How Many Revolutions Per Minute to Equal Free-Fall Acceleration?

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In summary: Just remember to always check your units and make sure they cancel out and give you the correct units for your final answer. In summary, the problem involves finding the number of revolutions per minute for a ball whirling in a horizontal circle with a radius of 0.8m, in order to have the same centripetal acceleration as the free-fall acceleration due to gravity. Using the equation a=v^2/r, we can determine that the ball must make 0.56 revolutions per second to achieve this.
  • #1
coconut88
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Homework Statement




Aboy whirls aball on a string in ahorizontal circle of radius 0.8m. How many revolutions per minute must the ball make if the magnitude of its centriptal acceleration is to be the same as the free-fall acceleration due to gravity g?

Homework Equations





The Attempt at a Solution

 
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  • #2
coconut88 said:

Homework Statement




Aboy whirls aball on a string in ahorizontal circle of radius 0.8m. How many revolutions per minute must the ball make if the magnitude of its centriptal acceleration is to be the same as the free-fall acceleration due to gravity g?

Homework Equations





The Attempt at a Solution


Give it a shot. We'll help you along.
 
  • #3
Hey,

Consider what acceleration is being referred to here, centripetal or linear?

Also, then consider that acceleration as it relates to the number of revolutions per minute.

Thanks,

-PFStudent
 
  • #4
the best way to solve these kinds of problems is to find the appropriate equations and determine which of the quantity has to be common in the equations(usually two equations).

then equate these equations to find the unknown, which should be in one of the relevant equations.
 
  • #5
a=v2/r this is the first equation that I used but I don't kow what is the secend will be ?
what does the revoluition mean ? what does the equation we need to use ?
pleace help me I me international student fot that I have a lot of diffeculties
 
  • #6
coconut88 said:
a=v2/r this is the first equation that I used but I don't kow what is the secend will be ?
what does the revoluition mean ? what does the equation we need to use ?
pleace help me I me international student fot that I have a lot of diffeculties

Yes, you can find v using a = v^2/r

number of revolutions means number of times it goes around in a circle...

let n = total number of revolutions. let d = distance. let t = time.

So

[tex]d = n*2{\pi}r[/tex]

[tex]\frac{d}{t}= \frac{n*2{\pi}r}{t}[/tex]

v = d/t

number of revolutions per second = n/t

Using these you should be able to get number of revolutions per second. What do you get?
 
  • #7
Thinks Learningphysics to your help
I used this v^2/r and the answer is 2.8 m/s after that I used 2rr and the answar is 5.0/1 rev that means 1 rev/5.0
aftet that I can't do any thing because I don't have the d !
??
 
  • #8
coconut88 said:
Thinks Learningphysics to your help
I used this v^2/r and the answer is 2.8 m/s after that I used 2rr and the answar is 5.0/1 rev that means 1 rev/5.0
aftet that I can't do any thing because I don't have the d !
??

you don't need d. let X = number of revolutions/second

you know that v = X*(distance per revolution) = X*5.0

so v = X*5.0

solve for X.
 
  • #9
The x will be .56
 
  • #10
coconut88 said:
The x will be .56

looks right to me. 0.56 revolutions/second
 
  • #11
Thinks learningphysic for your help
 
  • #12
coconut88 said:
Thinks learningphysic for your help

you're welcome.
 

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