Angular Velocity of a wind turbine

Click For Summary
The discussion revolves around calculating the initial angular velocity and the time elapsed for a wind turbine experiencing constant angular acceleration. The participant is uncertain about which equations to use for solving the problem but acknowledges the similarity between angular and linear motion equations. Key equations mentioned include the relationship between angular velocity, angular displacement, and angular acceleration. It is suggested that by eliminating time from the equations, one can solve for the initial angular velocity, and then use that value to determine the time for part b. Overall, the focus is on developing problem-solving skills through understanding the underlying equations.
neoncrazy101
Messages
22
Reaction score
0

Homework Statement


A wind turbine is initially spinning at a constant angular speed. As the wind's strength gradually increases, the turbine experiences a constant angular acceleration 0.127 rad/s2. After making 2790 revolutions, its angular speed is 155 rad/s. (a) What is the initial angular velocity of the turbine? (b) How much time elapses while the turbine is speeding up?



Homework Equations


I'm not sure. I'm given a lot of equations but I think w=w0 +\alphat


The Attempt at a Solution


Since I'm not sure of the equation I haven't really attempted. I know you don't do my homework and I don't want you to, but could you inform me as to what equation I should use for part A?
 
Physics news on Phys.org
The neat thing about angular velocity/acceleration etc.. is that if you have the linear velocity/acceleration equations down then the angular ones are pretty much the same thing.

So you have that the angular velocity is related to angular displacement by:
\omega = \frac{\theta}{t} (see how this is similar to v = d/t)
and also related to angular acceleration by:
\alpha = \frac{\Delta\omega}{t} (similar to a=v/t)

Other equations you might see are just derived from these. and just keep in mind that:
\Delta\omega = \omega_{final}-\omega_{initial}
 
When I first read your post I thought it made sense, and it does but I can't seem to get it to work. Any other advise or information to push me on?
 
Ya.
Actually the omega in the equation:
\omega=\frac{\theta}{t}
should be the average over the period of constant acceleration:
\omega_{avg}=\frac{\theta}{t}
this can then be rewritten as:
\frac{\omega_{initial}+\omega_{final}}{2}=\frac{ \theta}{t}

You are probably used to seeing equations more like d=(1/2)at^2 and stuff like that. I don't think that is the best way to learn this IMO... because when you learn calculus all of this gets replaced anyway. So the whole point here is for you to develop problem solving and algebraic skills. That's why I always like to tell people to do it this way. Even with linear acceleration/velocity/distance problems you only need 2 equations to solve everything... the tradeoff is that there is less memorization and these two equations are more intuitive, but you have to do a little more thinking.

So now with that equation I rewrote above and the one for angular acceleration in my first post you only have 2 unknowns. The time can be eliminated to make a single equation with only 1 unknown (the initial angular velocity) so you can solve for that.
Then to do part b just plug in your new value for the initial angular velocity to one of the equations and find time.

Hope that helps.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K