Rev/sec necessary to reach 20-g

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In summary: T = 1.3 secondsIn summary, the conversation discusses a physics problem involving calculating the revolutions per second required for a person to feel 20-g of acceleration. The equation Ac=((2pi*r/T)^2)/r is used, where T is the time period. After making a calculation, the answer of 1.3 seconds is achieved, but the question specifically asks for the answer in revolutions per second. The solution is to convert the unit of seconds to Hz or s-1, which can be done by dividing the number of seconds by the time period T. However, the correct answer is still unclear and more information is needed.
  • #1
hyde2042
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Homework Statement



http://imgur.com/Frn6v

Homework Equations



Ac=((2pi*r/T)^2)/r

Where T is the time period

The Attempt at a Solution



Since a g is 9.8m/s^2

20(9.8)=(4*pi^2*29)/T^2

1144.87/(20(9.8))=5.841

And solved for T getting...

sqrt(5.841)=2.417

The book's answer at the back reads 0.749 revs/sec.

Did I miss something or is there something else I'm supposed to do with the answer? Thank you for your time.
 
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  • #2
hyde2042 said:

Homework Statement



http://imgur.com/Frn6v


Homework Equations



Ac=((2pi*r/T)^2)/r

Where T is the time period

The Attempt at a Solution



Since a g is 9.8m/s^2

20(9.8)=(4*pi^2*29)/T^2

1144.87/(20(9.8))=5.841

sqrt(5.841)=2.417

THe book's answer at the back reads 0.749 revs/sec.

Did I miss something or is there something else I'm supposed to do with the answer? Thank you for your time.
and solved for T... ending up with

NASA scientists lost Mars probe which crashed into the planet heading for the calculated orbit height. In Their calculation they forgot that our SI units have metres for length, and set the orbital radius at xxx feet, when it should have been xxx metres.


You have used a radius of 29 in your calculation. The radius is 29 feet , NOT 29 metres.

Either express the 29 feet in metres or express g as 32.2 - the feet equivalent of our 9.8 ms-2
 
  • #3
Thanks for the laugh. It did seem weird reading the word "feet" in my Physics book, but didn't think too much of it.

However, after converting 29 feet to 8.84 meters, and substituting that into the equation I'm getting, after the sqrt, 1.3 for the answer.

Is there something else I may be overlooking?
 
  • #4
hyde2042 said:
Thanks for the laugh. It did seem weird reading the word "feet" in my Physics book, but didn't think too much of it.

However, after converting 29 feet to 8.84 meters, and substituting that into the equation I'm getting, after the sqrt, 1.3 for the answer.

Is there something else I may be overlooking?

What does you answer of 1.3 mean. An approximate check [like assuming pi = 3 etc] gives an answer very much like that so I am pretty sure it is correct.
Perhaps you have forgotten what quantity you have calculated and what quantity you were asked for.
 
  • #5
PeterO said:
What does you answer of 1.3 mean. An approximate check [like assuming pi = 3 etc] gives an answer very much like that so I am pretty sure it is correct.
Perhaps you have forgotten what quantity you have calculated.

I'm to be calculating how many revolutions per second are required for a person to feel 20-g of acceleration. I was actually going over the quantities used to see what I was left over with.

Which was 1.3 seconds. But they want the answer in Revolutions per second. I think this is where I'm going wrong as I don't know what to do to make it into rev/sec.
 
  • #6
hyde2042 said:
I'm to be calculating how many revolutions per second are required for a person to feel 20-g of acceleration. I was actually going over the quantities used to see what I was left over with.

Which was 1.3 seconds. But they want the answer in Revolutions per second. I think this is where I'm going wrong as I don't know what to do to make it into rev/sec.

The unit for revolutions per second is Hz or s-1

so how to you change s to s-1?

if T = 5 seconds, you get 12 per minute or 0.02 Hz
if T = 0.1 you will get 10 per second or 10 Hz

Soooo ...
 

1. What is the formula for calculating Rev/sec necessary to reach 20-g?

The formula for calculating Rev/sec necessary to reach 20-g is: Rev/sec = √(20 x g x R), where g is the acceleration due to gravity (9.8 m/s²) and R is the radius of rotation in meters.

2. How do you convert Rev/sec to RPM?

To convert Rev/sec to RPM (revolutions per minute), multiply the Rev/sec value by 60. This will give you the number of revolutions per minute necessary to reach 20-g.

3. What is the value of g in the equation for calculating Rev/sec?

The value of g in the equation for calculating Rev/sec is the acceleration due to gravity, which is approximately 9.8 m/s² on Earth. This value may vary depending on the location and altitude.

4. Can the radius of rotation affect the Rev/sec necessary to reach 20-g?

Yes, the radius of rotation does affect the Rev/sec necessary to reach 20-g. The larger the radius, the lower the Rev/sec required. This is because a larger radius provides a greater distance for the object to travel, resulting in a lower angular velocity needed to reach the desired g-force.

5. Is it possible to reach 20-g with a Rev/sec value of 0?

No, it is not possible to reach 20-g with a Rev/sec value of 0. This would mean that the object is not rotating at all, and without angular velocity, there is no centripetal force to create the necessary acceleration to reach 20-g.

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