Angular acceleration of a merry-go-round - no time given?

[murielglass]

Angular acceleration of a merry-go-round - no time given!?

Homework Statement

A merry-go-round accelerating uniformly from rest achieves its operating spped of 2.5rpm in 5rev. What is the magnitude of its angular acceleration?

2.5rpm=0.262rad/s

Homework Equations

angular acceleration = change in angular speed / time

The Attempt at a Solution

I've tried every possible thing I could think of. I actually know what the answer is (0.0011rad/s^2) but I can't find how they get to that. The teacher told me that if the final angular speed is 2.5rpm and it takes it 5rev to get that final angular speed, then t=2min.. but it doesn't make sense to me since the merrygoround is not moving at that final speed since t=0. plus, i don't get the right answer using t=120s. so I've tried calculating angular speeds for each revolution, assuming that in rev#1 the merry-go-round goes from 0 to 0.5rpm (=0.0524rad/s), from 0.5 to 1rpm, etc, averaging speeds and without averaging, but i keep getting it wrong. I'm guessing there's a conceptual issue I'm missing in this line of thought.

i figured out by the angular acceleration formula that t should equal 238.1s, but I want to know how they get to that number.

 

[rock.freak667]

Do you know the equations of motion for rotation?

Mainly

\omega_f^2 = \omega_i^2 + 2 \alpha \theta

 

[murielglass]

Do you know the equations of motion for rotation?

Mainly

\omega_f^2 = \omega_i^2 + 2 \alpha \theta

Yes! It's in my book too. could i use that one? theta is.. final position - initial position?

What would be my position here? final=10pi rad - initial=0?

that actually makes sense..

 

[murielglass]

Do you know the equations of motion for rotation?

Mainly

\omega_f^2 = \omega_i^2 + 2 \alpha \theta

i get 0.109 :D

thanks so much. God bless you, my friend.