Find Angular Speed: Airplane Propeller Blades & Constant Acceleration

[nesan]

Homework Statement

The propeller blades of an airplane are 4.0 m long. The plane is getting ready for takeoff, and the propeller starts turning from rest at a constant angular acceleration. The propeller blades go through two revolutions between the fifth and the eighth second of the rotation. Find the angular speed at the end of 8.2 s.

The Attempt at a Solution

v = r ω

It seems very easy but I'm stuck on how to find ω

I know there's a constant acceleration

so

ω = ω_ο + αt

Can someone point me in the right direction with how oto use

"The propeller blades go through two revolutions between the fifth and the eighth second of the rotation."

to get the acceleration.

Than you.

 

[Orodruin]

ω = ω_ο + αt

What is the angular velocity when t = 0, i.e., ω_ο? Given this, what is the total angle turned as a function of time?

 

[nesan]

What is the angular velocity when t = 0, i.e., ω_ο? Given this, what is the total angle turned as a function of time?

Since it says it starts at rest, when t = 0, ωο would be 0?

We use the other formula

θ = ω_ot + 1/2 αt^2

So θ(t) = 1/2αt^2

How would I figure out α?

 

[Orodruin]

There is some information about $\theta(t)$ given in the problem formulation. Can you decipher it?

 

[nesan]

There is some information about $\theta(t)$ given in the problem formulation. Can you decipher it?

Whoohoo, I got it.

"The propeller blades go through two revolutions between the fifth and the eighth second of the rotation."

So

θ(8) - θ(5) = 4 PI

- > α (0.5 * 8² - 0.5 * 5²) = 4 PI

Solve for α and times it by 8.2 to get angular speed.

I got approximately 5.28 which my textbook says is correct. :)

Thank you so much Orodruin.