Centripetal acceleration problem

[Juliusz]

Homework Statement

Hydroelectric power station's turbine has a diameter of 7.5m, and a rotation frequency of 94 rotations per minute. What is the turbine's blades acceleration?
P.S. This is translated from another language

Homework Equations

a=4π²Rn²
Where n=rotational frequency

The Attempt at a Solution

My understanding is that I need to convert rotations per minute, to rotations per second for the formula to work. Therefore: 94/60
Though when I plug that in the formula for n, I get a wrong answer.
I also divided the diameter by two to get the radius(R)
4*π²*3.75*(94/60)²≈363.37
The correct answer is 231.7m/s²

 

[haruspex]

What is the turbine's blades acceleration?

Which part of the blade? The tip? The middle? The mass centre? The radius of gyration?

 

[Juliusz]

Which part of the blade? The tip? The middle? The mass centre? The radius of gyration?

The tip. That's the problem with translating problems from a different language.

 

[PeroK]

4*π²*3.75*(94/60)²≈363.37
The correct answer is 231.7m/s²

Maybe you need to check that calculation.

PS That said, it looks to me like you are correct. I can't see where they get the other answer from.

 

[jbriggs444]

4*π²*3.75*(94/60)²≈363.37
The correct answer is 231.7m/s²

The ratio between your answer and the "correct" answer is 94/60 (to within 0.1%). It is possible that the supposedly correct answer is not, in fact, correct and that the answer key did not properly square the rotational frequency (as you have it) or the inverse time per rotation (as I usually remember the formula).

 

[Juliusz]

The ratio between your answer and the "correct" answer is 94/60 (to within 0.1%). It is possible that the supposedly correct answer is not, in fact, correct and that the answer key did not properly square the rotational frequency (as you have it) or the inverse time per rotation (as I usually remember the formula).

Maybe you need to check that calculation.

PS That said, it looks to me like you are correct. I can't see where they get the other answer from.

This is not the first time that a "correct" answer is actually wrong. Thank god for forums like these, otherwise I would break my head thinking about what I did wrong.

 

[Juliusz]

The ratio between your answer and the "correct" answer is 94/60 (to within 0.1%). It is possible that the supposedly correct answer is not, in fact, correct and that the answer key did not properly square the rotational frequency (as you have it) or the inverse time per rotation (as I usually remember the formula).

Is this the formula you are referring to?

 

[NotKepler]

I solved this problem another way.
I used the two equations:
v= 2(pi)r/T
a=v^2/r
where:
r = radius
T= period
v= velocity
a= circular acceleration

I first found v:
2pi(3.75)/(60/94)
v=36.9m/s

a= 36.9^2 / 3.75 = 363.4m/s^2

 

[PeroK]

I solved this problem another way. However, I think I did something terribly wrong because I am no where near what you got.
I used the two equations:
v= 2(pi)r/T
a=v^2/r
where:
r = radius
T= period
v= velocity
a= circular acceleration

I first found v:
2pi(3.75)/(94/60)
v=15.04m/s

a= 15.04^2 / 3.75 = 60.32m/s^2

What the period, $T$, if it's 94 revs per minute?

 

[NotKepler]

Since there is 94 revolutions per 60 seconds, then there are 1.57 revolutions per second (94/60) Right? Or do I not understand something?

 

[PeroK]

Since there is 94 revolutions per 60 seconds, then there are 1.57 revolutions per second (94/60) Right? Or do I not understand something?

So, what's the period? Which is the time for a revolution.

 

[NotKepler]

So, what's the period? Which is the time for a revolution.

One second

 

[PeroK]

One second

Is that your answer? Why would it be $1s$?

 

[NotKepler]

"In mathematics, a periodic function is a function that repeats its values in regular intervals or periods" The blade does a full rotation in 1.57s, I misunderstood your initial question.

 

[PeroK]

"In mathematics, a periodic function is a function that repeats its values in regular intervals or periods" The blade does a full rotation in 1.57s, I misunderstood your initial question.

And, if it did $1,000$ revs per minute, what would be the period?

 

[NotKepler]

And, if it did $1,000$ revs per minute, what would be the period?

16.16 revs per one second.

 

[PeroK]

16.16 revs per one second.

That's not a period. That's a frequency. The period is the length of time it takes for a revolution.

That ties in with how you tried to calculate your velocity, which was $2\pi r/T$, where $r$ is the radius of the motion.

 

[NotKepler]

That's not a period. That's a frequency. The period is the length of time it takes for a revolution.

That ties in with how you tried to calculate your velocity, which was $2\pi r/T$, where $r$ is the radius of the motion.

AHhhhhH I see, so the period would be 0.06s for 1000 revs per minute?

 

[PeroK]

AHhhhhH I see, so the period would be 0.06s for 1000 revs per minute?

Yes, in general, period = 1/frequency.

 

[NotKepler]

Yes, in general, period = 1/frequency.

Yes thank you!

 

[jbriggs444]

Is this the formula you are referring to?
$$\frac{\pi^2R}{T^2}=4$$
[rewrote attached formula using LaTeX]

No, that's not what I had in mind. The formula that I learned many years ago was$$a=\frac{4\pi^2r}{t^2}$$
Which is the same as $$a=4\pi^2rn^2$$The only difference is that one uses complete rotations per unit time (n) and the other uses time per complete rotation (t).

 

[Juliusz]

No, that's not what I had in mind. The formula that I learned many years ago was$$a=\frac{4\pi^2r}{t^2}$$
Which is the same as $$a=4\pi^2rn^2$$The only difference is that one uses complete rotations per unit time (n) and the other uses time per complete rotation (t).

I don't know what formula you quoted, but the one attached in my image is the same as you just wrote

 

[jbriggs444]

I don't know what formula you quoted, but the one attached in my image is the same as you just wrote

Take a look at the thumbnail presentation of the attached image from your post #7.

The attachment is fine. The thumbnail is not.