# Roational motion help

1. Dec 13, 2006

### fsm

1. The problem statement, all variables and given/known data
A magnetic computer disk 8.0 cm in diameter is initially at rest. A small dot is painted on the edge of the disk. The disk accelerates at 600 rad/s^2 for .5s, then coasts at a steady angular velocity for another .5s.

A.What is the speed of the dot at t = 1.0 s? 12m/s
b. Through how many revolutions has it turned?

2. Relevant equations
ummm rotational kinematics...this really isn't the issue

3. The attempt at a solution

I was thinking of find theta for the first and second part then adding them, but this did not work. Then i tried to find alpha for the second part 300rad/s^2 then using omega as 300rad/s to get theta at 450

I am stuck on this problem trying to find the revolutions. I tried the above attempt but it was wrong. I know this is pretty easy but the more and more I try I'm getting frustrated and all knowledge seems to fly out the window.

2. Dec 13, 2006

### Staff: Mentor

>> ummm rotational kinematics...this really isn't the issue

Sure it is. Write down the rotational kinematic equations that relate angular position, angular velocity and angular acceleration. Then show your work as you solve for a) and b). If you get stuck, we can offer suggestions by looking at your work.

3. Dec 13, 2006

I suggest you write down the equations for uniformly accelerated angular motion first.

Edit: late again.

4. Dec 13, 2006

### fsm

For part a:

w=w0+at
w=0+600*.5

v=w*r
v=300*0.04
v=12m/s

For part b:
theta=theta0 +w0t+.5at^2
=0+300+.5*300*1
=450
I used the average acceleration. Part b is really a guess.

5. Dec 13, 2006

### Staff: Mentor

Try breaking it up into the two parts (accelerating and then coasting) and see if you get the same answers. That would be a good thing to check.

6. Dec 13, 2006

### fsm

Isn't it 300 in both parts?

7. Dec 14, 2006

### fsm

ok for the accleration:
theta=theta0+wit+.5at^2
theta=0+0_.5*600*.5^2

For the coasting:
theta=.5(wf+wi)t
theta=.5(300+300).5

revolutions=225/2pi=35.8 revolutions

8. Dec 14, 2006

For the coasting you can use the expression $$\theta(t) = \omega \cdot t$$, where $$\omega$$ is the angular velocity after the 0.5 sec acceleration.

Edit: actually, your answer seems to be correct, although I can't understand how you got it, but nevermind.

9. Dec 14, 2006

### BobG

Right answer, but you really only need the first equation. Once for the interval from 0 to 0.5 sec. Once for the interval from .5 to 1 sec. In the second interval, your initial speed is 300 rad/sec for 0.5 sec while your acceleration is 0. It comes to the same answer either way.

10. Dec 14, 2006

### fsm

Coolness thanks for the help!!!!