Calculating Angular Velocity and Revolutions: Magnetic Disk Problem

[fsm]

Homework Statement

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?

Homework Equations

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

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.

 

[berkeman]

>> 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.

 

[radou]

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

Edit: late again.

 

[fsm]

For part a:

w=w0+at
w=0+600*.5
w=300rad/s

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.

 

[berkeman]

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.

 

[fsm]

Isn't it 300 in both parts?

 

[fsm]

ok for the accleration:
theta=theta0+wit+.5at^2
theta=0+0_.5*600*.5^2
theta=75rad

For the coasting:
theta=.5(wf+wi)t
theta=.5(300+300).5
theta=150rad

total distance=225rad
revolutions=225/2pi=35.8 revolutions

 

[radou]

For the coasting:
theta=.5(wf+wi)t
theta=.5(300+300).5
theta=150rad

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.

 

[BobG]

ok for the accleration:
theta=theta0+wit+.5at^2
theta=0+0_.5*600*.5^2
theta=75rad

For the coasting:
theta=.5(wf+wi)t
theta=.5(300+300).5
theta=150rad

total distance=225rad
revolutions=225/2pi=35.8 revolutions

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.

 

[fsm]

Coolness thanks for the help!