r

anyways im not sue how to find the tangential velocity and angular velocity whats the difference?

do i have to find the displacement (theta) and plug it into the formula V=Wr to find tangental velocity?

plz any help will do im so lost

the kind of questions im dealing with are the steriotypical one like

an airplane is on a 3m string (radius) spinning in a horizontal circular path makes one revolution every .51 seconds (time)

and they want to knwo the tangential velocity and the angular velocity and acceleration?

i think this applies to gravity in our other problems like with a satallite orbiting at4.50x10exp6
and they want to know force and acceleration?

2. Dec 11, 2009

Mindscrape

Okay, so imagine a record turning on a record player. As the record is turning you can definitely tell that the parts towards the center are going "slower" than at the outside. This is the velocity we are used to dealing with in every day life, the "tangential velocity." This should be more physically intuitive for you.

At the same time, let's say we stop the record player, we paint a line from the center of the record to the edge, and start the record player up again. We notice that the line as a whole takes a certain amount of time to go around the record, this is the angular velocity. Just as regular velocity is a description of the change in distance over a period in time, angular velocity is a description of the change in radians over a period of time.

For a perfect circle, we can relate tangential velocity to angular velocity by the formula $$v=\omega r$$. You don't need to know the displacement to relate the two. After all, as long as the record remains whole, we know from that line we've drawn that a spot a little bit away from the center travels as far as a spot on the edge, the angular displacement is the same. As far as the displacement goes, it's simply a matter of a ratio between the circumferences.

Hopefully that makes more sense now. If not, let the ideas marinate a bit, and then look it over again.

$$\omega = \frac{\Delta \theta}{\Delta t}$$