Can't figure it out

[doxigywlz]

Okay, I have this very EASY question (or so it appears) but I don't know how to figure out the time in this problem:

a coin with a diameter of 2.4 cm is dropped on edge onto a horizontal surface. the coin starts out with an initial angular speed of 18 rad/s and rolls in a straight line without slipping. if the rotation slows with an angular acceleration of magnitude 1.9 rad/s^2, how far does the coin roll before coming to rest?

I know w initial is 18 rad/s and w final is 0
alpha is 1.9rad/s^2 (or is it negative??)
we're looking for theta...

I could figure it out if I had the time.. Can someone help me out?? Thanks

[Doc Al]

I could figure it out if I had the time..
That's one way to do it. So figure out the time. Hint: What's the definition of acceleration?

[Pseudo Statistic]

I'm going to guess here...
Since v = rw, r = 2.4/2 = 1.2, and w = 18rads^-1, so 18(1.2) = v (Not going to calculate, too lazy and tired..)
a = Ar, and A = 1.9rads^-2, r = 1.2, so a = 1.9(1.2)...
Now, we'll consider v to be initial velocity, u...
Use the formula:
v^2 = u^2 + 2as
Where s = distance, plug it in..
0^2 = (18(1.2))^2 + 2(1.9(1.2))s
And solve for s.
I hope I gave the right method, tell me if it worked. :-\

[doxigywlz]

pseudo, no-- it didn't work (unless I did it wrong)...

Acceleration is meters per second squared or, in this case, radians per second squared... so how does that help me solve for time?
1.9=change in w over seconds squared.. i tried to solve for it, but i got it wrong..

please a little more help? i have to go to work now but i will definately check back later

[Pseudo Statistic]

Hmmm, I seem to have made a mistake writing that...
Change the acceleration to a negative value, thus giving the equation:
0^2 = (18(1.2))^2 + 2(-1.9(1.2))s
Try solving for s now and see if it works.
There's no way it's going to come to rest if it's not deceleration! (Unless I haven't learnt about something in Physics which causes something to come to rest, rather than net forces..)

[Doc Al]

Acceleration is meters per second squared or, in this case, radians per second squared... so how does that help me solve for time?
1.9=change in w over seconds squared.. i tried to solve for it, but i got it wrong..
The definition of angular acceleration is change of angular velocity (omega) per unit time. Writing it for rotational motion: \alpha = \Delta \omega / \Delta t. You know the change in \omega and the acceleration, so find the time.

By the way, Pseudo Statistic just solved the problem slightly differently. (That's why in my first response I said that your way is just one way of solving for the angle.) I recommend that you solve the problem both ways, just for the practice. Your way: find the time, then use it to find the distance. His way: Use the kinematic formula relating angular distance and speed to get the answer directly. (For some reason Pseudo Statistic converted from angular to linear speed in writing the kinematic equation--that's not wrong, just unnecessary. Also, as he realized, he made a slight error in signs. In any case, the kinematic equation he used is: \omega_f^2 = \omega_i^2 + 2\alpha \theta.)

[Pseudo Statistic]

The definition of angular acceleration is change of angular velocity (omega) per unit time. Writing it for rotational motion: \alpha = \Delta \omega / \Delta t. You know the change in \omega and the acceleration, so find the time.

By the way, Pseudo Statistic just solved the problem slightly differently. (That's why in my first response I said that your way is just one way of solving for the angle.) I recommend that you solve the problem both ways, just for the practice. Your way: find the time, then use it to find the distance. His way: Use the kinematic formula relating angular distance and speed to get the answer directly. (For some reason Pseudo Statistic converted from angular to linear speed in writing the kinematic equation--that's not wrong, just unnecessary. Also, as he realized, he made a slight error in signs. In any case, the kinematic equation he used is: \omega_f^2 = \omega_i^2 + 2\alpha \theta.)
Hey, atleast I got it right. ;)
We aren't even going to take Angular velocity/acceleration in the Physics course I'm in. :(