Angular velocity and angular acceleration of a turbine

In summary: You are asking for tangential and radial acceleration, which are two different types of acceleration.
  • #1
arhzz
266
52
Homework Statement
The angular speed of a turbine with a diameter of d = 1m increases steadily from 20rad / s to 40rad / s within 4s.

a) Calculate the angular acceleration

b) the total angle that the turbine has rotated during this time interval.

c) What is the normal acceleration that acts on the outermost tip of the turbine blade at the top speed?
Relevant Equations
a(t) = w/t
Okay so what I've done;

I've put the diammter d = 1m as r = 1m

The time interval of 4s is t = 4s

and the angular velocitys as;

ω1 = 20 rad/s

ω2 = 40 rad/s

Now to get the accelaration. Angular acceleration can be split into two parts tangetial acceleration and radial acceleration

What I think is what I am looking in this example (under a) is the radial acceleration.

I've found this formula to get angular acceleration

$$ a = \frac { \Delta \omega}{\Delta t} $$So for ## \Delta \omega ## I've done this

$$ \Delta \omega = \omega 2 - \omega 1 $$

That should be 20 rad/s, as for ## \Delta t## I'd reckon it remains 4, because 4 - 0 is 4

So a should be

$$ a = \frac {40 - 20} {4} $$

$$ a = 5 \frac {rad} {s^2} $$Now the problem is that although the unit should be rad/s^2 when I input the unit of omega and time in the calculation I'dont get rad/s^2,which means that my calculation is probably wrong. Could anyone confirm my thought process ? And if it is wrong what would be a good starting point, thank you!
 
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  • #2
arhzz said:
I'dont get rad/s^2
That is strange. What do you get ?
 
  • #3
arhzz said:
Angular acceleration can be split into two parts tangetial acceleration and radial acceleration
No, linear acceleration can be split into radial and tangential (in relation to some chosen axis).
arhzz said:
what I am looking in this example (under a) is the radial acceleration.
No, you are looking for angular acceleration.
arhzz said:
for Δt I'd reckon it remains 4
No, Δt= 4s, not 4.
Please show how you get units other than rad/s2 from rad/s divided by s.
 
  • #4
BvU said:
That is strange. What do you get ?
Well this is my calculation; A wise man said show as much of your work as you can so I will
$$a = \frac { \Delta \omega}{\Delta t} $$

Than I put the units in the equation

$$a = \frac {rad/s} {s}$$

So than we need to make it a double fraction multiply the outer one with the outer and inner with the inner,

Now if I did that correctl rad is multiply by s and s is multiplyed by 1. thus making it rads/s

Not rad/s^2.
 
  • #5
arhzz said:
Well this is my calculation; A wise man said show as much of your work as you can so I will
$$a = \frac { \Delta \omega}{\Delta t} $$

Than I put the units in the equation

$$a = \frac {rad/s} {s}$$

So than we need to make it a double fraction multiply the outer one with the outer and inner with the inner,

Now if I did that correctl rad is multiply by s and s is multiplyed by 1. thus making it rads/s

Not rad/s^2.
Show me how you calculate a half of a half: ##\frac{1/2}2##.
##\frac{rad/s}s=\frac{(rad/s)s}{(s)s}=\frac{rad}{s^2}##
 
  • #6
Yea actually when I was writing my post I've rechecked my calculation and realized I've made a mistake (an emberassing one at that). So does this mean that my calculation was right? the angular acceleration is 5? And also I'd just like to ask is there a radial acceleration and a tangetial acceleration, because I have found things like radial acceleration and tangetial acceleration when talking about circular motion. Also this question in particular also under a) is asking for angular acceleration, and under c) for "normal" acceleration. My way of seeing it was that under a) radial acceleration was asked and under c) tangetial. Have I missunderstood those concepts, is under c) asked a simple acceleration with m/s^2? Thank you!
 
  • #7
arhzz said:
Yea actually when I was writing my post I've rechecked my calculation and realized I've made a mistake (an emberassing one at that). So does this mean that my calculation was right? the angular acceleration is 5? And also I'd just like to ask is there a radial acceleration and a tangetial acceleration, because I have found things like radial acceleration and tangetial acceleration when talking about circular motion. Also this question in particular also under a) is asking for angular acceleration, and under c) for "normal" acceleration. My way of seeing it was that under a) radial acceleration was asked and under c) tangetial. Have I missunderstood those concepts, is under c) asked a simple acceleration with m/s^2? Thank you!
Assuming they are using the term in the standard way, normal acceleration is another name for centripetal acceleration, which is also known as radial acceleration. It is the component of linear acceleration which is normal to the instantaneous velocity.
The component parallel to the velocity is the tangential acceleration.

In polar coordinates (unit vectors ##\hat\theta## along the tangent and ##\hat r## along the radius), the tangential acceleration vector is ##(r\ddot\theta+2\dot r\dot\theta)\hat\theta##, and the radial acceleration vector is ##(\ddot r-r\dot\theta^2)\hat r##.
For 2D motion, the angular acceleration is ##\alpha=\ddot\theta##, while the angular velocity is ##\omega=\dot\theta##.

For the special case of circular motion, ##\dot r=\ddot r=0##, so the tangential acceleration vector is ##r\ddot\theta\hat\theta##, and the radial acceleration vector is ##-r\dot\theta^2\hat r##.
 
  • #8
haruspex said:
Assuming they are using the term in the standard way, normal acceleration is another name for centripetal acceleration, which is also known as radial acceleration. It is the component of linear acceleration which is normal to the instantaneous velocity.
The component parallel to the velocity is the tangential acceleration.

In polar coordinates (unit vectors ##\hat\theta## along the tangent and ##\hat r## along the radius), the tangential acceleration vector is ##(r\ddot\theta+2\dot r\dot\theta)\hat\theta##, and the radial acceleration vector is ##(\ddot r-r\dot\theta^2)\hat r##.
For 2D motion, the angular acceleration is ##\alpha=\ddot\theta##, while the angular velocity is ##\omega=\dot\theta##.

For the special case of circular motion, ##\dot r=\ddot r=0##, so the tangential acceleration vector is ##r\ddot\theta\hat\theta##, and the radial acceleration vector is ##-r\dot\theta^2\hat r##.
Ah, I see I'd assume my calculation was right?
 
  • #9
arhzz said:
Ah, I see I'd assume my calculation was right?
5 rad/s2 is right for a). What do you have for the other parts?
 
  • #10
haruspex said:
5 rad/s2 is right for a). What do you have for the other parts?

Okay so for b the total angle that the turbine has rotated during this time interval,was asked I solved it like this

φ=ω∗t

Since i have ω1 and ω2 I've done this

φ1=ω1∗t
φ2=ω2∗t

So for φ1=80 and for φ2=160

Than I have added those together and gotten φ = 240

As for the part c I've solved it like this

$$a_n = \frac {v^2} r$$

For v I've used this formula

$$ v = r * \omega $$

Again since I have 2 omega's I've calculated this speed twice, getting v1 = 20 m/s and v2 = 40m/s

Than I've put it back into ##a_n## like this;$$ a_n = \frac {v2^2 - v1^2} r $$

and after doing the calculations I've gotten ##a_n## = 1200 m/s^2
 
  • #11
arhzz said:
φ1=ω1∗t
ω1 is the initial rotation rate. What do you think you are calculating as φ1?
You can't just throw terms together because they are of the right dimension. You have to have some understanding of you are calculating at each step.
arhzz said:
Again since I have 2 omega's I've calculated this speed twice, getting v1 = 20 m/s ...
Part c) only concerns what is happening at the top speed. The history is no longer relevant.
 
  • #12
haruspex said:
ω1 is the initial rotation rate. What do you think you are calculating as φ1?
You can't just throw terms together because they are of the right dimension. You have to have some understanding of you are calculating at each step.

Part c) only concerns what is happening at the top speed. The history is no longer relevant.
Ah your right, it only asks what is happening at top speed, I've overseen that part.I'll keep that in mind when I redo my calculation.

Now for part b) my thought process was that the angle is changing with the speed (the faster the turbine rotates the more rotations ). So I've looked at the time interval,and the speeds within that time interval.

As for your question what I was calculating with φ1=ω1∗t ;

Well now that you said that it is the initial rotation I'd assume that at the initial rotation rate the angel was 0,because if it just started rotating it starts from an angle of 0 degrees.
 
  • #13
Rotation at constant angular acceleration is very like linear motion at constant acceleration. Equations just like the SUVAT equations can be applied.
For part b, you have an initial rotation speed, a final rotation speed and a time, and you want the rotational distance covered. Can you find a SUVAT equation with a corresponding four variables?
 
  • #14
Oh, we can look at the problem that way? Now if I'm not mistaken this would be the equation I need;

$$ s = \frac {(U+V)} 2 * T $$

Where U is the initial velocity, V the final velocity s the distance, and T the time.

So If I "adjust" it to my problem I should be getting this

φ = ## \frac {(\omega1 +\omega2)} 2 * t ##

φ = 160and for c) If we are only interested in the top speed;

a = 1600 m/s^2.

So whenever we have have a problem with an constant angular acceleration the motions of equation can be applied? What about when the angular acceleration is not constant, when it is changing over time?

Thank you for the insights!
 
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  • #15
arhzz said:
φ = ## \frac {(\omega1 +\omega2)} 2 * t ##
Yes.
arhzz said:
φ = 160
No.
arhzz said:
and for c) If we are only interested in the top speed;
a = 1600 m/s^2.
How are you calculating that? Looks like you just squared 40 rad/s, but that won't give you m/s^2.
arhzz said:
What about when the angular acceleration is not constant, when it is changing over time?
Then it is like linear acceleration that varies over time:
##v=\int a.dt##, ##s=\int v.dt##, ##\omega=\int \alpha.dt##, ##\theta=\int \omega.dt##.
 
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  • #16
arhzz said:
... So whenever we have have a problem with an constant angular acceleration the motions of equation can be applied?
Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html

Rotation at a constant radius means that we have angular displacement, time, angular velocity, and angular acceleration.
This rotational problem is equivalent to a mass accelerating at a constant rate from initial to final velocity on a straight path.
These rotation equations apply only in the case of constant angular acceleration.
 
Last edited:
  • #17
Okay I've made another quite emberassing mistake by b) it should be 120. I'll look into c again the units don't add up.

EDIT: Okay I think I've gotten it now

b) 120
c) 800 m/s^2
 
Last edited:
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  • #18
arhzz said:
Okay I've made another quite emberassing mistake by b) it should be 120. I'll look into c again the units don't add up.

EDIT: Okay I think I've gotten it now

b) 120
c) 800 m/s^2
Good.
 

Related to Angular velocity and angular acceleration of a turbine

1. What is angular velocity?

Angular velocity is the rate of change of angular displacement over time. It is a measure of how quickly an object is rotating around a fixed axis.

2. How is angular velocity measured?

Angular velocity is typically measured in radians per second (rad/s) or degrees per second (deg/s). It can also be expressed in revolutions per minute (RPM).

3. What factors affect the angular velocity of a turbine?

The angular velocity of a turbine is affected by the speed and direction of the fluid flow, the size and shape of the turbine blades, and the torque applied to the turbine shaft.

4. What is angular acceleration?

Angular acceleration is the rate of change of angular velocity over time. It is a measure of how quickly the angular velocity is changing.

5. How is angular acceleration related to angular velocity?

Angular acceleration and angular velocity are related by the equation α = Δω/Δt, where α is the angular acceleration, Δω is the change in angular velocity, and Δt is the change in time. In other words, angular acceleration is the change in angular velocity divided by the change in time.

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