Centripetal acceleration problem

In summary: Yes, in general, period = 60/fwhere ##f## is frequency in revolutions per minute. (Note... ##f## is not the same as ##n##.)In summary, the turbine's blades acceleration can be calculated using the formula a=4π2Rn2, where n is the rotational frequency. In this scenario, the rotational frequency is 94 rotations per minute, which is equivalent to 1.57 rotations per second. Using the formula, the correct answer for the turbine's blades acceleration is approximately 363.37m/s2. However, there may be discrepancies in the given "correct" answer, as it is possible that the answer key did not properly square the rotational frequency. Additionally, it
  • #1
Juliusz
13
3

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
 
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  • #2
Juliusz said:
What is the turbine's blades acceleration?
Which part of the blade? The tip? The middle? The mass centre? The radius of gyration?
 
  • #3
haruspex said:
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.
 
  • #4
Juliusz said:
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.
 
Last edited:
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  • #5
Juliusz said:
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).
 
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  • #6
jbriggs444 said:
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).
PeroK said:
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.
 
  • #7
jbriggs444 said:
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?
 

Attachments

  • rotational.PNG
    rotational.PNG
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  • #8
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
 
Last edited:
  • #9
Timothy Proudkii said:
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?
 
  • #10
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?
 
  • #11
Timothy Proudkii said:
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.
 
  • #12
PeroK said:
So, what's the period? Which is the time for a revolution.
One second
 
  • #13
Timothy Proudkii said:
One second

Is that your answer? Why would it be ##1s##?
 
  • #14
"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.
 
  • #15
Timothy Proudkii said:
"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?
 
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  • #16
PeroK said:
And, if it did ##1,000## revs per minute, what would be the period?
16.16 revs per one second.
 
  • #17
Timothy Proudkii said:
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.
 
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  • #18
PeroK said:
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?
 
  • #19
Timothy Proudkii said:
AHhhhhH I see, so the period would be 0.06s for 1000 revs per minute?

Yes, in general, period = 1/frequency.
 
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  • #20
PeroK said:
Yes, in general, period = 1/frequency.
Yes thank you!
 
  • #21
Juliusz said:
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).
 
  • #22
jbriggs444 said:
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
 
  • #23
Juliusz said:
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.
 

Related to Centripetal acceleration problem

What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and its magnitude is given by the formula a = v^2/r, where v is the velocity of the object and r is the radius of the circle.

What causes centripetal acceleration?

Centripetal acceleration is caused by a force acting on an object towards the center of a circular path. This force is often referred to as the centripetal force and it can be provided by various factors such as tension, gravity, or friction.

How can I calculate centripetal acceleration?

To calculate centripetal acceleration, you can use the formula a = v^2/r. First, determine the velocity of the object moving in a circular path. Then, find the radius of the circle. Finally, plug in these values into the formula to find the centripetal acceleration.

What is the difference between centripetal acceleration and centrifugal force?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path, while centrifugal force is the apparent outward force experienced by the object due to its inertia. Centrifugal force is not a real force, but rather a term used to describe the reaction to the centripetal force.

How does centripetal acceleration affect the motion of objects?

Centripetal acceleration affects the motion of objects by constantly changing the direction of their velocity. This causes the objects to move in a circular path instead of a straight line. Without centripetal acceleration, objects would continue to move in a straight line with constant velocity.

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