# Can't figure it out

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

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Doc Al
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doxigywlz said:
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?

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. :-\

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

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
Mentor
doxigywlz said:
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: $\omega_f^2 = \omega_i^2 + 2\alpha \theta$.)

Doc Al said:
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. :(